Difference between revisions of "2002 AMC 10P Problems/Problem 22"
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== Solution 1== | == Solution 1== | ||
− | We can solve this problem with an application of Legendre's | + | We can solve this problem with an application of [[Legendre's Formula|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, 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 <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> | ||
+ | |||
+ | As such, | ||
\begin{align*} | \begin{align*} | ||
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<math>e_5(1001!)=\frac{1001-S_5(1001)}{5-1}=\frac{1001-S_5(13001_5)]}{4}=\frac{1001-5}{4}=249.</math> | <math>e_5(1001!)=\frac{1001-S_5(1001)}{5-1}=\frac{1001-S_5(13001_5)]}{4}=\frac{1001-5}{4}=249.</math> | ||
− | In any case, our answer is <math>499-2(249)= | + | In any case, our answer is <math>499-2(249)= \boxed{\textbf{(B) } 1}.</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}} |
Latest revision as of 09:00, 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, 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:
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,
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,
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 |
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