Difference between revisions of "2015 AMC 10B Problems/Problem 23"
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===Solution 2=== | ===Solution 2=== | ||
− | By Legendre's Formula and the information given, we have that <math>3(\lfloor{\frac{n}{5}}\rfloor+\lfloor{\frac{n}{25}}\rfloor)=\lfloor{\frac{2n}{5}}\rfloor+\lfloor{\frac{2n}{25}}\rfloor</math>. | + | By Legendre's Formula and the information given, we have that <math>3\left(\left\lfloor{\frac{n}{5}}\right\rfloor+\left\lfloor{\frac{n}{25}}\right\rfloor\right)=\left\lfloor{\frac{2n}{5}}\right\rfloor+\left\lfloor{\frac{2n}{25}}\right\rfloor</math>. |
Trivially, it is obvious that <math>n<100</math> as there is no way that if <math>n>100</math>, <math>n!</math> would have <math>3</math> times as many zeroes as <math>(2n)!</math>. | Trivially, it is obvious that <math>n<100</math> as there is no way that if <math>n>100</math>, <math>n!</math> would have <math>3</math> times as many zeroes as <math>(2n)!</math>. |
Revision as of 11:07, 20 May 2018
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
Solution 1
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
Solution 2
By Legendre's Formula and the information given, we have that .
Trivially, it is obvious that as there is no way that if , would have times as many zeroes as .
First, let's plug in the number We get that , which is obviously not true. Hence,
After several attempts, we realize that the RHS needs to more "extra" zeroes than the LHS. Hence, is greater than a multiple of .
Very quickly, we find that the least are .
.
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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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