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Difference between revisions of "2002 AMC 10P Problems/Problem 22"

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(→‎Solution 2)
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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 15:40, 15 July 2024

Problem

In how many zeroes does the number $\frac{2002!}{(1001!)^2}$ end?

$\text{(A) }0 \qquad \text{(B) }1 \qquad \text{(C) }2 \qquad \text{(D) }200 \qquad \text{(E) }400$

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 $2$ compared to factors of $5,$ so finding the amount of factors of $5$ is equivalent to finding how many factors of $10$ 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 $2.$ Therefore, we plug in $p=5$ and $n=2002,$ then plug in $p=5$ and $n=1001$ and multiply by $2$ in:

\[e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}\]

As such,

e5(2002!)=⌊20025⌋+⌊200252⌋+⌊200253⌋+⌊200254⌋=400+80+16+3=499

or alternatively,

$e_5(2002!)=\frac{2002-S_5(2002)}{5-1}=\frac{2002-S_5(31002_5)]}{4}=\frac{2002-6}{4}=499.$

Similarly,

e5(2002!)=⌊10015⌋+⌊100152⌋+⌊100153⌋+⌊100154⌋=200+40+8+1=299

or alternatively,

$e_5(1001!)=\frac{1001-S_5(1001)}{5-1}=\frac{1001-S_5(13001_5)]}{4}=\frac{1001-5}{4}=249.$

In any case, our answer is $499-2(249)= \boxed{\textbf{(B) } 1}.$

Solution 2

In case we have forgotten Legendre's formula or haven't learned it, this solution is equally viable.

Cancel $1001!$ from both sides of the fraction. We get $\frac{2002!}{(1001)!^2}=\frac{1002 \cdot 1003 \cdot \; \dots \; \cdot 2002}{1 \cdot 2 \; \dots \; \cdot 1001}$

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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