Difference between revisions of "2015 AMC 10B Problems/Problem 23"
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We first note that the number of 0's in <math>n!</math> is determined by how many 5's are in the prime factorization. We use Legendre's Formula and split up into two cases: | We first note that the number of 0's in <math>n!</math> is determined by how many 5's are in the prime factorization. We use Legendre's Formula and split up into two cases: | ||
− | + | <math>\textbf{CASE ONE: } </math>5\leq 2n < 25.<math> | |
− | The only way we can fulfill the requirements is if <math>\lfloor \dfrac{n}{5} \rfloor = 1< | + | The only way we can fulfill the requirements is if </math>\lfloor \dfrac{n}{5} \rfloor = 1<math> and </math>\lfloor \dfrac{2n}{5} \rfloor=3<math> which means that </math>5\geq n <10<math> and </math>15\geq 2n 20<math>. The only way this works is if </math>n = 8 \text{ or } 9.<math> |
− | + | </math>\textbf{CASE TWO: } 25 \leq 2n<math>. | |
− | Since we want the smallest values of <math>n< | + | Since we want the smallest values of </math>n<math>, we first try it when </math>2n<30.<math> Thus </math>(2n)!<math> has 6 zeros, which implies that </math>n!<math> must have 2. The only way to do this while maintaining our restrictions for </math>2n<math> is if </math>n = 13 \text{ or } 14.<math> |
− | So the sum of the four values is <math>8+9+13+14=44< | + | So the sum of the four values is </math>8+9+13+14=44<math> so the digit is sum is </math>\boxed{\mathbf{(B)\ }8}.$ |
-ConfidentKoala4 | -ConfidentKoala4 |
Revision as of 19:18, 6 October 2022
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 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 .
We have 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
.
We find that the least four possible are
.
.
Solution 3
Let for some natural numbers
,
such that
. Notice that
. Thus
For smaller
, we temporarily let
To minimize
, we let
, then
Since
,
, the only integral value of
is
, from which we have
.
Now we let and
, then
Since
,
.
If , then
which is a contradiction.
Thus
Finally, the sum of the four smallest possible and
.
~ Nafer
Solution 4
We first note that the number of 0's in is determined by how many 5's are in the prime factorization. We use Legendre's Formula and split up into two cases:
5\leq 2n < 25.
\lfloor \dfrac{n}{5} \rfloor = 1
\lfloor \dfrac{2n}{5} \rfloor=3
5\geq n <10
15\geq 2n 20
n = 8 \text{ or } 9.$$ (Error compiling LaTeX. Unknown error_msg)\textbf{CASE TWO: } 25 \leq 2n$.
Since we want the smallest values of$ (Error compiling LaTeX. Unknown error_msg)n2n<30.
(2n)!
n!
2n
n = 13 \text{ or } 14.
8+9+13+14=44
\boxed{\mathbf{(B)\ }8}.$
-ConfidentKoala4
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.