Difference between revisions of "2002 AMC 10P Problems/Problem 22"
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or alternatively, | or alternatively, | ||
| − | <math>e_5( | + | <math>e_5(1001!)=\frac{2002-S_5(2002)}{5-1}=\frac{2002-S_5(31002_5)]}{4}=\frac{2002-6}{4}=499.</math> |
Similarly, | Similarly, | ||
\begin{align*} | \begin{align*} | ||
| − | e_5( | + | e_5(1001!)=&\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 \\ | =&200+40+8+1 \\ | ||
=&299 | =&299 | ||
Latest revision as of 04:20, 17 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,
![$e_5(1001!)=\frac{2002-S_5(2002)}{5-1}=\frac{2002-S_5(31002_5)]}{4}=\frac{2002-6}{4}=499.$](http://latex.artofproblemsolving.com/4/f/7/4f7105eda4d9d1319d6e85f8b04e28ce8d785157.png)
Similarly,
\begin{align*} e_5(1001!)=&\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. With similar reasoning to solution 1, all we need to find is the amount of multiples of in the problem.
Cancel 
from the top and bottom of the fraction. We get 
We can set a bijection between the two sets of terms with a multiple of 
Let 
be a number from that is a multiple of Its corresponding multiple of from 
will be 
For clarity, will group with 
will group with 
will group with 
Our goal is to find number(s) where 
for all 
will result in a multiple of Simplifying this equation, 
a multiple of We can conclude that 
is the only solution to this equation since is not divisible by any other number ending in a 
or 
(we can write the factors of to confirm this; 
and 
are the only numbers that do not end in or 
because all multiples of end in or ). A quick check reveals 
which has 
multiples of and 
which has multiples of Thus, is the only instance where there is an extra multiple of at the top, meaning 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. 
