AMC 10 · 2024 · #1

Grade 7 arithmeticalgebra
multi-digit-arithmeticplace-valuepattern-recognition identify-subproblemsconvert-to-algebra ↑ Prerequisites: multi-digit-arithmeticorder-of-operations
📏 Short solution 💡 2 insights
Problem
Compute the value of 9901 · 101 - 99 · 10101 and match it to one of the five answer choices.

Pick an answer.

(A)
2
(B)
20
(C)
200
(D)
202
(E)
2020
How to solve
Strategy Use Symmetry / Transform

Brute-force multiplication of 9901 · 101 and 99 · 10101 would give two five-digit numbers and an easy place for a careless slip. Instead, notice that 9901 = 99 · 100 + 1 and 10101 = 101 · 100 + 1 — both numbers split into a product of the other given factors plus 1. That symmetry is exactly what Tool #16 looks for: rewrite the expression so a large piece cancels. Tool #4 (let a = 99, b = 101) makes the symmetry visible at a glance and turns the whole problem into one line of algebra.

1STEP 1

Rewrite the big numbers

Both giants are a hundred times something, plus one.

a = 99, b = 101, 9901 = 99 · 100 + 1 = 100a + 1, 10101 = 101 · 100 + 1 = 100b + 1
2STEP 2

Substitute the letters

The whole expression becomes a formula in a and b.

9901 · 101 - 99 · 10101 = (100a + 1) · b - a · (100b + 1)
3STEP 3

Expand it

The same 100ab appears on both sides.

(100a + 1) · b - a · (100b + 1) = 100ab + b - 100ab - a
4STEP 4

Read the leftover

What is left is b minus a, namely 2.

100ab + b - 100ab - a = b - a = 101 - 99 = 2 → (A)
Answer
2
The answer choices jump from 2 to 20 to 200 to 202 to 2020, so they're testing whether the student noticed the cancellation. A direct check confirms it: 9901 · 101 = 1,000,001 and 99 · 10101 = 999,999, and 1,000,001 - 999,999 = 2. The two giant products are within 2 of each other, exactly matching b - a = 101 - 99, so (A) is the only consistent choice.
💡Key takeaway

Big-looking numbers often hide a small structure. Naming 99 and 101 as a and b exposes the matching 100ab piece that cancels — leaving the AMC 12 opener as a Grade 7 distributive-property exercise!