All lessons / Test-day Habits

Test-day Habits — Save the points you already know how to win.

3 chapters
About this topic

This is a different kind of lesson. Every other topic on this site (arithmetic, number theory, geometry, …) teaches math. This one teaches test-taking — how to actually capture the points you already deserve.

Why is it its own page? Because these habits apply to every problem on the AMC, regardless of topic. If you're trying to score 15+, the biggest single lift isn't learning a new technique — it's stopping the silly mistakes that drop already-solvable problems.

Two chapters: (1) the seven habits to build during practice, and (2) two tricks that pay off on the hardest problems (#20–25).

CHAPTER 1

Test-day habits — saving the points you already know

THEORY

Here's the part most kids miss: most of the points you'll lose on the AMC aren't from problems you can't solve. They're from problems you CAN solve but where you misread one word, mixed up cents and dollars, or bubbled the wrong letter at the end.

You don't have to learn new math to win those points back. You just have to build a few habits. We call them test-day habits. Practice them on every problem until they feel automatic — then on test day you don't even think about them.

THE SEVEN HABITS

  1. Read the question twice — what does it actually ask? Largest, smallest, how many, what value? Underline the question word.
  2. Circle the tiny words. “inclusive,” “except,” “positive integers,” “non-negative,” “at most,” “different.” These flip answers.
  3. Check the units. Cents vs dollars. Minutes vs hours. Inches vs feet. Square inches vs square feet. A unit slip turns 60 into 1.
  4. Plug the answer back in. If you got x = 5, walk it through every condition the problem gave. If anything breaks, you blew it — restart.
  5. Don't trust mental math past 3 digits. Write it on the scratch paper. Especially × and ÷.
  6. Re-read your final letter. You wrote “C = 12” on the scratch sheet but bubbled (B). It happens. Re-reading the bubble takes 2 seconds.
  7. If a problem is taking too long, mark it and move on. Bagging 18 easy ones is better than losing 3 of them while you wrestle with #22.

Tiny words that flip answers

Here are real traps from past AMCs — one word changes the answer:

“between 1 and 10”→ does it include 1 and 10?“positive integers”→ 0 doesn't count. 1, 2, 3, …“non-negative”→ 0 DOES count. 0, 1, 2, …“at most 4”→ 4 is allowed. “fewer than 4” means ≤3.“different digits”→ no repeats. 11 doesn't count.“distinct”→ same as “different”. No repeats.

The unit-slip trap

The problem says: “A pencil costs 25 cents. How many pencils can you buy with $5?” If you write 5 ÷ 25, you get 0.2 — nonsense. Translate first: $5 = 500 cents. Then 500 ÷ 25 = 20. The units have to match before you divide.

🎯 Try it
A bus travels at 30 miles per hour. How many minutes does it take to cover 10 miles?
Walkthrough: Distance ÷ speed = time = 10 / 30 = 1/3 of an HOUR. But the question asks for minutes. 1/3 hour × 60 = 20 minutes. The unit slip is reading “hours” off the speed and forgetting to convert.

The two-minute rule

If you've been on one problem for more than two minutes without making progress, mark it and move on. Come back later if there's time. The AMC isn't graded on which problems you tried — it's graded on how many you got right. A 20-minute battle with #22 that ends in a wrong guess just cost you all the easy points after it.

Easy points first. Hard points last. Always.
THE TRICK

Stuck on a hard problem? Before solving, look at the answers. Ask: what do they have in common? What's DIFFERENT about each?

  • If the answers are all whole numbers, the answer has to be a whole number — a fraction means you blew it.
  • If three answers are obviously too big or too small, you've narrowed five down to two with no math.
  • If the answers differ by units digit (3, 5, 7, 9, 11), find the units digit first — that's the easier sub-problem.
  • If the answers differ by parity (odd vs even), find the parity first.

This isn't “cheating.” The answer choices are part of the problem. The AMC writes them on purpose to reward kids who notice structure.

WORKED EXAMPLE
PROBLEM · 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 lost 3 games, how many games did he win?

A) 0 B) 1 C) 2 D) 3 E) 4

Peter won 4 games and lost 2. Emma won 3 and lost 3. Kyler lost 3. How many did Kyler win?

The instinct is to set up equations. Don't. Instead use the hidden total constraint: every chess game has exactly one winner and one loser, so

total wins = total losses

across all players. Add up the losses we know: 2 + 3 + 3 = 8. So total wins = 8 as well.

Peter and Emma's wins: 4 + 3 = 7. Kyler's wins = 8 − 7 = 1. Answer (B).

Habit check. The answer choices were 0, 1, 2, 3, 4 — all small whole numbers. We didn't need an equation; we needed the right sentence.

This is a “read the structure” problem. “Every game has a winner and a loser” is the hidden constraint — once you see it, the answer drops out in one line. Kids who try to write a system of equations end up with too many unknowns and panic. Always look for the structural fact first.

Answer: B — 1 win.
RULE OF THUMB

Read twice. Circle tiny words. Match units. Plug answers back. Re-read your bubble. If stuck two minutes, move on. Use the answer choices.

MORE LIKE THIS
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.
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.
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.
2024 · #9 All of the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many...

All of the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?

Show answer
Answer: E — 28 marbles.
Show hints
Hint 1 of 2
Pick one color to stand for a variable, then write the others in terms of it. The total reveals a hidden factor.
Still stuck? Show hint 2 →
Hint 2 of 2
Let r = red count. Green = 2r, blue = 4r. Total = 7r — it must be a multiple of 7.
Show solution
Approach: find the hidden multiple
  1. Let r be the number of red marbles. Half as many red as green → green = 2r. Twice as many blue as green → blue = 4r.
  2. Total = r + 2r + 4r = 7r — always a multiple of 7.
  3. Among the choices, only 28 = 7 × 4 is a multiple of 7.
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.
CHAPTER 2

Two tricks for the hardest problems (#20–#25)

THEORY

Once you have the seven habits down, what about the hardest problems on the test — #20 through #25? Those are where the kid going for a 20+ score actually battles. Two specific moves help here, and neither one requires more math knowledge. They’re ways of using the problem against itself.

Trick 1 — engineering induction

When a problem has a variable like “there are n people in a row” and you can’t compute the answer directly, try n = 1, then n = 2, then n = 3. Write the answers in a column. Look for the pattern. Assume it continues.

This isn’t a formal proof — but for AMC, it’s almost always right. The contest setters write problems where the pattern is honest.

Example. “How many regions does n straight lines divide the plane into, if no two are parallel and no three meet at a point?” Compute small cases: n=0 → 1 region. n=1 → 2. n=2 → 4. n=3 → 7. n=4 → 11. Differences: 1, 2, 3, 4 — each new line adds one more region than the previous. Pattern locked. For n=10, regions = 1 + (1+2+…+10) = 56.

Trick 2 — use the freedom in the problem

If the problem says “in any triangle” or “for any rectangle,” then ANY example you pick will give the right answer.

Pick the easiest one. Equilateral triangle. Square. Numbers like 1 or 10. The problem PROMISED the answer doesn’t depend on which one you pick — so use the simplest.

Example. “In an equilateral triangle ABC, point P is inside. The sum of distances from P to the three sides is constant. What is it?” The problem says “constant” — so the answer doesn’t depend on where P is. Pick P at the center. Now everything is symmetric and you can read off the answer in one line.

If the problem says “works for any X,” pick the easiest X. The setter has already done the heavy lifting.
THE TRICK

Both tricks share a philosophy: let the structure of the problem do work for you. Engineering induction lets the small cases reveal the pattern. The freedom trick lets you replace a hard configuration with an easy one. Neither one requires more math — they require recognizing that the problem allows you to simplify.

WORKED EXAMPLE
PROBLEM · 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 lost 3 games, how many games did he win?

A) 0 B) 1 C) 2 D) 3 E) 4

Same problem as ch.1 (Peter, Emma, Kyler chess). But notice another way to think about it — the ‘use the freedom’ angle.

The problem gives us just enough info: three losses, two players’ wins, and a structural constraint (every game has 1 winner and 1 loser). If we listed every possible game outcome, the casework would explode. But the problem promised the answer is determined — so there’s a structural fact that pins it down. Look for it. Find: total wins = total losses. Done.

Every well-formed contest problem has the just-enough property. If you feel like you’re drowning in possibilities, you’re missing the structural fact. Pause and look for it.

This is the meta-lesson: when you’re stuck, the problem is trying to give you a shortcut you haven’t spotted yet. Slow down. What constraint connects EVERY piece of given info? That’s usually the move.

Answer: B — 1 win.
RULE OF THUMB

Engineering induction: try n=1, 2, 3, spot the pattern, continue. Use the freedom: when the problem says “for any X,” pick the easiest X. And in general: if a problem feels too hard, you’re missing a structural fact. Slow down and look.

MORE LIKE THIS
2019 · #25 Alice has 24 apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?

Alice has 24 apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?

Show answer
Answer: C — 190 ways.
Show hints
Hint 1 of 2
Pre-give 2 apples to each — now you're distributing the remaining 18 with no constraints.
Still stuck? Show hint 2 →
Hint 2 of 2
Stars and bars: nonneg integer solutions to a + b + c = 18 is C(20, 2).
Show solution
Approach: subtract the floor, then stars and bars
  1. Give 2 apples to each person first: that uses 6, leaving 18 to distribute with no minimum (each person already has 2).
  2. Distribute 18 indistinguishable apples among 3 people with no constraint: C(18 + 2, 2) = C(20, 2) = 190.
2024 · #25 A small airplane has 4 rows of seats with 3 seats in each row. Eight passengers have boarded the plane and are distributed randomly...

A small airplane has 4 rows of seats with 3 seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be 2 adjacent seats in the same row for the couple?

Show answer
Answer: C — 20/33.
Show hints
Hint 1 of 2
Count the complement: when can the couple not sit together? Then subtract from 1.
Still stuck? Show hint 2 →
Hint 2 of 2
Per row of L-M-R, no adjacent pair is open iff M is occupied OR both L and R are occupied. Case-split on how many of the 4 middle seats are occupied.
Show solution
Approach: complementary counting on middle-seat occupancies
  1. Total arrangements: choose 8 of 12 seats for passengers, C(12, 8) = 495.
  2. For NO adjacent pair to be open in a row L–M–R: either M is occupied, or both L and R are occupied. Casework on k = number of rows with M occupied:
  3. k = 0: all four M's empty ⇒ all 8 edge seats filled. 1 way.
  4. k = 1: 4 choices of row, then 2 choices for the extra passenger in that row's edges. 8 ways.
  5. k = 2: C(4,2) = 6 row-choices × C(4,2) = 6 placements of remaining 2 passengers in the 4 unfilled edges. 36 ways.
  6. k = 3: C(4,3) = 4 row-choices × C(6,3) = 20 placements of remaining 3 passengers. 80 ways.
  7. k = 4: all middles filled (4 passengers); C(8,4) = 70 placements of the remaining 4 on edges. 70 ways.
  8. Total "no adjacent": 1 + 8 + 36 + 80 + 70 = 195. So adjacent count = 495 − 195 = 300.
  9. Probability = 300495 = 2033.
2023 · #25 Fifteen integers a1, a2, a3, …, a15 are arranged in order on a number line. The integers are equally spaced and have the property that1...

Fifteen integers a1, a2, a3, …, a15 are arranged in order on a number line. The integers are equally spaced and have the property that

1 ≤ a1 ≤ 10,   13 ≤ a2 ≤ 20,   and   241 ≤ a15 ≤ 250.

What is the sum of the digits of a14?

Show answer
Answer: A — 8.
Show hints
Hint 1 of 2
Equally spaced → arithmetic sequence with common difference d. From the bounds, get a tight range for 14d.
Still stuck? Show hint 2 →
Hint 2 of 2
a15 − a1 = 14d. With 241−10 ≤ 14d ≤ 250−1, the only multiple of 14 in [231, 249] is 238 = 14×17. So d = 17.
Show solution
Approach: nail d from bounds, then a1, then a14
  1. Equal spacing: a15 − a1 = 14d. Bounds give 231 ≤ 14d ≤ 249, and the only multiple of 14 in there is 238 → d = 17.
  2. Then a2 = a1 + 17 forces a1 ≤ 3, and a15 = a1 + 238 forces a1 ≥ 3. So a1 = 3.
  3. a14 = a15 − d = (3 + 238) − 17 = 224. Digit sum: 2 + 2 + 4 = 8.
2022 · #25 A cricket randomly hops between 4 leaves, on each turn hopping to one of the other 3 leaves with equal probability. After 4 hops, what...

A cricket randomly hops between 4 leaves, on each turn hopping to one of the other 3 leaves with equal probability. After 4 hops, what is the probability that the cricket has returned to the leaf where it started?

Show answer
Answer: E — 7/27.
Show hints
Hint 1 of 2
Track only one number: pn = probability the cricket is on the starting leaf after n hops.
Still stuck? Show hint 2 →
Hint 2 of 2
From start, the cricket must leave (contributes 0 to pn+1). From any other leaf, prob 1/3 of returning. So pn+1 = (1 − pn)/3.
Show solution
Approach: recursive probability on starting leaf
  1. Let pn = P(on starting leaf after n hops). From the start, the cricket can't stay; from any non-start, it returns with probability 1/3.
  2. pn+1 = (1 − pn) · 13.
  3. Iterate from p0 = 1: p1 = 0, p2 = 1/3, p3 = 2/9, p4 = (1 − 2/9)/3 = (7/9)/3 = 7/27.
CHAPTER 3

Eliminate wrong answers — make every guess a smart guess

THEORY

On the AMC 8 there is no penalty for a wrong answer, so you should put a letter on every problem. But a blind 1-in-5 guess is a waste — you can almost always do better.

The secret: you often don't need the whole answer. One small, cheap-to-compute property of the answer can knock out most of the choices.

units digit must be 1 →13579

Find one property of the answer, and four choices fall.

THE TRICK

Before guessing, run down this checklist. Each tactic is something you can compute in seconds:

TacticThe moveCrosses off choices that…
Units digitCompute only the last digitend in the wrong digit
ParityAsk: is the answer odd or even?have the wrong parity
Size / estimateGet the rough magnitudeare far too big or too small
Must divideThe answer has to be a multiple (or factor) of somethingaren't multiples of it
Plug in the choicesTest each option in the problem's conditionfail the condition
Odd one outIf four choices share a pattern, suspect the loner (use with care)— pick the one that breaks the pattern

Most problems hand you at least one of these for free. Two of them together usually leave a single survivor.

WORKED EXAMPLE
PROBLEM · 2012 #12

What is the units digit of 132012?

A) 1 B) 3 C) 5 D) 7 E) 9

What is the units digit of 132012?   (A) 1   (B) 3   (C) 5   (D) 7   (E) 9

You will never compute 132012. You don't have to — the question only asks for one property: the last digit.

Only the units digit of the base matters, so track powers of 3:

31=3, 32=9, 33=27, 34=81, then 3, 9, 7, 1, … — a cycle of 4: (3, 9, 7, 1).

2012 is a multiple of 4, so we land on the end of a cycle → units digit 1. That's choice (A) — and notice we never needed the actual number.

Even if the cycle had slipped your mind, the units digit of an odd power of an odd number is odd, and 5 only appears from a base ending in 5 — so (C) is out instantly, and a quick estimate of where the cycle lands clears the rest.

Answer: A — 1.
RULE OF THUMB

Never leave a blank. Before you guess, find one property of the answer — units digit, parity, size, a factor it must have — and delete every choice that fails it. A 1-in-2 guess is worth far more than a 1-in-5.

MORE LIKE THIS
2012 · #12 What is the units digit of 132012?

What is the units digit of 132012?

Show answer
Answer: A — 1.
Show hint
Hint 1
Only the units digit of 13 matters: same as 32012. Powers of 3 cycle: 3, 9, 7, 1, 3, 9, 7, 1, … (period 4).
Show solution
Approach: powers of 3 units-digit cycle (3, 9, 7, 1)
  1. Units digit of 13n equals units digit of 3n.
  2. Cycle length 4: 31→3, 32→9, 33→7, 34→1, then repeats.
  3. 2012 ≡ 0 (mod 4) ⇒ units digit = 1.
1997 · #25 All the even numbers from 2 to 98 inclusive, except those ending in 0, are multiplied together. What is the units digit of the product?

All the even numbers from 2 to 98 inclusive, except those ending in 0, are multiplied together. What is the units digit of the product?

Show answer
Answer: D — 6.
Show hints
Hint 1 of 2
Only units digits matter; each block of ten contributes the units 2, 4, 6, 8.
Still stuck? Show hint 2 →
Hint 2 of 2
2 × 4 × 6 × 8 ends in 4, and there are ten blocks — find the units digit of 4¹⁰.
Show solution
Approach: group by tens, then find the units of a power
  1. Each block of ten (like 2, 4, 6, 8) multiplies to a units digit of 4, and there are 10 such blocks, so we need the units digit of 4¹⁰.
  2. 4² ends in 6, and any power of 6 ends in 6, so 4¹⁰ = (4²)⁵ ends in 6.
1999 · #24 When 19992000 is divided by 5, the remainder is

When 19992000 is divided by 5, the remainder is

Show answer
Answer: D — 1.
Show hints
Hint 1 of 2
Only the units digit of 1999 matters for the units digit of the power.
Still stuck? Show hint 2 →
Hint 2 of 2
Powers of 9 end in 9, 1, 9, 1, … — and an even exponent lands on 1.
Show solution
Approach: track the units digit's cycle
  1. Powers of a number ending in 9 cycle 9, 1, 9, 1, …; since 2000 is even, 1999²⁰⁰⁰ ends in 1.
  2. A number ending in 1 leaves remainder 1 when divided by 5.
1996 · #15 The remainder when the product 1492 · 1776 · 1812 · 1996 is divided by 5 is

The remainder when the product 1492 · 1776 · 1812 · 1996 is divided by 5 is

Show answer
Answer: E — 4.
Show hints
Hint 1 of 2
Only the units digit of each factor matters for the remainder mod 5.
Still stuck? Show hint 2 →
Hint 2 of 2
Multiply the units digits and look at that product's units digit.
Show solution
Approach: work with units digits
  1. The units digits are 2, 6, 2, 6, and 2 · 6 · 2 · 6 = 144 ends in 4.
  2. A number ending in 4 leaves remainder 4 when divided by 5.
⬢ FINAL TEST

Habit-builder problems

Five mid-difficulty problems where the lightest path requires reading carefully and using structure, not brute-force computation.

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.
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.
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.
2024 · #9 All of the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many...

All of the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?

Show answer
Answer: E — 28 marbles.
Show hints
Hint 1 of 2
Pick one color to stand for a variable, then write the others in terms of it. The total reveals a hidden factor.
Still stuck? Show hint 2 →
Hint 2 of 2
Let r = red count. Green = 2r, blue = 4r. Total = 7r — it must be a multiple of 7.
Show solution
Approach: find the hidden multiple
  1. Let r be the number of red marbles. Half as many red as green → green = 2r. Twice as many blue as green → blue = 4r.
  2. Total = r + 2r + 4r = 7r — always a multiple of 7.
  3. Among the choices, only 28 = 7 × 4 is a multiple of 7.
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.
APPENDIX

Test-day habits quick-reference

Memorize these

THE SEVEN HABITS

  1. Read the question twice. Underline the question word. What does it actually ask?
  2. Circle the tiny words. “inclusive,” “except,” “positive,” “non-negative,” “at most,” “different,” “distinct.”
  3. Match the units. Cents vs dollars. Minutes vs hours. Inches vs feet.
  4. Plug your answer back in. Does it satisfy every condition the problem stated?
  5. Don’t trust mental math past 3 digits. Write it on scratch.
  6. Re-read your final letter. The bubble has to match your work.
  7. Two-minute rule. If a problem’s taking too long, mark it and move on.

USE THE ANSWERS

The answer choices are part of the problem. Ask: what do they have in common? What’s DIFFERENT about each?

  • All whole numbers → the answer is a whole number.
  • Three choices obviously too big or too small → cross them off without solving.
  • Choices differ by units digit → find the units digit; you’re done.
  • Choices differ by parity → find the parity first.
Common traps
  • Stopping at step 1. The question may have asked for X+Y, but you found just X.
  • Wrong unit. Answer is right in seconds but the question wanted minutes.
  • Off-by-one inclusive vs exclusive. “Pages 5 to 12” is 8 pages, not 7.
  • Bubbling the wrong letter. You wrote (C) but bubbled (B). A 2-second re-read prevents this.
  • Burning time on a hard problem. Every minute past 2 on one problem is two easy problems you didn’t reach.
Warm-ups

Practice these habits on every problem — they should feel automatic by test day:

  • Underline the question word on every problem you solve.
  • Before computing, scan the answer choices — what kind of number is the answer?
  • After you write your answer, plug it back in.
  • Set a 2-minute timer per problem in practice. Mark anything you can’t finish.