Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 12A Problems/Problem 23"

(Created page with '== Problem == The number obtained from the last two nonzero digits of <math>90!</math> is equal to <math>n</math>. What is <math>n</math>? <math>\textbf{(A)}\ 12 \qquad \textbf…')
 
 
(59 intermediate revisions by 26 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #23]] and [[2010 AMC 10A Problems|2010 AMC 10A #24]]}}
 +
 
== Problem ==
 
== Problem ==
  
Line 5: Line 7:
 
<math>\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68</math>
 
<math>\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68</math>
  
−
== Solution ==
+
== Hints and Method of Attack ==
 +
Let <math>P</math> be the result of dividing <math>90!</math> by tens such that <math>P</math> is not divisible by <math>10</math>. We want to consider <math>P \mod 100</math>. But because <math>100</math> is not prime, and because <math>P</math> is obviously divisible by <math>4</math> (if in doubt, look at the answer choices), we only need to consider <math>P \mod 25</math>.
 +
 
 +
However, <math>25</math> is a very particular number. <math>1 \cdot 2 \cdot 3 \cdot 4 \equiv -1 \text{ }(\text{mod }25)</math>, and so is <math>6 \cdot 7 \cdot 8 \cdot 9</math>. How can we group terms to take advantage of this fact?
 +
 
 +
There might be a problem when you cancel out the <math>10</math>s from <math>90!</math>. One method is to cancel out a factor of <math>2</math> from an existing number along with a factor of <math>5</math>. But this might prove cumbersome, as the grouping method will not be as effective. Instead, take advantage of ''inverses'' in modular arithmetic. Just leave the negative powers of <math>2</math> in a "storage base," and take care of the other terms first. Then, use Fermat's Little Theorem to solve for the power of <math>2</math>.
 +
 
 +
Video Solution: https://youtu.be/30CamkkifHM?t=766
 +
 
 +
== Solution 1==
  
 
We will use the fact that for any integer <math>n</math>,
 
We will use the fact that for any integer <math>n</math>,
Line 12: Line 23:
 
&=24\pmod{25}\equiv -1\pmod{25}.\end{align*}</cmath>
 
&=24\pmod{25}\equiv -1\pmod{25}.\end{align*}</cmath>
  
−
First, we find that the number of factors of <math>10</math> in <math>90!</math> is equal to <math>\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21</math>. Let <math>N=\frac{90!}{10^{21}}</math>. The <math>n</math> we want is therefore the last two digits of <math>N</math>, or <math>n\pmod{100}</math>. Since there is clearly an excess of factors of 2, we know that <math>N\equiv 0\pmod 4</math>, so it remains to find <math>n\pmod{25}</math>.
+
First, we find that the number of factors of <math>10</math> in <math>90!</math> is equal to <math>\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21</math>. Let <math>N=\frac{90!}{10^{21}}</math>. The <math>n</math> we want is therefore the last two digits of <math>N</math>, or <math>N\pmod{100}</math>. If instead we find <math>N\pmod{25}</math>, we know that <math>N\pmod{100}</math>, what we are looking for, could be <math>N\pmod{25}</math>, <math>N\pmod{25}+25</math>, <math>N\pmod{25}+50</math>, or <math>N\pmod{25}+75</math>. Only one of these numbers will be a multiple of four, and whichever one that is will be the answer, because <math>N\pmod{100}</math> has to be a multiple of 4.
  
−
If we take out all the factors of <math>5</math> in <math>N</math>, we can write <math>N</math> as <math>\frac M{2^{21}}</math> where
+
If we divide <math>90!</math> by <math>5^{21}</math> to create <math>M</math> by taking out all the factors of <math>5</math> in <math>90!</math>, we can write <math>N</math> as <math>\frac M{2^{21}}</math> where
 
<cmath>M=1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 6\cdot 7\cdot 8\cdot 9\cdot 2\cdots 89\cdot 18,</cmath>
 
<cmath>M=1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 6\cdot 7\cdot 8\cdot 9\cdot 2\cdots 89\cdot 18,</cmath>
−
where every factor of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form <math>5n</math> is replaced by <math>n</math>, and every number in the form <math>25n</math> is replaced by <math>n</math>.
+
where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form <math>5n</math> is replaced by <math>n</math>, and every number in the form <math>25n</math> is replaced by <math>n</math>.
  
 
The number <math>M</math> can be grouped as follows:
 
The number <math>M</math> can be grouped as follows:
Line 23: Line 34:
 
&\cdot (1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots (16\cdot 17\cdot 18) \\
 
&\cdot (1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots (16\cdot 17\cdot 18) \\
 
&\cdot (1\cdot 2\cdot 3).\end{align*}</cmath>
 
&\cdot (1\cdot 2\cdot 3).\end{align*}</cmath>
 +
 +
Where the first line is composed of the numbers in <math>90!</math> that aren't multiples of five, the second line is the multiples of five '''and not 25''' after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25.
  
 
Using the identity at the beginning of the solution, we can reduce <math>M</math> to
 
Using the identity at the beginning of the solution, we can reduce <math>M</math> to
Line 28: Line 41:
 
<cmath>\begin{align*}M&\equiv(-1)^{18} \cdot (-1)^3(16\cdot 17\cdot 18) \cdot (1\cdot 2\cdot 3) \\
 
<cmath>\begin{align*}M&\equiv(-1)^{18} \cdot (-1)^3(16\cdot 17\cdot 18) \cdot (1\cdot 2\cdot 3) \\
 
&= 1\cdot -21\cdot 6\\
 
&= 1\cdot -21\cdot 6\\
−
&= -1\pmod{25}=24\pmod{25}.\end{align*}</cmath>
+
&= -1\pmod{25} =24\pmod{25}.\end{align*}</cmath>
  
 
Using the fact that <math>2^{10}=1024\equiv -1\pmod{25}</math> (or simply the fact that <math>2^{21}=2097152</math> if you have your powers of 2 memorized), we can deduce that <math>2^{21}\equiv 2\pmod{25}</math>. Therefore <math>N=\frac M{2^{21}}\equiv \frac {24}2\pmod{25}=12\pmod{25}</math>.
 
Using the fact that <math>2^{10}=1024\equiv -1\pmod{25}</math> (or simply the fact that <math>2^{21}=2097152</math> if you have your powers of 2 memorized), we can deduce that <math>2^{21}\equiv 2\pmod{25}</math>. Therefore <math>N=\frac M{2^{21}}\equiv \frac {24}2\pmod{25}=12\pmod{25}</math>.
  
−
Finally, combining with the fact that <math>N\equiv 0\pmod 4</math> yields <math>n=\boxed{12\ \textbf{(A)}}</math>.
+
Finally, combining with the fact that <math>N\equiv 0\pmod 4</math> yields <math>n=\boxed{\textbf{(A)}\ 12}</math>.
 +
 
 +
==Solution 2==
 +
Let <math>P</math> be <math>90!</math> after we truncate its zeros. Notice that <math>90!</math> has exactly (floored) <math>\left\lfloor\frac{90}{5}\right\rfloor + \left\lfloor\frac{90}{25}\right\rfloor = 21</math> factors of 5; thus, <cmath>P = 2^{-21}\cdot 5^{-21}\cdot 90!.</cmath> We shall consider <math>P</math> modulo 4 and 25, to determine its residue modulo 100. It is easy to prove that <math>P</math> is divisible by 4 (consider the number of 2s dividing <math>90!</math> minus the number of 5s dividing <math>90!</math>), and so we only need to consider <math>P</math> modulo 25.
 +
 
 +
Now, notice that for integers <math>a, n</math> we have<cmath>(5n + a)(5n - a) \equiv -a^2 \mod 25.</cmath>
 +
 
 +
Thus, for integral a: <cmath>(10a + 1)(10a + 2)(10a + 3)(10a + 4)(10a + 6)(10a + 7)(10a + 8)(10a + 9) \equiv (-1)(-4)(-9)(-16) \equiv 576 \equiv 1 \mod 25.</cmath> Using this process, we can essentially remove all the numbers which had not formerly been a multiple of 5 in <math>90!</math> from consideration.
 +
 
 +
Now, we consider the remnants of the 5, 10, 15, 20, ..., 90 not yet eliminated. After factoring out a 2 from each of the 9 even numbers in this sequence, the 10, 20, 30, ..., 90 becomes 1, 2, 3, 4, 1, 6, 7, 8, 9, whose product is 1 mod 25. We have accounted for 9 of the 21 2's thus we still need to multiply by <math>2^{-12}</math>. Also, the 5, 15, 25, ..., 85 becomes 1, 3, 1, 7, 9, 11, 13, 3, 17. Multiplying out 1, 3, 1, 7, ..., 17 yields 2 modulo 25, and so we are left to compute <math>2^{-11}</math>. Also note that <math>\phi(25)=5^2-5^1=20</math>. Euler's Theorem states that <math>2^{\phi(25)} \equiv 1 \mod 25</math>, thus <math>2^{-11} \equiv 2^{-11+20} = 2^9 = 512 \equiv 12 \mod 25</math>. Because 12 is also a multiple of 4, we can utilize the Chinese Remainder Theorem to show that <math>P \equiv 12 \mod 100</math> and so the answer is <math>\boxed{12}</math>.
 +
 
 +
==Solution 3==
 +
We start of by truncating the <math>0</math>s off <math>90!</math>, just like Solution 2. Since there are <math>v_5(90!) = \left\lfloor\frac{90}{5}\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor = 21</math> terminating zeroes, we have the number we obtain from truncating the terminating zeroes at the end of <math>90!</math> will be <math>\frac{90!}{10^{21}}</math>.
 +
 
 +
By [[Chinese Remainder Theorem]], we can divide the mod <math>100</math> into mod <math>4</math> and mod <math>25</math>. We know that there are way more than <math>23</math> <math>2</math>s in <math>90!</math>, so we have
 +
 
 +
<cmath>\dfrac{90!}{10^{21}} \equiv 0 \text{ } \text{ (mod } 4 \text{)}</cmath>
 +
 
 +
Now, notice that <math>\frac{90!}{5^{21}}</math> is basically <math>90!</math> without any factors that are multiples of <math>5</math>.
 +
 
 +
<cmath>1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdots 23\cdot 24 \text{ } \text{ (mod } 25 \text{)}</cmath>
 +
 
 +
We can rewrite the expression as
 +
 
 +
<cmath>(1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdots 12)^2 \text{ } \text{ (mod } 25 \text{)}</cmath>
 +
 
 +
We also have
 +
 
 +
<cmath>1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdots 12 \equiv 8\cdot 8\cdot 9\cdot 11\cdot 12 \equiv 7 \text{ } \text{ (mod } 25 \text{)}</cmath>
 +
 
 +
Squaring the expression gets us
 +
 
 +
<cmath>(1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdots 12)^2 \equiv 49 \equiv -1 \text{ } \text{ (mod } 25 \text{)}</cmath>
 +
 
 +
Notice that there are <math>3</math> of this string of numbers multiplied together occurs four times in <math>\frac{90!}{10^{21}}</math>, the fourth being only a partial
 +
 
 +
<math>1\cdot 2\cdot 3\cdot 4\cdots 14 \equiv 24 \equiv -1 \text{ } \text{ (mod } 25 \text{)}</math>
 +
 
 +
Note that as in Solution 1, the residue from extracting the fives is
 +
<math>1\cdot 2\cdot 3\cdot 4\cdots 18 \cdot 1 \cdot 2 \cdot 3 \equiv -1 \text{ } \text{ (mod } 25 \text{)}</math>
 +
 
 +
Multiplying these together gives us
 +
 
 +
<cmath>\frac{90!}{10^{21}} \equiv \frac{\frac{90!}{5^{21}}}{2^{21}} \equiv \frac{(-1)^5}{2^{21}} \equiv \frac{-1}{2^{21}} \text{ } \text{ (mod } 25 \text{)}</cmath>
 +
 
 +
By Euler's Theorem, we have <math>\frac{-1}{2^{21}} \equiv -13^{21} \equiv -13^{\phi{(25)}}\cdot 13 \equiv -13 \equiv 12 \text{ } \text{ (mod } 25 \text{)}</math>
 +
 
 +
Thus, we want to solve <math>x \equiv 12 \text{ } \text{ (mod } 25 \text{)}</math> and <math>x \equiv 0 \text{ } \text{ (mod } 4 \text{)}</math>. An obvious solution is <math>x\equiv 12</math> and by Chinese Remainder Theorem, we have <math>x\equiv 12</math> is the only solution. So thus, <math>\frac{90!}{10^{21}} \equiv \boxed{12} \text{ } \text{ (mod } 100 \text{)}</math>
 +
 
 +
~sml1809
 +
 
 +
~minor edits by [https://artofproblemsolving.com/wiki/index.php/User:Kevinchen_yay KevinChen_Yay]
 +
 
 +
==Remark (Chinese Remainder Theorem)==
 +
 
 +
 
 +
Both solution 1 and solution 2 rely on <math>n \equiv 12 (\bmod { \quad 25})</math>, <math>n \equiv 0 (\bmod { \quad 4})</math> to get <math>n (\bmod { \quad 100})</math>
 +
 
 +
By [https://artofproblemsolving.com/wiki/index.php/Chinese_Remainder_Theorem Chinese Remainder Theorem], the general solution of the system of <math>2</math> linear congruences is:
 +
<math>n \equiv r_1 (\bmod { \quad m_1})</math>, <math>n \equiv r_2 (\bmod { \quad m_2})</math>, <math>(m_1, m_2) = 1</math>
 +
Find <math>k_1</math> and <math>k_2</math> such that <math>k_1 m_1 \equiv 1 (\bmod{ \quad m_2})</math>, <math>k_2 m_2 \equiv 1 (\bmod{ \quad m_1})</math>
 +
Then <math>n \equiv k_2 m_2 r_1 + k_1 m_1 r_2 (\bmod{ \quad m_1 m_2})</math>
 +
 
 +
In this problem, <math>n \equiv 12 (\bmod { \quad 25})</math>, <math>n \equiv 0 (\bmod { \quad 4})</math>:
 +
<math>n \equiv 12 (\bmod{ \quad 25})</math>, <math>n \equiv 0 (\bmod{ \quad 4})</math>, <math>(25, 4) = 1</math>
 +
<math>1 \cdot 25 \equiv 1 (\bmod{ \quad 4})</math>, <math>19 \cdot 4 \equiv 1 (\bmod{ \quad 25})</math>
 +
Then <math>n \equiv 19 \cdot 4 \cdot 12 + 1 \cdot 25 \cdot 0 \equiv 12 (\bmod{ \quad 100})</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
=
 +
== See also ==
 +
{{AMC12 box|year=2010|num-b=22|num-a=24|ab=A}}
 +
{{AMC10 box|year=2010|ab=A|num-b=23|num-a=25}}
 +
 
 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:12, 14 October 2024

The following problem is from both the 2010 AMC 12A #23 and 2010 AMC 10A #24, so both problems redirect to this page.

Problem

The number obtained from the last two nonzero digits of $90!$ is equal to $n$. What is $n$?

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68$

Hints and Method of Attack

Let $P$ be the result of dividing $90!$ by tens such that $P$ is not divisible by $10$. We want to consider $P \mod 100$. But because $100$ is not prime, and because $P$ is obviously divisible by $4$ (if in doubt, look at the answer choices), we only need to consider $P \mod 25$.

However, $25$ is a very particular number. $1 \cdot 2 \cdot 3 \cdot 4 \equiv -1 \text{ }(\text{mod }25)$, and so is $6 \cdot 7 \cdot 8 \cdot 9$. How can we group terms to take advantage of this fact?

There might be a problem when you cancel out the $10$s from $90!$. One method is to cancel out a factor of $2$ from an existing number along with a factor of $5$. But this might prove cumbersome, as the grouping method will not be as effective. Instead, take advantage of inverses in modular arithmetic. Just leave the negative powers of $2$ in a "storage base," and take care of the other terms first. Then, use Fermat's Little Theorem to solve for the power of $2$.

Video Solution: https://youtu.be/30CamkkifHM?t=766

Solution 1

We will use the fact that for any integer $n$, \begin{align*}(5n+1)(5n+2)(5n+3)(5n+4)&=[(5n+4)(5n+1)][(5n+2)(5n+3)]\\ &=(25n^2+25n+4)(25n^2+25n+6)\equiv 4\cdot 6\\ &=24\pmod{25}\equiv -1\pmod{25}.\end{align*}

First, we find that the number of factors of $10$ in $90!$ is equal to $\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21$. Let $N=\frac{90!}{10^{21}}$. The $n$ we want is therefore the last two digits of $N$, or $N\pmod{100}$. If instead we find $N\pmod{25}$, we know that $N\pmod{100}$, what we are looking for, could be $N\pmod{25}$, $N\pmod{25}+25$, $N\pmod{25}+50$, or $N\pmod{25}+75$. Only one of these numbers will be a multiple of four, and whichever one that is will be the answer, because $N\pmod{100}$ has to be a multiple of 4.

If we divide $90!$ by $5^{21}$ to create $M$ by taking out all the factors of $5$ in $90!$, we can write $N$ as $\frac M{2^{21}}$ where \[M=1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 6\cdot 7\cdot 8\cdot 9\cdot 2\cdots 89\cdot 18,\] where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form $5n$ is replaced by $n$, and every number in the form $25n$ is replaced by $n$.

The number $M$ can be grouped as follows:

\begin{align*}M= &(1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots(86\cdot 87\cdot 88\cdot 89)\\ &\cdot (1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots (16\cdot 17\cdot 18) \\ &\cdot (1\cdot 2\cdot 3).\end{align*}

Where the first line is composed of the numbers in $90!$ that aren't multiples of five, the second line is the multiples of five and not 25 after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25.

Using the identity at the beginning of the solution, we can reduce $M$ to

\begin{align*}M&\equiv(-1)^{18} \cdot (-1)^3(16\cdot 17\cdot 18) \cdot (1\cdot 2\cdot 3) \\ &= 1\cdot -21\cdot 6\\ &= -1\pmod{25} =24\pmod{25}.\end{align*}

Using the fact that $2^{10}=1024\equiv -1\pmod{25}$ (or simply the fact that $2^{21}=2097152$ if you have your powers of 2 memorized), we can deduce that $2^{21}\equiv 2\pmod{25}$. Therefore $N=\frac M{2^{21}}\equiv \frac {24}2\pmod{25}=12\pmod{25}$.

Finally, combining with the fact that $N\equiv 0\pmod 4$ yields $n=\boxed{\textbf{(A)}\ 12}$.

Solution 2

Let $P$ be $90!$ after we truncate its zeros. Notice that $90!$ has exactly (floored) $\left\lfloor\frac{90}{5}\right\rfloor + \left\lfloor\frac{90}{25}\right\rfloor = 21$ factors of 5; thus, \[P = 2^{-21}\cdot 5^{-21}\cdot 90!.\] We shall consider $P$ modulo 4 and 25, to determine its residue modulo 100. It is easy to prove that $P$ is divisible by 4 (consider the number of 2s dividing $90!$ minus the number of 5s dividing $90!$), and so we only need to consider $P$ modulo 25.

Now, notice that for integers $a, n$ we have\[(5n + a)(5n - a) \equiv -a^2 \mod 25.\]

Thus, for integral a: \[(10a + 1)(10a + 2)(10a + 3)(10a + 4)(10a + 6)(10a + 7)(10a + 8)(10a + 9) \equiv (-1)(-4)(-9)(-16) \equiv 576 \equiv 1 \mod 25.\] Using this process, we can essentially remove all the numbers which had not formerly been a multiple of 5 in $90!$ from consideration.

Now, we consider the remnants of the 5, 10, 15, 20, ..., 90 not yet eliminated. After factoring out a 2 from each of the 9 even numbers in this sequence, the 10, 20, 30, ..., 90 becomes 1, 2, 3, 4, 1, 6, 7, 8, 9, whose product is 1 mod 25. We have accounted for 9 of the 21 2's thus we still need to multiply by $2^{-12}$. Also, the 5, 15, 25, ..., 85 becomes 1, 3, 1, 7, 9, 11, 13, 3, 17. Multiplying out 1, 3, 1, 7, ..., 17 yields 2 modulo 25, and so we are left to compute $2^{-11}$. Also note that $\phi(25)=5^2-5^1=20$. Euler's Theorem states that $2^{\phi(25)} \equiv 1 \mod 25$, thus $2^{-11} \equiv 2^{-11+20} = 2^9 = 512 \equiv 12 \mod 25$. Because 12 is also a multiple of 4, we can utilize the Chinese Remainder Theorem to show that $P \equiv 12 \mod 100$ and so the answer is $\boxed{12}$.

Solution 3

We start of by truncating the $0$s off $90!$, just like Solution 2. Since there are $v_5(90!) = \left\lfloor\frac{90}{5}\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor = 21$ terminating zeroes, we have the number we obtain from truncating the terminating zeroes at the end of $90!$ will be $\frac{90!}{10^{21}}$.

By Chinese Remainder Theorem, we can divide the mod $100$ into mod $4$ and mod $25$. We know that there are way more than $23$ $2$s in $90!$, so we have

\[\dfrac{90!}{10^{21}} \equiv 0 \text{ } \text{ (mod } 4 \text{)}\]

Now, notice that $\frac{90!}{5^{21}}$ is basically $90!$ without any factors that are multiples of $5$.

\[1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdots 23\cdot 24 \text{ } \text{ (mod } 25 \text{)}\]

We can rewrite the expression as

\[(1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdots 12)^2 \text{ } \text{ (mod } 25 \text{)}\]

We also have

\[1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdots 12 \equiv 8\cdot 8\cdot 9\cdot 11\cdot 12 \equiv 7 \text{ } \text{ (mod } 25 \text{)}\]

Squaring the expression gets us

\[(1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdots 12)^2 \equiv 49 \equiv -1 \text{ } \text{ (mod } 25 \text{)}\]

Notice that there are $3$ of this string of numbers multiplied together occurs four times in $\frac{90!}{10^{21}}$, the fourth being only a partial

$1\cdot 2\cdot 3\cdot 4\cdots 14 \equiv 24 \equiv -1 \text{ } \text{ (mod } 25 \text{)}$

Note that as in Solution 1, the residue from extracting the fives is $1\cdot 2\cdot 3\cdot 4\cdots 18 \cdot 1 \cdot 2 \cdot 3 \equiv -1 \text{ } \text{ (mod } 25 \text{)}$

Multiplying these together gives us

\[\frac{90!}{10^{21}} \equiv \frac{\frac{90!}{5^{21}}}{2^{21}} \equiv \frac{(-1)^5}{2^{21}} \equiv \frac{-1}{2^{21}} \text{ } \text{ (mod } 25 \text{)}\]

By Euler's Theorem, we have $\frac{-1}{2^{21}} \equiv -13^{21} \equiv -13^{\phi{(25)}}\cdot 13 \equiv -13 \equiv 12 \text{ } \text{ (mod } 25 \text{)}$

Thus, we want to solve $x \equiv 12 \text{ } \text{ (mod } 25 \text{)}$ and $x \equiv 0 \text{ } \text{ (mod } 4 \text{)}$. An obvious solution is $x\equiv 12$ and by Chinese Remainder Theorem, we have $x\equiv 12$ is the only solution. So thus, $\frac{90!}{10^{21}} \equiv \boxed{12} \text{ } \text{ (mod } 100 \text{)}$

~sml1809

~minor edits by KevinChen_Yay

Remark (Chinese Remainder Theorem)

Both solution 1 and solution 2 rely on $n \equiv 12 (\bmod { \quad 25})$, $n \equiv 0 (\bmod { \quad 4})$ to get $n (\bmod { \quad 100})$

By Chinese Remainder Theorem, the general solution of the system of $2$ linear congruences is: 

$n \equiv r_1 (\bmod { \quad m_1})$, $n \equiv r_2 (\bmod { \quad m_2})$, $(m_1, m_2) = 1$
Find $k_1$ and $k_2$ such that $k_1 m_1 \equiv 1 (\bmod{ \quad m_2})$, $k_2 m_2 \equiv 1 (\bmod{ \quad m_1})$
Then $n \equiv k_2 m_2 r_1 + k_1 m_1 r_2 (\bmod{ \quad m_1 m_2})$

In this problem, $n \equiv 12 (\bmod { \quad 25})$, $n \equiv 0 (\bmod { \quad 4})$: 

$n \equiv 12 (\bmod{ \quad 25})$, $n \equiv 0 (\bmod{ \quad 4})$, $(25, 4) = 1$
$1 \cdot 25 \equiv 1 (\bmod{ \quad 4})$, $19 \cdot 4 \equiv 1 (\bmod{ \quad 25})$
Then $n \equiv 19 \cdot 4 \cdot 12 + 1 \cdot 25 \cdot 0 \equiv 12 (\bmod{ \quad 100})$

~isabelchen =

See also

2010 AMC 12A (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 12 Problems and Solutions
2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_23&oldid=229849"