Difference between revisions of "2015 AMC 10B Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | Let <math>n</math> be a positive integer greater than 4 such that the decimal representation of <math>n!</math> ends in <math>k</math> zeros and the decimal representation of <math>(2n)!</math> ends in <math>3k</math> zeros. Let <math>s</math> denote the sum of the four least possible values of <math>n</math>. What is the sum of the digits of <math>s</math>? | + | Let <math>n</math> be a positive integer greater than <math>4</math> such that the decimal representation of <math>n!</math> ends in <math>k</math> zeros and the decimal representation of <math>(2n)!</math> ends in <math>3k</math> zeros. Let <math>s</math> denote the sum of the four least possible values of <math>n</math>. What is the sum of the digits of <math>s</math>? |
<math> \textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11 </math> | <math> \textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11 </math> | ||
− | ==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 17: | Line 17: | ||
==Solution 2== | ==Solution 2== | ||
− | By Legendre's Formula and the information given, we have that | + | By [[Legendre's Formula]] and the information given, we have that <math>3\left(\left\lfloor{\frac{n}{5}}\right\rfloor+\left\lfloor{\frac{n}{25}}\right\rfloor\right)=\left\lfloor{\frac{2n}{5}}\right\rfloor+\left\lfloor{\frac{2n}{25}}\right\rfloor</math>. |
− | <math> | + | We have <math>n<100</math> as there is no way that if <math>n>100</math>, <math>(2n)!</math> would have <math>3</math> times as many zeroes as <math>n!</math>. |
− | + | First, let's plug in the number <math>5</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> | 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>. | 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>. | ||
− | + | We find that the least four possible <math>n</math> are <math>8,9,13,14</math>. | |
<math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{B}</math>. | <math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{B}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | 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(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor+\left\lfloor\frac{2n}{125}\right\rfloor</cmath> | ||
+ | For smaller <math>n</math>, we temporarily let <math>\left\lfloor\frac{2n}{125}\right\rfloor=0</math> | ||
+ | <cmath>3(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor</cmath> | ||
+ | <cmath>3(\left\lfloor\frac{5m+k}{5}\right\rfloor+\left\lfloor\frac{5m+k}{25}\right\rfloor)=\left\lfloor\frac{2(5m+k)}{5}\right\rfloor+\left\lfloor\frac{2(5m+k)}{25}\right\rfloor</cmath> | ||
+ | <cmath>3(\left\lfloor\frac{5m+k}{5}\right\rfloor+\left\lfloor\frac{5m+k}{25}\right\rfloor)=\left\lfloor\frac{10m+2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor</cmath> | ||
+ | <cmath>3m+3\left\lfloor\frac{5m+k}{25}\right\rfloor=2m+\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor</cmath> | ||
+ | <cmath>m+3\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor</cmath> | ||
+ | To minimize <math>n</math>, we let <math>\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{10m+2k}{25}\right\rfloor=0</math>, then | ||
+ | <cmath>m=\left\lfloor\frac{2k}{5}\right\rfloor</cmath> | ||
+ | Since <math>k<5</math>, <math>m>0</math>, the only integral value of <math>m</math> is <math>1</math>, from which we have <math>k=3,4\Longrightarrow n=8,9</math>. | ||
+ | |||
+ | Now we let <math>\left\lfloor\frac{5m+k}{25}\right\rfloor=0</math> and <math>\left\lfloor\frac{10m+2k}{25}\right\rfloor=1</math>, then | ||
+ | <cmath>m=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor</cmath> | ||
+ | Since <math>k<5</math>, <math>10m>15\Longrightarrow m\ge2</math>. | ||
+ | |||
+ | If <math>m>2</math>, then | ||
+ | <cmath>m>\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor</cmath> | ||
+ | which is a contradiction. | ||
+ | |||
+ | Thus <math>m=2\Longrightarrow\left\lfloor\frac{2k}{5}\right\rfloor=1\Longrightarrow n=13,14</math> | ||
+ | |||
+ | Finally, the sum of the four smallest possible <math>n=8+9+13+14=44</math> and <math>4+4=8</math>. <math>\boxed{\mathrm{(B)}}</math> | ||
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | == Solution 4== | ||
+ | |||
+ | 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: } 5\leq 2n < 25.</math> | ||
+ | |||
+ | 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\leq n <10</math> and <math>15\leq 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</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</math> so the digit is sum is <math>\boxed{\mathbf{(B)\ }8}.</math> | ||
+ | |||
+ | -ConfidentKoala4 | ||
+ | |||
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | We will use trial and error to determine the answer to this problem. If <math>n = 5</math>, then <math>n!</math> has <math>1</math> zero, and <math>2n!</math> will have <math>2</math> zeros. But <math>3 \cdot 1 \neq 2</math> so <math>n = 5</math> does not work. Similarly <math>n = 6, 7</math> do not work either. But <math>n = 8</math> works because <math>8!</math> has <math>1</math> zero and <math>16!</math> has <math>3</math> zeros. Note that <math>n = 9</math> also works because <math>9!</math> has <math>1</math> zero and <math>18!</math> has <math>3</math> zeros. After performing trial and error several times, we find that <math>n = 10, 11, 12</math> do not work but <math>n = 13, 14</math> do work. Therefore, the four smallest values of <math>n</math> are <math>8, 9, 13, 14</math>. Therefore adding them together gives <math>44</math> and our answer is <math>\boxed{\mathbf{(B)\ }8}.</math> | ||
+ | |||
+ | -[[User: Yiyj1|Yiyj1]] | ||
==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}} |
Latest revision as of 01:16, 19 July 2024
Problem
Let be a positive integer greater than
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:
The only way we can fulfill the requirements is if and
which means that
and
. The only way this works is if
.
Since we want the smallest values of , we first try it when
Thus
has 6 zeros, which implies that
must have 2. The only way to do this while maintaining our restrictions for
is if
So the sum of the four values is so the digit is sum is
-ConfidentKoala4
Solution 5
We will use trial and error to determine the answer to this problem. If , then
has
zero, and
will have
zeros. But
so
does not work. Similarly
do not work either. But
works because
has
zero and
has
zeros. Note that
also works because
has
zero and
has
zeros. After performing trial and error several times, we find that
do not work but
do work. Therefore, the four smallest values of
are
. Therefore adding them together gives
and our answer is
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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