2015 AMC 10B Problems/Problem 23
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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