# 2007 iTest Problems/Problem 13

## Problem

What is the smallest positive integer $k$ such that the number ${{2k}\choose k}$ ends in two zeros?

$\text{(A) } 3 \quad \text{(B) } 4 \quad \text{(C) } 5 \quad \text{(D) } 6 \quad \text{(E) } 7 \quad \text{(F) } 8 \quad \text{(G) } 9 \quad \text{(H) } 10 \quad \text{(I) } 11 \quad \text{(J) } 12 \quad \text{(K) } 13 \quad \text{(L) } 14 \quad \text{(M) } 2007\quad$

## Solution

When writing out $\tbinom{2k}{k}$, the numerator has the numbers from $k+1$ to $2k$ being multiplied, and the denominator has the numbers from $1$ to $k$ being multiplied. In order for $\tbinom{2k}{k}$ to have two zeroes, the numerator must have two more factors of $2$ and $5$ than the denominator.

Going through the options from lowest to highest, the first value of $k$ that satisfies the conditions is $13$ because there are $4$ factors of five and $12$ factors of two in the numerator, while there are $2$ factors of five and $10$ factors of two in the denominator. The answer is $\boxed{\textbf{(K)}}$.