Algebra & Patterns — Turn the story into a letter.

Algebra word problems are word problems with one extra step: name the unknown. Once you've written "let x = the number of apples," every phrase in the problem becomes a fact you can write using x.

Common translations:

• "twice as many as" → 2x
• "5 more than" → x + 5
• "one less than half of" → x/2 − 1
• "three times the sum of x and 7" → 3(x + 7)
• "the difference between A and B" → A − B (order matters!)
• "product" → multiply. "quotient" → divide.

Practice this until it's reflexive. Most algebra problems get easier once you've written one equation. The trick is having the courage to write the equation before you 'know how' to solve it.

The bar-model approach (a Singapore-style trick) works for problems where one quantity is described in terms of another. Draw a bar for the unknown. Then draw any related quantity as a longer or shorter bar.

Example. Tom has $40 more than Jerry. Together they have $120. How much does each have?

Cover the gold $40 piece with your finger. What's left is two equal bars summing to $120 − $40 = $80. So each bar = $40 (Jerry). Tom = $40 + $40 = $80.

The bar showed you the equation visually: 2(Jerry) + 40 = 120.

Every AMC contest has at least one problem with a strange new symbol — ★, ♦, ♠, △, ◆ — and a definition. Don’t panic. The symbol is just a machine: it eats two inputs and spits out a number using the rule the problem gives you.

Put the LEFT number on the left side of the recipe, the RIGHT number on the right side, then compute. That’s it.

Order can matter

If the rule treats a and b differently (one gets squared, one gets multiplied), then swapping inputs changes the answer. Read carefully which is on the left.

Nested operations — work inside out

A sequence is a list of numbers following a rule. Two AMC favorites:

Same step (+3) every time. From a₁ to a₅, the step is taken 4 times, not 5.

The off-by-one trap. The first term has zero steps from itself; the 100th term has 99 steps. nth term = start + (n−1) × step. Always check on small cases.

Walkthrough. Find the 50th term of 7, 11, 15, 19, …

• First term a₁ = 7. Common difference d = 4.
• a₅₀ = 7 + (50 − 1) × 4 = 7 + 196 = 203.

Verify: a₁ = 7 + 0·4 = 7 ✓.

Two equations, two unknowns — collapse to ONE unknown using either of these:

The sum-and-difference shortcut

When the system is x + y = S and x − y = D:

• Add them: 2x = S + D, so x = (S + D)/2 — the AVERAGE.
• Subtract them: 2y = S − D, so y = (S − D)/2 — the half-difference.

Memorize this. AMC uses sum-and-difference constantly.

“Five positive numbers add to 50.” That’s a sum constraint. The total is fixed, so the numbers can’t all grow — if one goes up, another has to come down.

Imagine the sum as a fixed-size pizza divided into 5 slices. If you make one slice bigger, every other slice has to shrink to keep the total pizza the same.

The “distinct” trap

If the problem says distinct positive integers, you can’t use four 1’s. The smallest four distinct positive integers are 1, 2, 3, 4 — sum 10. So the largest would be 20 − 10 = 10. Big drop! Always re-read for the word “distinct.”

When AMC asks for the 50th, 100th, or 1000th term of a pattern, it's almost never asking you to compute that many terms. There's a cycle. Once you spot it, the question reduces to: 'which position in the cycle does the Nth term fall into?'

How to find the cycle:

1. Compute the first several terms (4 to 6 usually).
2. Look for when the sequence (or a key feature) starts repeating.
3. The cycle length = the number of terms between two consecutive repeats.

Then compute n mod cycle_length to find the position.

Cycle examples on AMC:

• Units digits of powers (covered in NT chapter 7).
• Days of the week (cycle 7).
• Repeating decimals (e.g., 1/7 = 0.142857142857… cycle 6).
• Rotations/folds that return to start.
• 'Apply a function repeatedly' — what does it return to?

When a problem describes a sequence of operations done to a mystery number and tells you the result, the cleanest approach is to undo each step in reverse order.

Think of it like getting dressed and undressed: if you put on socks, then shoes, then ties — to get back to bare feet you take off ties first, then shoes, then socks. Last on, first off.

Three algebraic identities the AMC tests repeatedly:

Why difference of squares is huge: if you see a² − b², immediately rewrite as (a+b)(a−b). This turns nasty subtraction into easy multiplication.

Examples:

• 99² − 98² = (99+98)(99−98) = 197 × 1 = 197. Computing 99² and 98² separately would be brutal.
• 51² − 49² = (51+49)(51−49) = 100 × 2 = 200.
• 1000² − 999² = 1999.

The reverse direction is also useful: when you see a product like (big)(small_diff), sometimes it's a hidden difference of squares.

Sum of consecutive odd numbers has a difference-of-squares flavor: 1 + 3 + 5 + … + (2n − 1) = n². (Why? Each new odd number turns an (n−1)² square into an n² square by adding an L-shaped border.)

Sequences of consecutive integers are everywhere on AMC. They have ONE structural fact that collapses most problems:

Why it works — pair from the ends

Pair the smallest with the largest, the 2nd-smallest with the 2nd-largest, and so on. Each pair has the SAME sum — twice the middle. The middle pairs with itself (it’s the fulcrum).

For 5 consecutive integers 8, 9, 10, 11, 12: two pairs of 20 (= 2·middle) plus the middle 10 = 50. Same as 5 × 10. Sum-equals-count-times-middle ✓.

What if n is even? The middle is a half-integer.

For 4 consecutive integers 7, 8, 9, 10: middle sits between 8 and 9, at 8.5. Sum = 4 × 8.5 = 34. Verify: 7+8+9+10 = 34 ✓.

For 6 consecutive integers summing to 2013: middle = 2013/6 = 335.5. So 3rd term = 335, 4th = 336. Integers: 333, 334, 335, 336, 337, 338. Largest = 338.

Other consecutive-integer facts

Page-number problems — a special flavor

“How many digits to number pages 1 through 200?” Break by digit-length blocks:

For pages 1–200: 9 + 180 + 3·(200−99) = 9 + 180 + 303 = 492 digits.

Two consecutive numbers multiplied = X

If n(n+1) = 1056, take the square root: √1056 ≈ 32.5. Try 32 × 33 = 1056 ✓. The integer just under √X is your n.

So far the sequences you've seen had clean closed-form rules — arithmetic (add d each time), geometric (multiply by r), or pattern (every k-th term repeats). But AMC also tests recursive sequences, where each term is built from several previous terms.

The strategy is always the same: start from the given terms and compute forward, one term at a time. Don't try to find a closed formula — for AMC 8, you almost never need one. Just build a table.

Walkthrough. The sequence starts 1, 2 and each term after is the product of the previous two. Find the 6th term.

• a₁ = 1, a₂ = 2.
• a₃ = 1 · 2 = 2.
• a₄ = 2 · 2 = 4.
• a₅ = 2 · 4 = 8.
• a₆ = 4 · 8 = 32.

The table method is robust and almost mistake-proof. The only place students slip is on the indexing — "6th term" means a₆, not a₅. Count carefully.

Working backwards from a known term

Some problems give you a middle or late term and ask you to find an earlier one. If the rule is reversible (like 'each is the product of the previous two'), you can divide backwards.

If a_n = a_{n-1} · a_{n-2}, then a_{n-2} = a_n / a_{n-1}. So once you know two consecutive terms, you can step backwards or forwards.

Function-iteration: apply the same rule over and over

A close cousin of recursion is function-iteration: start with a number, apply a rule, then apply the rule again to the result, and so on.

Example. The rule is 'double, then add 1'. Start at 3:

• Step 1: 2·3 + 1 = 7.
• Step 2: 2·7 + 1 = 15.
• Step 3: 2·15 + 1 = 31.
• Step 4: 2·31 + 1 = 63.

This is the same as the recursive rule a_n = 2·a_{n-1} + 1 with a₀ = 3. Once you see the equivalence, function-iteration problems are just recursion in disguise.

The Fibonacci-stairs story — recursion in real life

The most famous recursive sequence is Fibonacci: 1, 1, 2, 3, 5, 8, 13, 21, … where each term is the sum of the previous two. Where does it show up?

Here’s a story: you’re climbing a staircase. You can take 1 stair or 2 stairs at a time. How many different ways are there to climb n stairs?

Why does each f(n) = f(n−1) + f(n−2)? Look at the last step of any path to stair n. Either it’s a 1-step (so before that the climber was at stair n−1), or it’s a 2-step (so before that they were at stair n−2). Every way to reach stair n falls into one of these two camps. So total ways = ways to reach n−1 + ways to reach n−2.

This pattern shows up in any counting problem where you can take 1 or 2 steps and have to reach a target. Spot it once and you’ll always recognize it.