AMC 10 · 2020 · #19
Grade 6 arithmeticPick an answer.
AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #7 (Subproblems): the big calculation C(52, 10) has 10 factors on top and 10 on the bottom — break the cancellation into bite-size pieces. Tool #3 (Eliminate): with only 5 candidate values for A, test each against a divisibility rule (digit sum modulo 9). Tool #5 (Pattern): the digit sum 18 + 5A varies with A and modulo 9 gives a different residue for every candidate, so one divisibility check pins down the answer.
Write C(52, 10) as ten falling factors over 10!; cancel 10!'s small factors against the top, leaving denominator 252.
Cancel small factors against 10! first to shrink the arithmetic.
6.NS.B.4Identify SubproblemsCancel 252 against the top (51=3·17, 49=7·7, the 6, 52=2·26), leaving C(52, 10) = 26 · 17 · 5 · 7 · 47 · 46 · 11 · 43.
Match each factor of 10! to a piece of the top — what's left is the answer.
6.NS.B.4Identify SubproblemsChain the eight products (26·17=442, ·5, ·7, ·47, ·46, ·11, ·43) to reach C(52, 10) = 15,820,024,220.
Chain the multiplications — eight short products.
5.NBT.B.5Identify SubproblemsLine up 15,820,024,220 with 158A00A4AA0 digit by digit; every A slot holds the same value, giving A = 2, choice (A).
Line up the digits and read off A.
4.NBT.A.2Eliminate PossibilitiesCross-check by 9s: the template's digit sum is 18 + 4A, and 15,820,024,220 sums to 26 = 18 + 4·2, consistent with A = 2.
Digit sum is a quick cross-check that the answer matches the pattern.
4.NBT.A.2Look For A PatternThis AMC 10 problem only needs Grade 6 cancel-and-multiply you already know — simplify C(52, 10) by cancelling factors of 10!, then multiply through to get 15,820,024,220. Lining it up with 158A00A4AA0 shows every A is 2. The answer is (A).