# 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 $\boxed{\textbf{(K)}}$.

## See Also

 2007 iTest (Problems) Preceded by:Problem 12 Followed by:Problem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
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