Track / Stage 4 / 87 of 340 #347 of 2000

Problem 347

AMC 12 late, AIME early

Number theory Difficulty 4.7 Find the answer

4. The maximum two-digit prime factor of the integer $n=\mathrm{C}_{200}^{100}$ is ( ).
(A) 61
(B) 67
(C) 83
(D) 97

Your answer

Official solution

4. (A).

Let $p$ be the largest two-digit prime factor of $n=\mathrm{C}_{200}^{100}=\frac{200!}{(100!)^{2}}$ , then $0<p<100$ .

Since $(100!)^{2}$ contains $p^{2}$ , 200! must contain at least $p^{3}$ . Therefore, $3 p<200$ , which means $p<\frac{200}{3}<67$ , yielding $p=61$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.