1983 AIME Problems/Problem 8

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Problem

What is the largest 2-digit prime factor of the integer ${200\choose 100}$?

Solution

Expanding the binomial coefficient, we get ${200 \choose 100}=\frac{200!}{100!100!}$. Let the prime be $p$; then $10 \le p < 100$. If $p > 50$, then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor three times in the numerator, or $3p<200$. The largest such prime is $\boxed{061}$, which is our answer.

See also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions