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Logic & Word Problems — Turn the story into a table.

9 chapters Practice problems →
About this topic

Logic problems on the AMC 8 come in a few shapes: 'Who's in seat #N?', 'Find a value of n that disproves this claim', truth-teller/liar puzzles, and ordering puzzles. They look intimidating because they're written as English sentences, but every one of them is a constraint satisfaction problem in disguise.

The general approach: turn the story into a table (or a diagram, or a list of conditions), then deduce row by row.

This lesson covers nine ideas: (1) reading a story carefully and inventorying the unknowns, (2) using contrapositives ('if A then B' ≡ 'if not B then not A'), (3) negating statements correctly, (4) ordering and ranking puzzles, (5) casework on logical possibilities, (6) finding counterexamples, (7) following implication chains, (8) truth-teller/liar puzzles, (9) optimization within constraints.

CHAPTER 1

Turn the story into a table

THEORY

The first move in any logic puzzle: draw a table. Stop trying to keep names and clues in your head. Every logic problem on the AMC is a constraint puzzle dressed up in English — once you draw the grid, the answer almost falls out.

Match the puzzle to the table shape

Puzzle smells like …Draw this
Seating (“A is next to B”)A row of chairs with name slots
Schedule (“Mike plays soccer on Tuesday”)People × activities grid
Two-way attributes (people ↔ hobbies, jobs, etc.)People × attributes grid
Truth-tellers / liarsStatement table with truth values
Ordering (“A is taller than B”)A vertical line, place each name on it

The 3×3 grid in action

Three people (Alex, Beth, Carl) have three different jobs (Lawyer, Doctor, Teacher). Three clues:

  • Alex is NOT the doctor.
  • Alex is NOT the teacher.
  • Carl is NOT the teacher.
Step 1 — put a × in every cell a clue rules outLawyerDoctorTeacherAlexmaybe××BethmaybemaybemaybeCarlmaybemaybe×Step 2 — Alex’s row has only ONE option left. Lock it in, then cascade.LawyerDoctorTeacherAlexLAWYER××Beth××TEACHERCarl×DOCTOR×
  1. Alex’s row had two ×’s after applying clues, so Alex is forced to be Lawyer.
  2. Once Alex = Lawyer, the rest of the Lawyer column gets ×’s (only one person can be lawyer).
  3. Carl can’t be Lawyer (Alex has it) or Teacher (clue) ⇒ Carl = Doctor.
  4. Beth gets the leftover: Teacher.
Without the grid: confusing word soup. With the grid: 30 seconds to the answer.
THE TRICK

If a clue is given as 'NOT X', restate it as 'IS one of [everything except X]'. This turns negative clues into positive ones, easier to apply.

WORKED EXAMPLE
PROBLEM · 1994 #11

Last summer 100 students attended basketball camp. Of those, 52 were boys and 48 were girls. Also, 40 students were from Jonas Middle School and 60 were from Clay Middle School. Twenty of the girls were from Jonas Middle School. How many of the boys were from Clay Middle School?

A) 20 B) 32 C) 40 D) 48 E) 52

100 students at basketball camp: 52 boys, 48 girls; 40 from Jonas, 60 from Clay. Twenty of the girls were from Jonas. How many boys were from Clay?

Draw a 2×2 table with rows {boys, girls} and columns {Jonas, Clay}, filling in row and column totals on the margins:

JonasClayTotal
Boys??52
Girls20?48
Total4060100

One interior cell is given (girls × Jonas = 20). Every other cell is forced by a row or column total:

  • Girls row: 20 + (girls from Clay) = 48 ⇒ girls from Clay = 28.
  • Clay column: 28 + (boys from Clay) = 60 ⇒ boys from Clay = 32.

The whole problem reduces to one move: put the four totals on the margins and the single given cell in place, then each remaining cell is just a subtraction. The table does the work.

Answer: B — 32.
RULE OF THUMB

Draw the table first. Then apply clues. Don't try to keep it all in your head.

MORE LIKE THIS
2013 · #19 Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget...

Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest score?

Show answer
Answer: D — Cassie, Hannah, Bridget.
Show hint
Hint 1
Both Cassie and Bridget saw Hannah's score. Why can each be sure of her own ranking only after seeing Hannah's?
Show solution
Approach: use what each girl can deduce after seeing Hannah's score
  1. Cassie's claim is only certain if her score > Hannah's (otherwise Hannah's lower score below Cassie's could be the lowest, and Cassie wouldn't know). So Cassie > Hannah.
  2. Bridget's claim is only certain if her score < Hannah's. So Hannah > Bridget.
  3. Combine: Cassie > Hannah > Bridget.
1993 · #14 (figure problem)
ajhsme-1993-14
Show answer
Answer: C — 4.
Show hints
Hint 1 of 2
Every row and every column must contain 1, 2, 3 exactly once — like a mini Sudoku.
Still stuck? Show hint 2 →
Hint 2 of 2
Fill in forced cells one at a time.
Show solution
Approach: deduce each cell from the row/column rule
  1. The top row already has 1, so its other cells are 2 and 3; the column with 2 and the diagonal force the middle column to read 3, 2, 1.
  2. Working through, A = 1 and B = 3, so A + B = 4.
2003 · #17 The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair....

The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?

ChildEye ColorHair Color
BenjaminBlueBlack
JimBrownBlond
NadeenBrownBlack
AustinBlueBlond
TevynBlueBlack
SueBlueBlond
Show answer
Answer: E — Austin and Sue.
Show hints
Hint 1 of 2
Write down Jim's two traits, then see who could even be in his family.
Still stuck? Show hint 2 →
Hint 2 of 2
Three of the other children share one trait among themselves — that locks in both families.
Show solution
Approach: group by a shared trait
  1. Jim has brown eyes and blond hair, so a sibling must share one of those: only Nadeen (brown eyes), Austin (blond), or Sue (blond) qualify.
  2. But Benjamin, Nadeen, and Tevyn all have black hair, so those three already form a valid family.
  3. That forces Jim's family to be Jim, Austin, and Sue (all blond) — his siblings are Austin and Sue.
2004 · #18 Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual's...

Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual's score is the sum of the scores in the target regions that are hit. The scores for the target regions are the whole numbers 1 through 10. Each throw hits the target in a region with a different value. The scores are: Alice 16, Ben 4, Cindy 7, Dave 11, Ellen 17. Who hits the region worth 6 points?

Show answer
Answer: A — Alice.
Show hint
Hint 1
Each region is used exactly once. Start with the smallest score (Ben = 4): the only pair from 1–10 is {1, 3}.
Show solution
Approach: uniqueness forces each pair in turn
  1. Ben (4): only 1 + 3.
  2. Cindy (7): from remaining digits {2, 4, 5, 6, 7, 8, 9, 10}, the only pair is 2 + 5.
  3. Dave (11): only 4 + 7.
  4. Remaining for Alice and Ellen: {6, 8, 9, 10}. Alice 16 = 6 + 10. Ellen 17 = 8 + 9.
  5. Alice hits 6.
2001 · #20 Kaleana shows her test score to Quay, Marty, and Shana, but the others keep theirs hidden. Quay thinks, "At least two of us have the...

Kaleana shows her test score to Quay, Marty, and Shana, but the others keep theirs hidden. Quay thinks, "At least two of us have the same score." Marty thinks, "I didn't get the lowest score." Shana thinks, "I didn't get the highest score." List the scores from lowest to highest for Marty (M), Quay (Q), and Shana (S).

Show answer
Answer: A — S, Q, M.
Show hints
Hint 1 of 2
Quay can only see Kaleana's score — what must be true for him to be sure two scores match?
Still stuck? Show hint 2 →
Hint 2 of 2
That fixes Quay = Kaleana; then read Marty's and Shana's thoughts relative to Kaleana.
Show solution
Approach: turn each statement into an inequality
  1. Quay only knows Kaleana's score, so to be certain two match he must equal her: Q = K.
  2. Marty isn't lowest, so M > K; Shana isn't highest, so S < K. Replacing K with Q gives S < Q < M.
  3. Lowest to highest: S, Q, M.
CHAPTER 2

Contrapositive — 'if A then B' = 'if not B then not A'

THEORY

The most useful identity in elementary logic:

CONTRAPOSITIVE

“If A, then B” means the SAME thing as “If not B, then not A.”

If it’s raining,the ground is wet.originalsameideaIf the ground is NOT wet,it’s NOT raining.contrapositive (same)

The four ways to flip an “if-then” — only TWO mean the same

From any rule “If A, then B,” you can make three other sentences by swapping or negating. Only one of them is equivalent to the original. The other two are AMC traps.

NameShapeExample (“rain → wet”)Same as original?
OriginalIf A → BIf raining, then wet.— (it IS the original)
ContrapositiveIf NOT B → NOT AIf NOT wet, then NOT raining.YES ✓
ConverseIf B → AIf wet, then raining.NO × (could be a hose)
InverseIf NOT A → NOT BIf NOT raining, then NOT wet.NO × (could be a hose)

Notice the wet ground could be from a sprinkler — so “wet → raining” isn’t guaranteed. That’s why the converse is wrong. The contrapositive escapes this trap because it just flips the same statement around.

Memorize: original = contrapositive.  Converse & inverse = traps.

Use it when the “not” side is easier to think about. “All cats are mammals” can be flipped to “if it’s NOT a mammal, it’s NOT a cat.” Sometimes thinking in negatives unlocks a problem instantly.

THE TRICK

If a problem gives 'If X then Y' and you want to know whether Y must hold, look for evidence of NOT-Y in the problem. If NOT-Y holds, then by contrapositive NOT-X must hold too.

WORKED EXAMPLE
PROBLEM · 1985 #25

Five cards are lying on a table as shown.

P Q
3 4 6

Each card has a letter on one side and a whole number on the other side. Jane said, “If a vowel is on one side of any card, then an even number is on the other side.” Mary showed Jane was wrong by turning over one card. Which card did Mary turn over?

A) 3 B) 4 C) 6 D) P E) Q

Jane's claim: 'If a vowel is on one side of any card, then an even number is on the other side.'

Mary wants to disprove this. The contrapositive: 'If the number is odd, the letter is consonant.' Disproving Jane's claim means finding a card where the rule fails — i.e., a vowel paired with an odd number.

Of the visible cards (3, 4, 6, P, Q): only the cards showing ODD numbers can possibly disprove the claim (we'd need to check what's on the other side for a vowel). Turning over 4 or 6 (even) doesn't help — the claim is consistent with anything behind an even number. Turning over P or Q (consonants) doesn't help — the claim only restricts vowels.

So Mary turned over the card showing the odd number 3.

This is the classic Wason selection task in disguise. To disprove 'A → B', you need a case with A true and B false. Don't waste turns on cards that can't possibly produce that case.

Answer: A — 3.
RULE OF THUMB

To disprove 'if A then B': find a case with A true and B false. Use the contrapositive: 'not B → not A' is equivalent to the original.

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1987 · #20 "If a whole number n is not prime, then the whole number n − 2 is not prime." A value of n which shows this statement to be false is

"If a whole number n is not prime, then the whole number n − 2 is not prime." A value of n which shows this statement to be false is

Show answer
Answer: A — 9.
Show hints
Hint 1 of 2
Look for n that is itself not prime but n − 2 IS prime.
Still stuck? Show hint 2 →
Hint 2 of 2
Check each composite choice's n − 2.
Show solution
Approach: find a composite n whose n−2 is prime
  1. Try n = 9: 9 is not prime, and 9 − 2 = 7 is prime.
  2. That falsifies the claim, so n = 9.
2004 · #13 Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true. I. Bill is the oldest. II. Amy is...

Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true. I. Bill is the oldest. II. Amy is not the oldest. III. Celine is not the youngest. Rank the friends from the oldest to the youngest.

Show answer
Answer: E — Amy, Celine, Bill.
Show hints
Hint 1 of 2
If I is true, II is also true (Bill oldest ⇒ Amy isn't); that's two trues, forbidden. So I is false.
Still stuck? Show hint 2 →
Hint 2 of 2
Then Bill is not oldest. If II is true (Amy not oldest), Celine must be oldest ⇒ III also true (since the youngest can't be the oldest); forbidden again.
Show solution
Approach: rule out which single statement is true
  1. I true ⇒ II also true. Two trues forbidden ⇒ I false.
  2. II true ⇒ Celine oldest, making III true. Forbidden ⇒ II false.
  3. So III is the only true one and I, II are false. Bill not oldest (I false); Amy is oldest (II false). Then by III, Celine is not youngest, so Bill is youngest.
  4. Order: Amy, Celine, Bill.
CHAPTER 3

Negating clues — when statements are FALSE

THEORY

When a puzzle says “these clues are all FALSE,” you have to negate each clue before applying it. Negating sounds simple but has traps. Keep this table handy:

The clue says …The negation means …Trap to avoid
X is next to YX is at least 2 seats away from YNot “X is far from Y” — just not adjacent.
X is between Y and ZX is outside the Y–Z range (or sits exactly on Y or Z)“Not between” means “to one side,” not the opposite.
A is taller than BA is shorter than OR equal to BNOT just “shorter than.” The “or equal” matters.
A is greater than 5A is ≤ 5Not just “less than 5.”
All X have property PAt least one X does NOT have PNOT “no X has P.”
Some X has property PNo X has P—
A and B (both happen)NOT A or NOT B (at least one fails)De Morgan: “and” flips to “or.”
A or B (at least one)NOT A and NOT B (neither happens)De Morgan again.
Always include the “or equal” in ≤ or ≥ negations. The boundary matters.
🎯 Try it
Negate this: “Every second-grader read fewer than 5 books.”
Walkthrough: “Every X has P” negates to “at least one X does NOT have P.” And “fewer than 5” negates to “5 or more.” So: at least one second-grader read 5 or more books.
THE TRICK

For 'all NOT' puzzles, negating each clue gives a system you can apply directly. Don't try to reason while keeping the negation in your head — write it down.

WORKED EXAMPLE
PROBLEM · 1987 #17

Abby, Bret, Carl, and Dana are seated in a row of four seats numbered #1 to #4. Joe looks at them and says:

"Bret is next to Carl."
"Abby is between Bret and Carl."

However each one of Joe's statements is false. Bret is actually sitting in seat #3. Who is sitting in seat #2?

A) Abby B) Bret C) Carl D) Dana E) There is not enough information to be sure.

Abby, Bret, Carl, and Dana sit in seats #1–#4. Joe says: 'Bret is next to Carl' AND 'Abby is between Bret and Carl.' Both statements are FALSE. Bret is actually in seat #3. Who's in seat #2?

Negate each clue, then apply:

  • 'Bret next to Carl' is FALSE ⇒ Carl is NOT adjacent to Bret (seat #3). So Carl ≠ #2 and Carl ≠ #4. With #3 taken by Bret, the only remaining spot for Carl is #1.
  • 'Abby between Bret and Carl' is FALSE ⇒ Abby is NOT strictly between #1 and #3. The only seat strictly between is #2, so Abby ≠ #2. With #1 and #3 taken, Abby must be in #4.

That leaves Dana in seat #2.

Both clues being FALSE is doing the heavy lifting. Each negation is a 'not adjacent' or 'not between' condition that lets you cross off cells one at a time.

Answer: D — Dana.
RULE OF THUMB

Convert FALSE clues to their negations before applying. Be precise about 'not'.

MORE LIKE THIS
2004 · #13 Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true. I. Bill is the oldest. II. Amy is...

Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true. I. Bill is the oldest. II. Amy is not the oldest. III. Celine is not the youngest. Rank the friends from the oldest to the youngest.

Show answer
Answer: E — Amy, Celine, Bill.
Show hints
Hint 1 of 2
If I is true, II is also true (Bill oldest ⇒ Amy isn't); that's two trues, forbidden. So I is false.
Still stuck? Show hint 2 →
Hint 2 of 2
Then Bill is not oldest. If II is true (Amy not oldest), Celine must be oldest ⇒ III also true (since the youngest can't be the oldest); forbidden again.
Show solution
Approach: rule out which single statement is true
  1. I true ⇒ II also true. Two trues forbidden ⇒ I false.
  2. II true ⇒ Celine oldest, making III true. Forbidden ⇒ II false.
  3. So III is the only true one and I, II are false. Bill not oldest (I false); Amy is oldest (II false). Then by III, Celine is not youngest, so Bill is youngest.
  4. Order: Amy, Celine, Bill.
★ MINI-QUIZ

Tables, negations, contrapositive

Three logic puzzles that fall to careful clue-application and good negation.

2013 · #19 Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget...

Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest score?

Show answer
Answer: D — Cassie, Hannah, Bridget.
Show hint
Hint 1
Both Cassie and Bridget saw Hannah's score. Why can each be sure of her own ranking only after seeing Hannah's?
Show solution
Approach: use what each girl can deduce after seeing Hannah's score
  1. Cassie's claim is only certain if her score > Hannah's (otherwise Hannah's lower score below Cassie's could be the lowest, and Cassie wouldn't know). So Cassie > Hannah.
  2. Bridget's claim is only certain if her score < Hannah's. So Hannah > Bridget.
  3. Combine: Cassie > Hannah > Bridget.
1985 · #25 Five cards are lying on a table as shown. P Q 3 4 6Each card has a letter on one side and a whole number on the other side. Jane said,...

Five cards are lying on a table as shown.

P Q
3 4 6

Each card has a letter on one side and a whole number on the other side. Jane said, “If a vowel is on one side of any card, then an even number is on the other side.” Mary showed Jane was wrong by turning over one card. Which card did Mary turn over?

Show answer
Answer: A — 3.
Show hints
Hint 1 of 2
The claim is "vowel → even". A counterexample needs a vowel paired with an odd number.
Still stuck? Show hint 2 →
Hint 2 of 2
Turning over a card with a consonant or an even number can never disprove the claim — focus on the odd-numbered cards.
Show solution
Approach: test only the cards that could falsify the implication
  1. To find a vowel paired with an odd number, check the odd cards (only one is showing). If 3's reverse is a vowel, the claim is false.
  2. So Mary turned over 3.
2014 · #23 Three members of the Euclid Middle School girls' softball team had the following conversation.Ashley: I just realized that our uniform...

Three members of the Euclid Middle School girls' softball team had the following conversation.
Ashley: I just realized that our uniform numbers are all 2-digit primes.
Bethany: And the sum of your two uniform numbers is the date of my birthday earlier this month.
Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.
Ashley: And the sum of your two uniform numbers is today's date.
What number does Caitlin wear?

Show answer
Answer: A — 11.
Show hints
Hint 1 of 3
All three pairwise sums of the uniform numbers are dates (1–31), so the three primes are small two-digit primes. List candidates.
Still stuck? Show hint 2 →
Hint 2 of 3
Pairwise sums ≤ 31 force the primes to be in {11, 13, 17, 19}. Find a triple whose three pairwise sums are all distinct.
Still stuck? Show hint 3 →
Hint 3 of 3
Then Bethany's date is the smallest sum (earlier in the month), Caitlin's is the largest (later), today is between.
Show solution
Approach: list primes, find a workable triple, assign by date order
  1. Each pairwise sum is a date 1–31, so each is ≤ 31. Two-digit primes: 11, 13, 17, 19, 23, 29. Pairs that sum to ≤ 31: only those from {11, 13, 17, 19} (any pair including 23 or 29 exceeds 31 for the larger sums).
  2. Among the triples in {11, 13, 17, 19} we need three distinct pairwise sums. {11, 13, 17}: 24, 28, 30 — distinct. ✓
  3. Bethany's birthday = smallest sum = 24, so Ashley + Caitlin = 24 ⇒ {A, C} = {11, 13}.
  4. Caitlin's birthday = largest sum = 30, so Ashley + Bethany = 30 ⇒ {A, B} = {13, 17}.
  5. Ashley is in both sets, so Ashley = 13. Then Caitlin = 11, Bethany = 17. Today = Bethany + Caitlin = 28 (between 24 and 30 — consistent).
  6. Caitlin wears 11.
CHAPTER 4

Ordering and ranking puzzles

THEORY

Whenever you see “A is taller than B, B is taller than C, …” draw one vertical line and place each person on it. Tall = up, short = down. The line does all the bookkeeping.

Clues: Alice > Bob,   Bob > Carla,   Diana > Alicetaller ↑Diana1st (tallest)Alice2ndBob3rdCarla4th (shortest)Once the four are placed, the answer to any “Nth tallest” question is one glance.

TRANSITIVITY — the chain-link rule

If A > B and B > C, then automatically A > C. You don’t need a separate clue for it. Use this to spread thin clues into a full ordering.

The anchor trick

When clues use words like “3 minutes ahead,” “5 minutes behind,” pick ONE person as the anchor and measure everyone’s gap from them. The arithmetic becomes one addition each.

THE TRICK

If only a partial ordering is determined, check if the question's specific position has a unique answer. Sometimes 'second-tallest' is forced even when the full order isn't.

WORKED EXAMPLE
PROBLEM · 2026 #10

Five runners finished a race: Luke, Melina, Nico, Olympia, and Pedro. Nico finished 11 minutes behind Pedro. Olympia finished 2 minutes ahead of Melina but 3 minutes behind Pedro. Olympia finished 6 minutes ahead of Luke. Which runner finished fourth?

A) Luke B) Melina C) Nico D) Olympia E) Pedro

Five runners finished a race. Nico finished 11 min behind Pedro. Olympia finished 2 min ahead of Melina but 3 min behind Pedro. Olympia finished 6 min ahead of Luke. Who finished fourth?

Use Pedro as the anchor and measure everyone's gap behind him:

  • Pedro: 0 (anchor)
  • Olympia: +3 (3 min behind Pedro)
  • Melina: +5 (2 min behind Olympia, so 3 + 2)
  • Luke: +9 (6 min behind Olympia, so 3 + 6)
  • Nico: +11 (11 min behind Pedro)

Sorted by finish time: Pedro, Olympia, Melina, Luke, Nico. Fourth is Luke.

Picking Pedro as the anchor (since multiple clues are relative to him) avoids juggling relative offsets. Every other runner becomes a simple addition.

Answer: A — Luke.
RULE OF THUMB

Draw the ordering line. Place each pair. Use transitivity. Read off the position the question asks about.

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1999 · #6 Bo, Coe, Flo, Jo, and Moe have different amounts of money. Neither Jo nor Bo has as much money as Flo. Both Bo and Coe have more than...

Bo, Coe, Flo, Jo, and Moe have different amounts of money. Neither Jo nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Jo has more than Moe, but less than Bo. Who has the least amount of money?

Show answer
Answer: E — Moe.
Show hints
Hint 1 of 2
You don't need the full ranking — just find who everyone beats.
Still stuck? Show hint 2 →
Hint 2 of 2
Every clue that names Moe places someone above him.
Show solution
Approach: find who sits below everyone
  1. Bo, Coe, and Jo are each said to have more than Moe, and Flo has more than Bo, so Flo beats Moe too.
  2. Everyone has more than Moe, so Moe has the least.
1993 · #23 Five runners, P, Q, R, S, T, have a race. P beats Q, P beats R, Q beats S, and T finishes after P and before Q. Who could NOT have...

Five runners, P, Q, R, S, T, have a race. P beats Q, P beats R, Q beats S, and T finishes after P and before Q. Who could NOT have finished third in the race?

Show answer
Answer: C — P and S.
Show hints
Hint 1 of 2
Combine the clues into chains: P is ahead of Q, R, and T, and S is behind Q.
Still stuck? Show hint 2 →
Hint 2 of 2
Count how many runners must be ahead of each person.
Show solution
Approach: see who is forced too high or too low
  1. P beats Q, R, and T, and P < T < Q < S, so P is always first — it can't be third.
  2. S comes after Q, which comes after both P and T, so at least three runners beat S, putting it 4th or later — also never third. So the answer is P and S.
CHAPTER 5

Casework — when constraints don't determine uniquely

THEORY

Sometimes the clues leave you with several possibilities and no way to narrow further. The fix: try each one separately and see which survive every other clue. That’s casework.

Think of it as a tree: the trunk is the choice you can’t resolve, the branches are the cases. Walk down each branch on its own. Mark the dead ends with ×. What’s left is the answer.

Bob in seat 2 or 4?case A: seat 2case B: seat 4Alice in 3(forced)Alice in 5(forced)check ALL clues× fails — deadcheck ALL clues✓ survives!

Case A dies because of some later clue. Case B survives. That’s the answer.

CLEAN CASEWORK RECIPE

  1. Label the cases: Case A, Case B, …
  2. For each case, apply EVERY constraint (not just the ones used to set up the case).
  3. Mark dead ends with ×.
  4. Count what survives. For “how many possible” questions, that’s your answer.

Two warnings: cases must be mutually exclusive (no overlap, or you double-count) and exhaustive (no gaps, or you miss a case). When in doubt, list the cases out loud and ask: “Does every scenario fit into exactly one?”

THE TRICK

If a problem has 3+ unknowns and several constraints, fix one unknown by cases and let the constraints determine the rest. Then check consistency with the case's range.

WORKED EXAMPLE
PROBLEM · 2020 #6

Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?

A) Aaron B) Darren C) Karen D) Maren E) Sharon

Five friends ride a train with 5 one-person cars. Maren is in car 5. Aaron sits directly behind Sharon (Sharon-Aaron are a glued pair). Darren sits ahead of Aaron. At least one person sits between Karen and Darren. Who's in the middle car (car 3)?

Anchor on the Sharon−Aaron pair (call their cars k and k+1). Case on k:

  • k = 1 (cars 1, 2): Darren needs to be ahead of car 2 — only car 1, already taken. Fails.
  • k = 2 (cars 2, 3): Darren must be in car 1. Karen fills car 4. Spacing: Karen (car 4) and Darren (car 1) are 3 apart — satisfies 'at least one between' easily. Valid. Arrangement: Darren, Sharon, Aaron, Karen, Maren. Middle = Aaron.
  • k = 3 (cars 3, 4): Darren needs cars 1 or 2. Karen gets the other. They'd be in cars 1 and 2 — no one between them. Fails the spacing rule.

Only the middle case survives, so Aaron sits in car 3.

The glued Sharon−Aaron pair only fits in three places (cars 1−2, 2−3, or 3−4). Casework on those three placements knocks out two and leaves one consistent answer.

Answer: A — Aaron.
RULE OF THUMB

List all cases. Apply ALL constraints to each. Count consistent ones.

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2023 · #8 (figure problem)
amc8-2023-08
Show answer
Answer: A — 000101.
Show hints
Hint 1 of 2
Each round of the tournament has exactly two winners and two losers across the four players. So each column of the table sums to 2.
Still stuck? Show hint 2 →
Hint 2 of 2
Add Lola + Lolo + Tiya for each of the 6 rounds. If that sum is 2, Tiyo lost; if 1, Tiyo won.
Show solution
Approach: each round's wins sum to 2
  1. Per round, four players play two pairings, so exactly 2 wins and 2 losses are distributed each round.
  2. Sum Lola + Lolo + Tiya per round: 1+1+0=2, 1+0+1=2, 1+1+0=2, 0+0+1=1, 1+1+0=2, 1+0+0=1.
  3. Tiyo's slot = 2 − that sum each round: 0, 0, 0, 1, 0, 1 = 000101.
1996 · #14 (figure problem)
ajhsme-1996-14
Show answer
Answer: B — 29.
Show hints
Hint 1 of 2
The two lines cross at one shared square, which is counted in both sums.
Still stuck? Show hint 2 →
Hint 2 of 2
Sum of all six digits = 23 + 12 − (shared square), so pin down the shared digit.
Show solution
Approach: the shared square is double-counted
  1. The column (sum 23) needs three large distinct digits — only 6, 8, 9 work — and the shared digit is one of these.
  2. The row (sum 12) then needs three more digits adding to 12 − shared; this is possible only when the shared digit is 6 (with 1, 2, 3).
  3. So the six digits sum to 23 + 12 − 6 = 29.
2003 · #14 In this addition problem, each letter stands for a different digit. T W O + T W O ------- F O U RIf T = 7 and the letter O represents an...

In this addition problem, each letter stands for a different digit.

    T W O
  + T W O
  -------
  F O U R

If T = 7 and the letter O represents an even number, what is the only possible value for W?

Show answer
Answer: D — W = 3.
Show hints
Hint 1 of 2
Two 3-digit numbers add to a 4-digit number, so F must be 1.
Still stuck? Show hint 2 →
Hint 2 of 2
Look at the hundreds column with T = 7: it forces O to be 4 or 5, and "O is even" picks one.
Show solution
Approach: work column by column
  1. Two 3-digit numbers sum to the 4-digit number FOUR, so the leading carry makes F = 1.
  2. Hundreds column: 7 + 7 = 14 (plus any carry from the tens) makes the hundreds digit O equal to 4 or 5; since O is even, O = 4, and nothing carried out of the tens.
  3. Units: 4 + 4 = 8 gives R = 8 with no carry, so the tens column is simply W + W = U, again with no carry.
  4. So 2W = U must stay below 10 and avoid the digits already used (7, 1, 4, 8): W = 3 gives U = 6, the only option. W = 3.
2003 · #17 The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair....

The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?

ChildEye ColorHair Color
BenjaminBlueBlack
JimBrownBlond
NadeenBrownBlack
AustinBlueBlond
TevynBlueBlack
SueBlueBlond
Show answer
Answer: E — Austin and Sue.
Show hints
Hint 1 of 2
Write down Jim's two traits, then see who could even be in his family.
Still stuck? Show hint 2 →
Hint 2 of 2
Three of the other children share one trait among themselves — that locks in both families.
Show solution
Approach: group by a shared trait
  1. Jim has brown eyes and blond hair, so a sibling must share one of those: only Nadeen (brown eyes), Austin (blond), or Sue (blond) qualify.
  2. But Benjamin, Nadeen, and Tevyn all have black hair, so those three already form a valid family.
  3. That forces Jim's family to be Jim, Austin, and Sue (all blond) — his siblings are Austin and Sue.
2004 · #5 Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each...

Ms. Hamilton's eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?

Show answer
Answer: D — 15 games.
Show hint
Hint 1
Each game eliminates exactly one team. 15 must be eliminated.
Show solution
Approach: one elimination per game
  1. Need 15 teams eliminated (everyone except the winner).
  2. Games: 15.
2018 · #3 Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so...

Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?

Show answer
Answer: D — Dan.
Show hints
Hint 1 of 2
List the early 'bad' numbers: 7, 14, 17, 21, 27, 28, 35, … Track who lands on each as the circle shrinks.
Still stuck? Show hint 2 →
Hint 2 of 2
After each elimination, continue counting from the next person in the shrunken circle.
Show solution
Approach: step through the counting, eliminating people
  1. Counts that eliminate: 7, 14, 17, 21, 27, 28, … (7 itself, 14, multiples of 7, or any number with a 7 digit).
  2. Pass 1 (6 people): A=1, B=2, C=3, D=4, E=5, F=6, A=7 ⇒ A out.
  3. Pass 2 (5 people: B, C, D, E, F): B=8, C=9, D=10, E=11, F=12, B=13, C=14 ⇒ C out.
  4. Pass 3 (4: B, D, E, F): D=15, E=16, F=17 ⇒ F out.
  5. Pass 4 (3: B, D, E): B=18, D=19, E=20, B=21 ⇒ B out.
  6. Pass 5 (2: D, E): D=22, E=23, D=24, E=25, D=26, E=27 ⇒ E out.
  7. Last remaining: Dan.
CHAPTER 6

Find a counterexample

THEORY

To disprove a claim that says “FOR ALL X, blah,” you just need to find one single X where blah is false. That one X is called a counterexample. One is enough to kill the whole claim.

Claim: “All swans are white.”🪵white ✓🪵white ✓🪵white ✓🪵white ✓…🪵BLACK! ×1000 white swans don’t PROVE the claim.But ONE black swan DISPROVES it.

THE ASYMMETRY OF “FOR ALL”

To …You need …
PROVE “For all X, P(X)”To check EVERY X (or a general argument)
DISPROVE “For all X, P(X)”ONE counterexample — that’s it
PROVE “There exists X with P(X)”ONE example — that’s it
DISPROVE “There exists X with P(X)”To check EVERY X (or a general argument)

How to hunt a counterexample

AMC counterexamples almost always live in the first 10 small values. Scan the answer choices systematically. For each candidate, check the TWO conditions:

  1. Does the candidate satisfy the “if” part (the hypothesis)?
  2. If yes, does it FAIL the “then” part (the conclusion)?

If both happen, you’ve found one. (If the candidate doesn’t even satisfy the “if,” it can’t disprove anything — skip it.)

THE TRICK

For 'if n is not prime, then [property]' counterexamples, try small composites: 4, 6, 8, 9, 10. One of these will likely break the property.

WORKED EXAMPLE
PROBLEM · 1987 #20

"If a whole number n is not prime, then the whole number n − 2 is not prime." A value of n which shows this statement to be false is

A) 9 B) 12 C) 13 D) 16 E) 23

Claim: 'if n is not prime, then n − 2 is not prime'. The choices are 9, 12, 13, 16, 23. Find a counterexample.

A counterexample needs n composite (so the hypothesis 'not prime' holds) AND n − 2 prime (so the conclusion fails). Scan the choices:

  • n = 9: composite (3·3), 9 − 2 = 7 prime. Counterexample.
  • n = 13: prime, so hypothesis fails — can't disprove anything.
  • n = 12: 12 − 2 = 10, not prime.
  • n = 16: 16 − 2 = 14, not prime.
  • n = 23: prime, irrelevant.

Answer: n = 9.

For a counterexample to 'A ⇒ B', you need A true AND B false — both. Skip any choice where A is already false; on the rest, test B.

Answer: A — 9.
RULE OF THUMB

To disprove a universal claim, try small/special values. Usually the smallest counterexample is among the first 10.

MORE LIKE THIS
2011 · #21 Students guess that Norb's age is 24, 28, 30, 32, 36, 38, 41, 44, 47, and 49. Norb says, "At least half of you guessed too low, two of...

Students guess that Norb's age is 24, 28, 30, 32, 36, 38, 41, 44, 47, and 49. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?

Show answer
Answer: C — 37.
Show hints
Hint 1 of 2
"At least half too low" with 10 guesses means age > 5th-smallest guess = 36, so age ≥ 37.
Still stuck? Show hint 2 →
Hint 2 of 2
"Two off by one" means age is squeezed between two guesses that differ by 2. The only such pair above 36 is 36 and 38, or 47 and 49.
Show solution
Approach: apply each clue in turn
  1. Sorted guesses: 24, 28, 30, 32, 36, 38, 41, 44, 47, 49. "At least half too low" ⇒ age > 36.
  2. "Two are off by one" ⇒ age sits between two guesses 2 apart. Candidates: 37 (between 36 and 38) or 48 (between 47 and 49).
  3. Age is prime ⇒ 37 (prime) wins; 48 isn't prime.
  4. Norb is 37.
2000 · #20 You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02, with at least one coin of each...

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02, with at least one coin of each type. How many dimes must you have?

Show answer
Answer: A — 1 dime.
Show hints
Hint 1 of 2
Start with one of each coin (41¢) and work out what the other 5 coins make.
Still stuck? Show hint 2 →
Hint 2 of 2
The leftover must end in a 1, which pins the number of pennies.
Show solution
Approach: set aside one of each, then place the rest
  1. One of each coin uses 1 + 5 + 10 + 25 = 41¢, leaving 61¢ in 5 more coins. To end in a 1, exactly one more is a penny, leaving 60¢ in 4 coins.
  2. Those 4 must include quarters (4 dimes reach only 40¢); two quarters leave 10¢ = two nickels.
  3. No extra dimes are needed, so there is just the original 1 dime.
2025 · #21 (figure problem)
amc8-2025-21
Show answer
Answer: A — 12.
Show hints
Hint 1 of 2
Look at the largest group of pods that are all directly connected to each other — their grades are forced into a small set.
Still stuck? Show hint 2 →
Hint 2 of 2
Pods A, B, C, F form a 4-clique. The only 4-element subset of {1,…,7} with every pair differing by at least 2 is {1, 3, 5, 7}.
Show solution
Approach: find the clique, then work from the most constrained pods outward
  1. Among A, B, C, F every pair is directly connected, so all four pairs differ by ≥ 2. The only way to pick 4 numbers from 1–7 with that property is the set {1, 3, 5, 7}.
  2. G connects to A and F. If G = 2, then A, F ≠ 1 or 3, forcing {A, F} ⊂ {5, 7} and {B, C} = {1, 3}.
  3. D and E only touch C and F. The extreme grades 1 and 7 each have just one neighbor in this clique, so place 1 at C and 7 at F. The remaining {4, 6} go to D, E.
  4. Filling in: D = 6 (avoids 7 enough), E = 4 (avoids 1 and 7), B = 3 (next to C = 1), A = 5. All constraints hold.
  5. C + E + F = 1 + 4 + 7 = 12.
★ MINI-QUIZ

Ordering, casework, counterexamples

Three logic puzzles requiring more involved reasoning.

2019 · #19 In a tournament there are six teams that play each other twice. A team earns 3 points for a win, 1 point for a draw, and 0 points for a...

In a tournament there are six teams that play each other twice. A team earns 3 points for a win, 1 point for a draw, and 0 points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?

Show answer
Answer: C — 24 points each.
Show hints
Hint 1 of 2
Maximize top 3 by having them sweep the bottom 3.
Still stuck? Show hint 2 →
Hint 2 of 2
Each top team plays 6 games vs the bottom 3 (3 opponents × 2 games), earning 6 × 3 = 18 points. Among themselves, balance wins.
Show solution
Approach: sweep bottom 3, balance among top 3
  1. Have each top-3 team win all 6 of its games against the bottom 3 (3 opponents × 2 games each) → 6 × 3 = 18 points per top team.
  2. Among the top 3, each pair plays twice. To tie all three, give each pair a 1-win, 1-loss split — each team in a pair gains 3 points (one win) and loses 3 points worth (one loss = 0).
  3. Each top team is in 2 pairs and wins one game in each → +3 + 3 = 6 more points.
  4. Maximum: 18 + 6 = 24.
2000 · #20 You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02, with at least one coin of each...

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02, with at least one coin of each type. How many dimes must you have?

Show answer
Answer: A — 1 dime.
Show hints
Hint 1 of 2
Start with one of each coin (41¢) and work out what the other 5 coins make.
Still stuck? Show hint 2 →
Hint 2 of 2
The leftover must end in a 1, which pins the number of pennies.
Show solution
Approach: set aside one of each, then place the rest
  1. One of each coin uses 1 + 5 + 10 + 25 = 41¢, leaving 61¢ in 5 more coins. To end in a 1, exactly one more is a penny, leaving 60¢ in 4 coins.
  2. Those 4 must include quarters (4 dimes reach only 40¢); two quarters leave 10¢ = two nickels.
  3. No extra dimes are needed, so there is just the original 1 dime.
2025 · #21 (figure problem)
amc8-2025-21
Show answer
Answer: A — 12.
Show hints
Hint 1 of 2
Look at the largest group of pods that are all directly connected to each other — their grades are forced into a small set.
Still stuck? Show hint 2 →
Hint 2 of 2
Pods A, B, C, F form a 4-clique. The only 4-element subset of {1,…,7} with every pair differing by at least 2 is {1, 3, 5, 7}.
Show solution
Approach: find the clique, then work from the most constrained pods outward
  1. Among A, B, C, F every pair is directly connected, so all four pairs differ by ≥ 2. The only way to pick 4 numbers from 1–7 with that property is the set {1, 3, 5, 7}.
  2. G connects to A and F. If G = 2, then A, F ≠ 1 or 3, forcing {A, F} ⊂ {5, 7} and {B, C} = {1, 3}.
  3. D and E only touch C and F. The extreme grades 1 and 7 each have just one neighbor in this clique, so place 1 at C and 7 at F. The remaining {4, 6} go to D, E.
  4. Filling in: D = 6 (avoids 7 enough), E = 4 (avoids 1 and 7), B = 3 (next to C = 1), A = 5. All constraints hold.
  5. C + E + F = 1 + 4 + 7 = 12.
CHAPTER 7

Implication chains

THEORY

When several “if-then” statements share a middle, they chain:

AlanBethCarlosDianagets A→ she gets A toogets A→ he gets A toogets A→ she gets A too

If Alan gets an A, the chain forces all four. Light up one bulb, the whole string lights up.

CHAINS FIRE FORWARD

If A → B → C → D and you know A is true, then B, C, AND D are ALL forced true. You can’t stop the chain partway.

So if the problem also tells you “only K people got an A,” the chain has to START LATE enough so it doesn’t over-light. Specifically:

Start the chain at …Number that light upOK if K = 2?
Alan (first link)4 (Alan, Beth, Carlos, Diana)× too many
Beth3 (Beth, Carlos, Diana)× too many
Carlos2 (Carlos, Diana)✓ just right
Diana (last link)1 (Diana only)× too few
To get exactly K lit at the end, start at position (chain length) − K + 1.
THE TRICK

For 'Alan gets A → Beth gets A → Carlos gets A → Diana gets A' chains, the constraint 'only K students got A' forces the chain to start at position |chain| − K + 1.

WORKED EXAMPLE
PROBLEM · 1986 #22

Alan, Beth, Carlos, and Diana were discussing their possible grades in mathematics class this grading period. Alan said, "If I get an A, then Beth will get an A." Beth said, "If I get an A, then Carlos will get an A." Carlos said, "If I get an A, then Diana will get an A." All of these statements were true, but only two of the students received an A. Which two received A's?

A) Alan, Beth B) Beth, Carlos C) Carlos, Diana D) Alan, Diana E) Beth, Diana

Three implications: Alan→Beth, Beth→Carlos, Carlos→Diana. All true. Only 2 got A.

If Alan got an A, the chain forces all 4 (too many — only 2 allowed). If Beth, the chain forces 3 (too many). If Carlos, the chain forces Diana — exactly 2 ✓.

So Carlos and Diana got A's.

Implication chains amplify. To limit the number of consequences, start the chain late.

Answer: C — Carlos, Diana.
RULE OF THUMB

Implication chains fire forward. To get N elements at the end, start at position |chain| − N + 1.

MORE LIKE THIS
2013 · #19 Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget...

Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest score?

Show answer
Answer: D — Cassie, Hannah, Bridget.
Show hint
Hint 1
Both Cassie and Bridget saw Hannah's score. Why can each be sure of her own ranking only after seeing Hannah's?
Show solution
Approach: use what each girl can deduce after seeing Hannah's score
  1. Cassie's claim is only certain if her score > Hannah's (otherwise Hannah's lower score below Cassie's could be the lowest, and Cassie wouldn't know). So Cassie > Hannah.
  2. Bridget's claim is only certain if her score < Hannah's. So Hannah > Bridget.
  3. Combine: Cassie > Hannah > Bridget.
2014 · #23 Three members of the Euclid Middle School girls' softball team had the following conversation.Ashley: I just realized that our uniform...

Three members of the Euclid Middle School girls' softball team had the following conversation.
Ashley: I just realized that our uniform numbers are all 2-digit primes.
Bethany: And the sum of your two uniform numbers is the date of my birthday earlier this month.
Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.
Ashley: And the sum of your two uniform numbers is today's date.
What number does Caitlin wear?

Show answer
Answer: A — 11.
Show hints
Hint 1 of 3
All three pairwise sums of the uniform numbers are dates (1–31), so the three primes are small two-digit primes. List candidates.
Still stuck? Show hint 2 →
Hint 2 of 3
Pairwise sums ≤ 31 force the primes to be in {11, 13, 17, 19}. Find a triple whose three pairwise sums are all distinct.
Still stuck? Show hint 3 →
Hint 3 of 3
Then Bethany's date is the smallest sum (earlier in the month), Caitlin's is the largest (later), today is between.
Show solution
Approach: list primes, find a workable triple, assign by date order
  1. Each pairwise sum is a date 1–31, so each is ≤ 31. Two-digit primes: 11, 13, 17, 19, 23, 29. Pairs that sum to ≤ 31: only those from {11, 13, 17, 19} (any pair including 23 or 29 exceeds 31 for the larger sums).
  2. Among the triples in {11, 13, 17, 19} we need three distinct pairwise sums. {11, 13, 17}: 24, 28, 30 — distinct. ✓
  3. Bethany's birthday = smallest sum = 24, so Ashley + Caitlin = 24 ⇒ {A, C} = {11, 13}.
  4. Caitlin's birthday = largest sum = 30, so Ashley + Bethany = 30 ⇒ {A, B} = {13, 17}.
  5. Ashley is in both sets, so Ashley = 13. Then Caitlin = 11, Bethany = 17. Today = Bethany + Caitlin = 28 (between 24 and 30 — consistent).
  6. Caitlin wears 11.
2004 · #13 Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true. I. Bill is the oldest. II. Amy is...

Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true. I. Bill is the oldest. II. Amy is not the oldest. III. Celine is not the youngest. Rank the friends from the oldest to the youngest.

Show answer
Answer: E — Amy, Celine, Bill.
Show hints
Hint 1 of 2
If I is true, II is also true (Bill oldest ⇒ Amy isn't); that's two trues, forbidden. So I is false.
Still stuck? Show hint 2 →
Hint 2 of 2
Then Bill is not oldest. If II is true (Amy not oldest), Celine must be oldest ⇒ III also true (since the youngest can't be the oldest); forbidden again.
Show solution
Approach: rule out which single statement is true
  1. I true ⇒ II also true. Two trues forbidden ⇒ I false.
  2. II true ⇒ Celine oldest, making III true. Forbidden ⇒ II false.
  3. So III is the only true one and I, II are false. Bill not oldest (I false); Amy is oldest (II false). Then by III, Celine is not youngest, so Bill is youngest.
  4. Order: Amy, Celine, Bill.
CHAPTER 8

Exactly-one-true and truth-status puzzles

THEORY

“Alex says X is true. Bobbi says Alex is lying. One is a truth-teller, the other a liar. Which is which?”

These puzzles look slippery. They’re actually mechanical — just two rules and a TABLE.

THE TWO RULES

  • A truth-teller’s statement is TRUE.
  • A liar’s statement is FALSE.

Strategy: try each “who is what” assignment and check for contradictions. Build a table like this:

CaseAlex’s typeBobbi’s typeAlex says “X is true”Bobbi says “Alex is lying”Verdict
1truth-tellerliarX really IS true ✓Bobbi’s claim must be FALSE ⇒ Alex is NOT lying ✓ (he’s a truth-teller)OK
2liartruth-tellerX really is FALSE ✓Bobbi’s claim must be TRUE ⇒ Alex IS lying ✓OK

Both cases pass! So we can’t decide who’s lying just from those clues. (In a real AMC problem there’d be a third clue that kills one of the rows.)

The agreement shortcut

A says about B …Same typeDifferent type
“B is a truth-teller”consistentcontradiction
“B is lying”contradictionconsistent

Memorize this 2×2: saying-truth-of-someone only works if you’re the SAME type; calling-a-liar only works if you’re a DIFFERENT type. Cuts your casework in half.

THE TRICK

If statement 'A says S' is being analyzed: there are exactly two cases. 'A is truth-teller AND S is true' OR 'A is liar AND S is false'. These are exhaustive.

WORKED EXAMPLE
PROBLEM · 2004 #13

Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true. I. Bill is the oldest. II. Amy is not the oldest. III. Celine is not the youngest. Rank the friends from the oldest to the youngest.

A) Bill, Amy, Celine B) Amy, Bill, Celine C) Celine, Amy, Bill D) Celine, Bill, Amy E) Amy, Celine, Bill

Amy, Bill, and Celine have different ages. Exactly one of these statements is true:

  • I. Bill is the oldest.
  • II. Amy is not the oldest.
  • III. Celine is not the youngest.

Rank the friends from oldest to youngest.

This is truth-teller logic with three statements: try each as 'the true one' and check consistency.

  • Suppose I is true (Bill is the oldest). Then Amy is also not the oldest — so II is true too. That's two trues. Forbidden. So I is false.
  • Suppose II is true (Amy is not the oldest). Combined with I being false (Bill is not the oldest), the oldest must be Celine. Then III ('Celine is not the youngest') is also true. Two trues. Forbidden. So II is false.
  • So III is the only true statement. I false ⇒ Bill is not the oldest. II false ⇒ Amy IS the oldest. III true ⇒ Celine is not the youngest, so Bill is the youngest.

Order from oldest: Amy, Celine, Bill.

'Exactly one true' is the structure that makes this a truth-teller puzzle. Test each statement as the true one; the trap is that statements often imply each other's truth, so picking 'I is true' might silently force II true as well — ruling out that case.

Answer: E — Amy, Celine, Bill.
RULE OF THUMB

For exactly-one-true puzzles: try each statement as the true one, derive the others' truth values, eliminate cases that force more than one true.

MORE LIKE THIS
2013 · #19 Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget...

Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest score?

Show answer
Answer: D — Cassie, Hannah, Bridget.
Show hint
Hint 1
Both Cassie and Bridget saw Hannah's score. Why can each be sure of her own ranking only after seeing Hannah's?
Show solution
Approach: use what each girl can deduce after seeing Hannah's score
  1. Cassie's claim is only certain if her score > Hannah's (otherwise Hannah's lower score below Cassie's could be the lowest, and Cassie wouldn't know). So Cassie > Hannah.
  2. Bridget's claim is only certain if her score < Hannah's. So Hannah > Bridget.
  3. Combine: Cassie > Hannah > Bridget.
2014 · #23 Three members of the Euclid Middle School girls' softball team had the following conversation.Ashley: I just realized that our uniform...

Three members of the Euclid Middle School girls' softball team had the following conversation.
Ashley: I just realized that our uniform numbers are all 2-digit primes.
Bethany: And the sum of your two uniform numbers is the date of my birthday earlier this month.
Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.
Ashley: And the sum of your two uniform numbers is today's date.
What number does Caitlin wear?

Show answer
Answer: A — 11.
Show hints
Hint 1 of 3
All three pairwise sums of the uniform numbers are dates (1–31), so the three primes are small two-digit primes. List candidates.
Still stuck? Show hint 2 →
Hint 2 of 3
Pairwise sums ≤ 31 force the primes to be in {11, 13, 17, 19}. Find a triple whose three pairwise sums are all distinct.
Still stuck? Show hint 3 →
Hint 3 of 3
Then Bethany's date is the smallest sum (earlier in the month), Caitlin's is the largest (later), today is between.
Show solution
Approach: list primes, find a workable triple, assign by date order
  1. Each pairwise sum is a date 1–31, so each is ≤ 31. Two-digit primes: 11, 13, 17, 19, 23, 29. Pairs that sum to ≤ 31: only those from {11, 13, 17, 19} (any pair including 23 or 29 exceeds 31 for the larger sums).
  2. Among the triples in {11, 13, 17, 19} we need three distinct pairwise sums. {11, 13, 17}: 24, 28, 30 — distinct. ✓
  3. Bethany's birthday = smallest sum = 24, so Ashley + Caitlin = 24 ⇒ {A, C} = {11, 13}.
  4. Caitlin's birthday = largest sum = 30, so Ashley + Bethany = 30 ⇒ {A, B} = {13, 17}.
  5. Ashley is in both sets, so Ashley = 13. Then Caitlin = 11, Bethany = 17. Today = Bethany + Caitlin = 28 (between 24 and 30 — consistent).
  6. Caitlin wears 11.
CHAPTER 9

Optimization word problems

THEORY

Optimization = “What’s the biggest / smallest / fewest / most X allowed by all the rules?”

The recipe is the same every time: push toward the extreme until a rule says STOP.

“What’s the LARGEST allowed value of x?”0∞startpush x bigger…RULE: x ≤ 9.4STOP hereIf the answer must be a whole number, the biggest integer that fits is 9 (not 10).

OPTIMIZATION RECIPE

  1. List EVERY constraint.
  2. Decide which one is binding — the rule that says STOP first as you push.
  3. Push to that boundary.
  4. If the answer has to be a whole number, round to the nearest legal one.

The push-to-extremes trick for sums

If a total is fixed and you want one piece to be HUGE, shrink every other piece to its minimum. Whatever’s left in the total budget goes to your target.

GoalWhat to do with the OTHERS
Maximize one of n numbers (sum fixed)Push others to their minimum
Minimize one of n numbers (sum fixed)Push others to their maximum
Maximize the median of n sorted numbersPush the lower half down; push the upper half up to a common ceiling
Integer trap: if x must be ≤ 9.4, the biggest whole-number x is 9. Round DOWN for maxes, UP for mins.
THE TRICK

For 'fewest/most X' problems: identify the binding constraint, push to its boundary, check feasibility.

WORKED EXAMPLE
PROBLEM · 2014 #24

One day the Beverage Barn sold 252 cans of soda to 100 customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?

A) 2.5 B) 3.0 C) 3.5 D) 4.0 E) 4.5

The Beverage Barn sold 252 cans to 100 customers, every customer buying at least one. What's the largest possible median number of cans per customer?

Sort the 100 counts ascending. The median is the average of the 50th and 51st values, so we want both as large as possible. To free up cans, push the first 49 customers down to the minimum (1 can each): that uses 49 cans, leaving 252 − 49 = 203 cans for the last 51 people.

Let the 50th value be a and the 51st–100th all equal b (with a ≤ b). Constraint: a + 50·b ≤ 203.

  • Try b = 5: needs a + 250 ≤ 203. Impossible.
  • Try b = 4: a ≤ 3. Take a = 3. Median = (3 + 4)/2 = 3.5 (choice C).

For maximize-the-median problems: shove everyone below the median down to the floor, push everyone from the median up to a common ceiling, then read off what the sum constraint allows. The binding constraint is the total; everything else is just how to spend it.

Answer: C — 3.5.
RULE OF THUMB

Push to extremes. If a per-value bound is hit, cap and adjust.

MORE LIKE THIS
1990 · #19 There are 120 seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?

There are 120 seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?

Show answer
Answer: B — 40.
Show hints
Hint 1 of 2
You want every empty seat to touch an occupied one.
Still stuck? Show hint 2 →
Hint 2 of 2
Each occupied seat can 'cover' itself plus its two neighbors — three seats per person.
Show solution
Approach: each occupied seat covers a block of 3
  1. Each occupied seat covers itself and its two neighbors — at most 3 seats. So 120 ÷ 3 = 40 people are necessary, and 39 leaves at least one empty seat with two empty neighbors.
  2. 40 is also enough: seat one person in every block of (occupied, empty, empty). Every empty seat touches an occupied one. So the answer is 40.
2017 · #13 Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler...

Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?

Show answer
Answer: B — 1 win.
Show hint
Hint 1
Every game has one winner and one loser, so total wins across all players = total losses.
Show solution
Approach: wins = losses across all three
  1. Total losses: 2 + 3 + 3 = 8.
  2. Total wins: 4 + 3 + K = 8 ⇒ K = 1.
2019 · #19 In a tournament there are six teams that play each other twice. A team earns 3 points for a win, 1 point for a draw, and 0 points for a...

In a tournament there are six teams that play each other twice. A team earns 3 points for a win, 1 point for a draw, and 0 points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?

Show answer
Answer: C — 24 points each.
Show hints
Hint 1 of 2
Maximize top 3 by having them sweep the bottom 3.
Still stuck? Show hint 2 →
Hint 2 of 2
Each top team plays 6 games vs the bottom 3 (3 opponents × 2 games), earning 6 × 3 = 18 points. Among themselves, balance wins.
Show solution
Approach: sweep bottom 3, balance among top 3
  1. Have each top-3 team win all 6 of its games against the bottom 3 (3 opponents × 2 games each) → 6 × 3 = 18 points per top team.
  2. Among the top 3, each pair plays twice. To tie all three, give each pair a 1-win, 1-loss split — each team in a pair gains 3 points (one win) and loses 3 points worth (one loss = 0).
  3. Each top team is in 2 pairs and wins one game in each → +3 + 3 = 6 more points.
  4. Maximum: 18 + 6 = 24.
1999 · #11 (figure problem)
amc8-1999-11
Show answer
Answer: D — 24.
Show hints
Hint 1 of 2
The horizontal and vertical lines share the middle square — it counts in both sums.
Still stuck? Show hint 2 →
Hint 2 of 2
So put the largest number in the center to make the totals as big as possible.
Show solution
Approach: the shared center counts twice — maximize it
  1. The two line-sums together use all five numbers once, plus the center one extra time: 35 + center.
  2. Putting 13 in the center gives 35 + 13 = 48, and each line is half of that: 48 ÷ 2 = 24.
2005 · #16 A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color....

A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?

Show answer
Answer: D — 13.
Show hint
Hint 1
Worst case: 4 of each color (12 socks) without yet getting 5 of one color.
Show solution
Approach: pigeonhole on the worst case
  1. After 12 socks, possible to have 4 red, 4 white, 4 blue (no color has 5).
  2. 13th sock must give some color 5 ⇒ minimum = 13.
2022 · #5 Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned 6 years old, she received a newborn kitten as...

Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned 6 years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is 30. How many years older than Bella is Anna?

Show answer
Answer: C — 3 years older.
Show hints
Hint 1 of 2
Don't solve for ages directly — figure out who is how old today first.
Still stuck? Show hint 2 →
Hint 2 of 2
Work out each one's age today first. Bella was 6 five years ago; the kitten was newborn five years ago. The leftover part of 30 is Anna's age.
Show solution
  1. Bella today: 6 + 5 = 11. Kitten today: 0 + 5 = 5.
  2. The three ages sum to 30, so Anna = 30 − 11 − 5 = 14.
  3. Anna − Bella = 14 − 11 = 3 years.
⬢ FINAL TEST

Stretch test

Five harder logic problems combining sum-constraints, casework, pigeonhole, and optimization.

2000 · #20 You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02, with at least one coin of each...

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02, with at least one coin of each type. How many dimes must you have?

Show answer
Answer: A — 1 dime.
Show hints
Hint 1 of 2
Start with one of each coin (41¢) and work out what the other 5 coins make.
Still stuck? Show hint 2 →
Hint 2 of 2
The leftover must end in a 1, which pins the number of pennies.
Show solution
Approach: set aside one of each, then place the rest
  1. One of each coin uses 1 + 5 + 10 + 25 = 41¢, leaving 61¢ in 5 more coins. To end in a 1, exactly one more is a penny, leaving 60¢ in 4 coins.
  2. Those 4 must include quarters (4 dimes reach only 40¢); two quarters leave 10¢ = two nickels.
  3. No extra dimes are needed, so there is just the original 1 dime.
2001 · #24 (figure problem)
amc8-2001-24
Show answer
Answer: B — 5 white pairs.
Show hints
Hint 1 of 2
Track what's left on each half after the red and blue pairs are used up.
Still stuck? Show hint 2 →
Hint 2 of 2
Each half starts with 3 red, 5 blue, 8 white; subtract the coinciding pairs and the red-white pairs.
Show solution
Approach: account for each color on one half
  1. Each half has 3 red, 5 blue, 8 white. The 2 red pairs use 2 reds per half (1 red left); the 3 blue pairs use 3 blues per half (2 blue left).
  2. The 2 red-white pairs use that last red and 1 white per half, and the 2 leftover blues must pair with whites (no more blue-blue allowed), using 2 more whites.
  3. That leaves 8 − 1 − 2 = 5 whites per half, which coincide as 5 white pairs.
2019 · #19 In a tournament there are six teams that play each other twice. A team earns 3 points for a win, 1 point for a draw, and 0 points for a...

In a tournament there are six teams that play each other twice. A team earns 3 points for a win, 1 point for a draw, and 0 points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?

Show answer
Answer: C — 24 points each.
Show hints
Hint 1 of 2
Maximize top 3 by having them sweep the bottom 3.
Still stuck? Show hint 2 →
Hint 2 of 2
Each top team plays 6 games vs the bottom 3 (3 opponents × 2 games), earning 6 × 3 = 18 points. Among themselves, balance wins.
Show solution
Approach: sweep bottom 3, balance among top 3
  1. Have each top-3 team win all 6 of its games against the bottom 3 (3 opponents × 2 games each) → 6 × 3 = 18 points per top team.
  2. Among the top 3, each pair plays twice. To tie all three, give each pair a 1-win, 1-loss split — each team in a pair gains 3 points (one win) and loses 3 points worth (one loss = 0).
  3. Each top team is in 2 pairs and wins one game in each → +3 + 3 = 6 more points.
  4. Maximum: 18 + 6 = 24.
2005 · #16 A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color....

A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?

Show answer
Answer: D — 13.
Show hint
Hint 1
Worst case: 4 of each color (12 socks) without yet getting 5 of one color.
Show solution
Approach: pigeonhole on the worst case
  1. After 12 socks, possible to have 4 red, 4 white, 4 blue (no color has 5).
  2. 13th sock must give some color 5 ⇒ minimum = 13.
2014 · #24 One day the Beverage Barn sold 252 cans of soda to 100 customers, and every customer bought at least one can of soda. What is the...

One day the Beverage Barn sold 252 cans of soda to 100 customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?

Show answer
Answer: C — 3.5.
Show hints
Hint 1 of 3
Order the counts smallest to largest. Median = average of 50th and 51st values. To make those big, push the first 49 down to the minimum (which is 1).
Still stuck? Show hint 2 →
Hint 2 of 3
After setting the first 49 to 1, 203 cans remain for 51 customers; each from 51st onward must be ≥ 51st value.
Still stuck? Show hint 3 →
Hint 3 of 3
Let 50th = a, 51st = b, with a ≤ b. Make the last 50 all equal to b. Constraint: a + 50b ≤ 203.
Show solution
Approach: minimize the first 49, push the median pair as high as possible
  1. Set customers 1–49 to 1 can each: uses 49 cans, leaves 252 − 49 = 203 for the last 51.
  2. Let the 50th value be a and the 51st value be b (a ≤ b). Make customers 51–100 all equal to b: total of last 51 = a + 50b ≤ 203.
  3. Maximize (a + b)/2 over integers with a ≤ b. Try b = 4: a + 200 ≤ 203 ⇒ a ≤ 3, so a = 3 (still ≤ b). Median = (3 + 4)/2 = 3.5.
  4. Try b = 5: a + 250 ≤ 203 — impossible. So 3.5 is the max.
APPENDIX

Logic & word problems quick-reference

Memorize these

FACTS TO KNOW

  • Contrapositive: 'If A then B' ≡ 'If not B then not A'.
  • Converse (not equivalent): 'If B then A'.
  • Inverse (not equivalent): 'If not A then not B'.
  • De Morgan: NOT(A and B) = (NOT A) or (NOT B). NOT(A or B) = (NOT A) and (NOT B).
  • Counterexample disproves a 'for all'. One specific example with A and not-B kills 'A → B'.
  • One example doesn't prove a 'for all'. You'd need a general argument.
  • For 'there exists' claims: one example proves; you'd need to check every case to disprove.
Common traps
  • Confusing converse with original. 'All cats are mammals' does NOT mean 'all mammals are cats'.
  • Negating 'all' to 'none'. The negation of 'all X are Y' is 'some X is not Y', not 'no X is Y'.
  • 'Or' includes 'both' (inclusive).
  • Skipping casework when statements leave multiple possibilities.
  • Forgetting to verify your solution satisfies ALL the original constraints, not just the ones you used to derive it.
Warm-ups

Drill these:

  • Contrapositive of 'If x is even, then x² is even' is 'If x² is odd, then x is odd'.
  • 'Not (A and B)' = 'Not A or Not B' (De Morgan).
  • Negation of 'I will go if it rains' is 'It rains but I don't go'.
  • For 'all natural numbers > 1 are prime': counterexample n=4 (composite).
→ Practice Logic & Word Problems problems