Difference between revisions of "2015 AMC 10B Problems/Problem 23"
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Let <math>n=5m+k</math> for some natural numbers <math>m</math>, <math>k</math> such that <math>k\in\{0,1,2,3,4\}</math>. Notice that <math>n<5^3=125</math>. Thus | Let <math>n=5m+k</math> for some natural numbers <math>m</math>, <math>k</math> such that <math>k\in\{0,1,2,3,4\}</math>. Notice that <math>n<5^3=125</math>. Thus | ||
− | <cmath>3(\lfloor\frac{n}{5}\rfloor+\lfloor\frac{n}{25})=\lfloor\frac{2n}{5}\rfloor+\lfloor\frac{2n}{25}</cmath> | + | <cmath>3(\lfloor\frac{n}{5}\rfloor+\lfloor\frac{n}{25}\rfloor)=\lfloor\frac{2n}{5}\rfloor+\lfloor\frac{2n}{25}\rfloor+\rfloor\frac{2n}{125}\rfloor</cmath> |
==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 14:50, 9 August 2019
Contents
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
.
.
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
Let for some natural numbers
,
such that
. Notice that
. Thus
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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