Difference between revisions of "2002 AMC 10P Problems/Problem 22"
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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 <math>5</math> in the problem. | 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 <math>5</math> in the problem. | ||
− | Cancel <math>1001!</math> from the top and bottom 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> We can set a bijection between the two sets | + | Cancel <math>1001!</math> from the top and bottom 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> We can set a bijection between the two sets of terms with a multiple of <math>5.</math> Let <math>n</math> be a number from <math>1001!</math> that is a multiple of <math>5.</math> Its corresponding multiple of <math>5</math> from <math>\frac{2002!}{1001!}</math> will be <math>n+1000.</math> For clarity, <math>5</math> will group with <math>1005,</math> <math>10</math> will group with <math>1010</math> <math> \; \dots \; </math> <math>1000</math> will group with <math>2000.</math> Our goal is to find number(s) where <math>\frac{n+1000}{n}</math> for all <math>n \leq 1000, n \in 5\mathbb{Z}^{+}</math> will result in a multiple of <math>5.</math> Simplifying this equation, <math>1+\frac{1000}{n}=</math> a multiple of <math>5.</math> We can conclude that <math>n=250</math> is the only solution to this equation since <math>1000</math> is not divisible by any other number ending in a <math>4</math> or <math>9</math> (we can write the factors of <math>1000</math> to confirm this; <math>2^1=2,</math> <math>2^2=4,</math> and <math>2^3=8</math> are the only numbers that do not end in <math>5</math> or <math>0</math> because all multiples of <math>5</math> end in <math>5</math> or <math>0</math>). A quick check reveals <math>\frac{1250}{5}=250,</math> which has <math>3</math> multiples of <math>5</math> and <math>\frac{250}{5}=50,</math> which has <math>2</math> multiples of <math>5.</math> Thus, <math>n=250</math> is the only instance where there is an extra multiple of <math>5</math> at the top, meaning our answer is <math>\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}} |
Revision as of 21:26, 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:
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
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 |
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