Difference between revisions of "2007 iTest Problems/Problem 13"
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==Solution== | ==Solution== | ||
− | When writing out <math>\ | + | 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 14:35, 4 August 2018
Problem
What is the smallest positive integer such that the number ends in two zeros?
Solution
When writing out , the numerator has the numbers from to being multiplied, and the denominator has the numbers from to being multiplied. In order for to have two zeroes, the numerator must have two more factors of and than the denominator.
Going through the options from lowest to highest, the first value of that satisfies the conditions is because there are factors of five and factors of two in the numerator, while there are factors of five and factors of two in the denominator. The answer is .
See Also
2007 iTest (Problems) | ||
Preceded by: Problem 12 |
Followed by: Problem 14 | |
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