Difference between revisions of "2002 AMC 10P Problems/Problem 22"
(→Solution 1) |
(→Solution 2) |
||
| Line 52: | Line 52: | ||
In case we have forgotten Legendre's formula or haven't learned it, this solution is equally viable. | In case we have forgotten Legendre's formula or haven't learned it, this solution is equally viable. | ||
| − | Cancel <math>1001!</math> from both sides of the fraction. We get <math>\frac{2002!}{(1001)!^2)=\frac{1002 \cdot 1003 \cdot \; \dots \; \cdot 2002}{1 \cdot 2 \; \dots \; \cdot 1001</math> | + | Cancel <math>1001!</math> from both sides of the fraction. We get <math>\frac{2002!}{(1001)!^2)=\frac{1002 \cdot 1003 \cdot \; \dots \; \cdot 2002}{1 \cdot 2 \; \dots \; \cdot 1001}</math> |
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=P|num-b=21|num-a=23}} | {{AMC10 box|year=2002|ab=P|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:40, 15 July 2024
Contents
Problem
In how many zeroes does the number 
end?
Solution 1
We can solve this problem with an application of Legendre's Formula.
We know that there will be an abundance of factors of 
compared to factors of 
so finding the amount of factors of 
is equivalent to finding how many factors of 
there are, which is equivalent to how many zeroes there are at the end of the number. Additionally, squaring a number will multiply the exponent of each factor by 
Therefore, we plug in 
and 
then plug in and 
and multiply by in:
![\[e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}\]](http://latex.artofproblemsolving.com/e/3/2/e321e2207deff3c5288006b566cc40a14eb2a944.png)
As such,
\begin{align*} e_5(2002!)=&\left\lfloor\frac{2002}{5}\right\rfloor+\left\lfloor\frac{2002}{5^2}\right\rfloor+\left\lfloor\frac {2002}{5^3}\right\rfloor+\left\lfloor\frac{2002}{5^4}\right\rfloor\\ =&400+80+16+3 \\ =&499 \end{align*}
or alternatively,
![$e_5(2002!)=\frac{2002-S_5(2002)}{5-1}=\frac{2002-S_5(31002_5)]}{4}=\frac{2002-6}{4}=499.$](http://latex.artofproblemsolving.com/d/c/1/dc121ebe3e5181feb2fbb033dc786a163424d83d.png)
Similarly,
\begin{align*} e_5(2002!)=&\left\lfloor\frac{1001}{5}\right\rfloor+\left\lfloor\frac{1001}{5^2}\right\rfloor+\left\lfloor\frac {1001}{5^3}\right\rfloor+\left\lfloor\frac{1001}{5^4}\right\rfloor\\ =&200+40+8+1 \\ =&299 \end{align*}
or alternatively,
![$e_5(1001!)=\frac{1001-S_5(1001)}{5-1}=\frac{1001-S_5(13001_5)]}{4}=\frac{1001-5}{4}=249.$](http://latex.artofproblemsolving.com/6/e/e/6ee0ad839c0857e2ca464310bf109cfd651e816d.png)
In any case, our answer is 
Solution 2
In case we have forgotten Legendre's formula or haven't learned it, this solution is equally viable.
Cancel 
from both sides of the fraction. We get $\frac{2002!}{(1001)!^2)=\frac{1002 \cdot 1003 \cdot \; \dots \; \cdot 2002}{1 \cdot 2 \; \dots \; \cdot 1001}$ (Error compiling LaTeX. Unknown error_msg)
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
