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2021 AMC 12A Problem 15

Problem 15 of 25IntermediateAlgebraCounting & Probability

A choir director must select a group of singers from among his 66 tenors and 88 basses. The only requirements are that the difference between the number of tenors and basses must be a multiple of 4,4, and the group must have at least one singer. Let NN be the number of groups that can be selected. What is the remainder when NN is divided by 100?100?

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Solution

Choosing tt tenors and bb basses is weighted by (6t)(8b).\binom{6}{t}\binom{8}{b}. To keep only tb0(mod4),t - b \equiv 0 \pmod 4, apply a roots of unity filter with ω=i:\omega = i: N+1=14j=03(1+ij)6(1+ij)8. \begin{aligned} &N + 1 \\ &= \frac14\sum_{j=0}^{3}(1 + i^{j})^6\,(1 + i^{-j})^8. \end{aligned} The j=0j = 0 term is 2628=16384.2^6\cdot 2^8 = 16384. The j=2j = 2 term has factor (1+i2)6=0.(1 + i^2)^6 = 0. The j=1j = 1 and j=3j = 3 terms are 128i-128i and 128i,128i, which cancel. So the sum is 16384,16384, and 163844=4096.\dfrac{16384}{4} = 4096. This count includes the empty group, so N=40961=4095,N = 4096 - 1 = 4095, and N95(mod100).N \equiv 95 \pmod{100}. Thus, the correct answer is D.

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Concepts: roots of unity · combinations · binomial theorem

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