Difference between revisions of "2002 AMC 10P Problems/Problem 22"
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We can solve this problem with an application of Legendre's formula. | We can solve this problem with an application of Legendre's formula. | ||
| − | We know that there will be an abundance of factors of <math>2</math> compared to factors of <math>5,</math> so finding the amount of factors of <math>5</math> is equivalent to finding how many factors of <math>10</math> there are. Therefore, we plug in p=5 and n=2002, then plug in p=5 and n=1001 in: | + | We know that there will be an abundance of factors of <math>2</math> compared to factors of <math>5,</math> so finding the amount of factors of <math>5</math> is equivalent to finding how many factors of <math>10</math> there are. Additionally, squaring a number will multiply the exponent of each factor by <math>2.</math> Therefore, we plug in <math>p=5</math> and <math>n=2002,</math> then plug in <math>p=5</math> and <math>n=1001</math> and multiply by <math>2</math> in: |
<cmath>e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}</cmath> | <cmath>e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}</cmath> | ||
Revision as of 06:38, 15 July 2024
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. 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)
\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 
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. 
