Difference between revisions of "2015 AMC 10B Problems/Problem 23"
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We first look at the case when <math>k!</math> has <math>1</math> zero and <math>(2k)!</math> has <math>3</math> zeros. If <math>k=5,6,7</math>, <math>(2k)!</math> has only <math>2</math> zeros. But for <math>k=8,9</math>, <math>(2k)!</math> has <math>3</math> zeros. Thus, <math>k=8</math> and <math>k=9</math> work. | We first look at the case when <math>k!</math> has <math>1</math> zero and <math>(2k)!</math> has <math>3</math> zeros. If <math>k=5,6,7</math>, <math>(2k)!</math> has only <math>2</math> zeros. But for <math>k=8,9</math>, <math>(2k)!</math> has <math>3</math> zeros. Thus, <math>k=8</math> and <math>k=9</math> work. | ||
− | Secondly, we look at the case when <math>k!</math> has <math>2</math> zeros and <math>(2k)!</math> has <math>6</math> zeros. If <math>k=10,11,12</math>, <math>(2k)!</math> has only <math>4</math> zeros. But for <math>k=13,14</math>, <math>(2k)!</math> has <math>6</math> zeros. Thus, the smallest four values of <math>k</math> that work are <math>k=8,9,13,14</math>, which sum to <math>44</math>. The sum of the digits of <math>44</math> is <math>\boxed{\mathbf{(B)\8}}</math> | + | Secondly, we look at the case when <math>k!</math> has <math>2</math> zeros and <math>(2k)!</math> has <math>6</math> zeros. If <math>k=10,11,12</math>, <math>(2k)!</math> has only <math>4</math> zeros. But for <math>k=13,14</math>, <math>(2k)!</math> has <math>6</math> zeros. Thus, the smallest four values of <math>k</math> that work are <math>k=8,9,13,14</math>, which sum to <math>44</math>. The sum of the digits of <math>44</math> is <math>\boxed{\mathbf{(B)\ 8}}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|before=Problem 3|num-a=5}} | {{AMC10 box|year=2015|ab=B|before=Problem 3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:50, 4 March 2015
Problem
Let be a positive integer greater than 4 such that the decimal representation of ends in zeros and the decimal representation of ends in zeros. Let denote the sum of the four least possible values of . What is the sum of the digits of ?
Solution
A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have:
We first look at the case when has zero and has zeros. If , has only zeros. But for , has zeros. Thus, and work.
Secondly, we look at the case when has zeros and has zeros. If , has only zeros. But for , has zeros. Thus, the smallest four values of that work are , which sum to . The sum of the digits of is
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 | ||
All AMC 10 Problems and Solutions |
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