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

(Solution to Problem 13)
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==Solution==
 
==Solution==
  
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When writing out <math>\binom{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>\binom{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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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.
  
 +
<br>
 
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>.
 
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>.
  

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
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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