Difference between revisions of "2015 AMC 10B Problems/Problem 23"
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===Solution 3 (Bashing)=== | ===Solution 3 (Bashing)=== | ||
We notice that for a <math>0</math> to be at the end of a factorial, one multiple of five must be there. Therefore, it is intuitive to start at <math>5</math> and work up. If you bash enough you get <math>8</math>, <math>9</math>, <math>13</math>, and <math>14</math>. Going any higher will give too many zeros, and then we can stop going higher. <math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{\textbf{(B) }8}</math>. | We notice that for a <math>0</math> to be at the end of a factorial, one multiple of five must be there. Therefore, it is intuitive to start at <math>5</math> and work up. If you bash enough you get <math>8</math>, <math>9</math>, <math>13</math>, and <math>14</math>. Going any higher will give too many zeros, and then we can stop going higher. <math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{\textbf{(B) }8}</math>. | ||
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| + | == Solution 4 == | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2015|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:19, 9 August 2019
Contents
[hide]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:
![\[\begin{array}{c|c|c|c|c|c|c} \mathrm{Factorial}&0!-4!&5!-9!&10!-14!&15!-19!&20!-24!&25!-29!\\\hline \mathrm{Zeros}&0&1&2&3&4&6 \end{array}\]](http://latex.artofproblemsolving.com/0/0/4/00411b340b9a09060a6751fbc9f261972418994d.png)
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 
.
.
Solution 3 (Bashing)
We notice that for a 
to be at the end of a factorial, one multiple of five must be there. Therefore, it is intuitive to start at and work up. If you bash enough you get 
, 
, 
, and 
. Going any higher will give too many zeros, and then we can stop going higher. 
.
Solution 4
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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