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Difference between revisions of "2007 iTest Problems/Problem 13"

(Created page with "== Problem == What is the smallest positive integer <math>k</math> such that the number <math>{{2k}\choose k}</math> ends in two zeros? <math>\text{(A) } 3 \quad \text{(B) } 4 ...")
 
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\text{(M) } 2007\quad </math>
 
\text{(M) } 2007\quad </math>
  
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== Solution ==
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==Solution==
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When writing out <math>\tbinom{2k}{k}</math>, the numerator has the numbers from <math>k+1</math> to <math>2k</math> being multiplied, and the denominator has the numbers from <math>1</math> to <math>k</math> being multiplied.  In order for <math>\tbinom{2k}{k}</math> to have two zeroes, the numerator must have two more factors of <math>2</math> and <math>5</math> than the denominator.
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Going through the options from lowest to highest, the first value of <math>k</math> that satisfies the conditions is <math>13</math> because there are <math>4</math> factors of five and <math>12</math> factors of two in the numerator, while there are <math>2</math> factors of five and <math>10</math> factors of two in the denominator.  The answer is <math>\boxed{\textbf{(K)}}</math>.
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==See Also==
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{{iTest box|year=2007|num-b=12|num-a=14}}
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[[Category:Introductory Number Theory Problems]]
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[[Category:Introductory Combinatorics Problems]]

Latest revision as of 13:35, 4 August 2018

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, Answer Key)
Preceded by:
Problem 12 		Followed by:
Problem 14
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