AMC 10 · 2024 · #1
Grade 7 arithmeticalgebraPick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Let a = 99 and b = 101, so the big numbers become 100a + 1 and 100b + 1.
Giving the repeated numbers short names is the Grade 6 "letters stand for numbers" move; it makes the hidden symmetry pop out.
6.EE.A.2Use Matrix LogicRewrite the whole expression as (100a + 1) · b - a · (100b + 1), lining the two 100 factors up in parallel.
Same expression, new clothing — the substitution doesn't change the value, but it lines up matching pieces.
6.EE.A.3Count The ComplementDistribute both products to get 100ab + b - 100ab - a, where 100ab shows up twice with opposite signs.
Distribute carefully, especially the minus sign across the second parenthesis.
7.EE.A.1Count The ComplementThe two 100ab terms cancel, leaving b - a = 101 - 99, which is answer (A).
Once the heavy term cancels, only the two tiny extras +b and -a survive — and their difference is just 2.
7.EE.A.1Count The ComplementBig-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 10 opener as a Grade 7 distributive-property exercise!