2020 AMC 10B Problem 19

The number is $N = \binom{52}{10} = \dfrac{52!}{10!\,42!}$. Instead of computing it, test divisibility by small primes using the exponent of each prime in the factorials.

Prime 3. Exponent in $52!$: $17 + 5 + 1 = 23$. In $10!$: $3 + 1 = 4$. In $42!$: $14 + 4 + 1 = 19$. Net: $23 - 23 = 0$, so $3 \nmid N$. The digit sum of $158A00A4AA0$ is $18 + 4A$, which must not be a multiple of $3$: this rules out $A = 3$ and $A = 6$.

Prime 2. Exponent in $52!$: $26 + 13 + 6 + 3 + 1 = 49$. In $10!$: $5 + 2 + 1 = 8$. In $42!$: $21 + 10 + 5 + 2 + 1 = 39$. Net: $49 - 47 = 2$, so $N$ is divisible by $4$ but not by $8$.

Divisible by $4$: the last two digits $A0$ must form a multiple of $4$, so $A$ is even; this rules out $7$. Not divisible by $8$: the last three digits $AA0$ must not be a multiple of $8$. For $A = 4$, $440 = 8 \cdot 55$ fails; for $A = 2$, $220 = 8 \cdot 27.5$ is fine.

Only $A = 2$ survives (indeed $\binom{52}{10} = 15\,820\,024\,220$).

The answer is $\boxed{\textbf{(A)}\ 2}$.

A number given with unknown digits is a divisibility puzzle: you cannot compute it, but you can know exactly which small primes divide it via Legendre's formula $\sum \lfloor n/p^k \rfloor$ applied to the factorials, and the digit rules for $3$, $4$ and $8$ translate that into constraints on $A$. The "not divisible" facts are as useful as the "divisible" ones.

• Trying to compute C(52,10) exactly, or using divisibility by 11 (which every choice satisfies) and getting nowhere.
• Assuming $N$ is divisible by $9$ or by $3$ because the factorials contain many threes, without subtracting the exponents in the denominator.