Difference between revisions of "2015 AMC 10B Problems/Problem 23"
m (→Solution 2) |
m (→Solution) |
||
Line 5: | Line 5: | ||
==Solution== | ==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: | 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: | ||
Line 15: | Line 17: | ||
Secondly, we look at the case when <math>n!</math> has <math>2</math> zeros and <math>(2n)!</math> has <math>6</math> zeros. If <math>n=10,11,12</math>, <math>(2n)!</math> has only <math>4</math> zeros. But for <math>n=13,14</math>, <math>(2n)!</math> has <math>6</math> zeros. Thus, the smallest four values of <math>n</math> that work are <math>n=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>n!</math> has <math>2</math> zeros and <math>(2n)!</math> has <math>6</math> zeros. If <math>n=10,11,12</math>, <math>(2n)!</math> has only <math>4</math> zeros. But for <math>n=13,14</math>, <math>(2n)!</math> has <math>6</math> zeros. Thus, the smallest four values of <math>n</math> that work are <math>n=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> | ||
+ | |||
+ | ===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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | First, let's plug in the number <math>5</math> | ||
+ | We get that <math>3(1)=1</math>, which is obviously not true. Hence, <math>n>5</math> | ||
+ | |||
+ | After several attempts, we realize that the RHS needs <math>1</math> to <math>2</math> more "extra" zeroes than the LHS. Hence, <math>n</math> is greater than a multiple of <math>5</math>. | ||
+ | |||
+ | Very quickly, we find that the least <math>4 n's</math> are <math>8,9,13,14</math>. | ||
+ | |||
+ | <math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{B}</math>. | ||
==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 11:06, 26 July 2017
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.