Difference between revisions of "2006 AMC 12B Problems/Problem 22"

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== Problem ==
 
== Problem ==
 +
Suppose <math>a</math>, <math>b</math> and <math>c</math> are positive integers with <math>a+b+c=2006</math>, and <math>a!b!c!=m\cdot 10^n</math>, where <math>m</math> and <math>n</math> are integers and <math>m</math> is not divisible by <math>10</math>. What is the smallest possible value of <math>n</math>?
  
== Solution ==
+
<math>
 +
\mathrm{(A)}\ 489
 +
\qquad
 +
\mathrm{(B)}\ 492
 +
\qquad
 +
\mathrm{(C)}\ 495
 +
\qquad
 +
\mathrm{(D)}\ 498
 +
\qquad
 +
\mathrm{(E)}\ 501
 +
 
 +
</math>
 +
 
 +
== Solution 1==
 +
The power of <math>10</math> for any factorial is given by the well-known algorithm
 +
<cmath>\left\lfloor \frac
 +
n{5}\right\rfloor +
 +
\left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \cdots</cmath>
 +
It is rational to guess numbers right before powers of <math>5</math> because we won't have any extra numbers from higher powers of <math>5</math>. As we list out the powers of 5, it is clear that <math>5^{4}=625</math> is less than 2006 and  <math>5^{5}=3125</math> is greater. Therefore, set <math>a</math> and <math>b</math> to be 624. Thus, c is <math>2006-(624\cdot 2)=758</math>. Applying the algorithm, we see that our answer is <math>152+152+188= \boxed{492}</math>.
 +
 
 +
== Solution 2==
 +
Clearly, the power of <math>2</math> that divides <math>n!</math> is larger or equal than the power of <math>5</math> which divides
 +
it. Hence we are trying to minimize the power of <math>5</math> that will divide <math>a!b!c!</math>.
 +
 
 +
Consider <math>n! = 1\cdot 2 \cdot \dots \cdot n</math>. Each fifth term is divisible by <math>5</math>, each <math>25</math>-th one
 +
by <math>25</math>, and so on. Hence the total power of <math>5</math> that divides <math>n</math> is <math>\left\lfloor \frac
 +
n{5}\right\rfloor +
 +
\left\lfloor \frac n{25}\right\rfloor + \cdots</math>. (For any <math>n</math> only finitely many terms in the sum
 +
are
 +
non-zero.)
 +
 
 +
In our case we have <math>a<2006</math>, so the largest power of <math>5</math> that will be less than <math>a</math> is at most
 +
<math>5^4 = 625</math>. Therefore the power of <math>5</math> that divides <math>a!</math> is equal to <math>\left\lfloor \frac
 +
a{5}\right\rfloor +
 +
\left\lfloor \frac a{25}\right\rfloor + \left\lfloor \frac a{125}\right\rfloor + \left\lfloor \frac
 +
a{625}\right\rfloor </math>. The same
 +
is true for <math>b</math> and <math>c</math>.
 +
 
 +
Intuition may now try to lure us to split <math>2006</math> into <math>a+b+c</math> as evenly as possible, giving
 +
<math>a=b=669</math> and <math>c=668</math>. However, this solution is not optimal.
 +
 
 +
To see how we can do better, let's rearrange the terms as follows:
 +
 
 +
<cmath>
 +
\begin{align*}
 +
\text{result}
 +
& = \Big\lfloor \frac a{5}\Big\rfloor + \Big\lfloor \frac b{5}\Big\rfloor + \Big\lfloor \frac
 +
c{5}\Big\rfloor
 +
\\
 +
& + \Big\lfloor \frac a{25}\Big\rfloor + \Big\lfloor \frac b{25}\Big\rfloor + \Big\lfloor \frac
 +
c{25}\Big\rfloor
 +
\\
 +
& + \Big\lfloor \frac a{125}\Big\rfloor + \Big\lfloor \frac b{125}\Big\rfloor + \Big\lfloor \frac
 +
c{125}\Big\rfloor
 +
\\
 +
& + \Big\lfloor \frac a{625}\Big\rfloor + \Big\lfloor \frac b{625}\Big\rfloor + \Big\lfloor \frac
 +
c{625}\Big\rfloor
 +
\end{align*}
 +
</cmath>
 +
 
 +
The idea is that the rows of the above equation are roughly equal to <math>\left\lfloor \frac
 +
n{5}\right\rfloor</math>, <math>\left\lfloor \frac n{25}\right\rfloor</math>, etc.
 +
 
 +
More precisely, we can now notice that for any positive integers <math>a,b,c,k</math> we can write <math>a,b,c</math> in the form <math>a=a_0k
 +
+ a_1</math>, <math>b=b_0k+b_1</math>, <math>c=c_0k + c_1</math>, where all <math>a_i,b_i,c_i</math> are integers and <math>0\leq
 +
a_1,b_1,c_1<k</math>.
 +
 
 +
It follows that
 +
<cmath>\Big\lfloor \frac a{k}\Big\rfloor + \Big\lfloor \frac b{k}\Big\rfloor + \Big\lfloor \frac
 +
c{k}\Big\rfloor = a_0+b_0+c_0</cmath>
 +
and
 +
<cmath>\Big\lfloor \frac {a+b+c}k\Big\rfloor = a_0 + b_0 + c_0 + \Big\lfloor \frac
 +
{a_1+b_1+c_1}k\Big\rfloor \leq a_0 + b_0 + c_0 + 2 </cmath>
 +
 
 +
Hence we get that for any positive integers <math>a,b,c,k</math> we have
 +
<cmath>
 +
\Big\lfloor \frac a{k}\Big\rfloor + \Big\lfloor \frac b{k}\Big\rfloor + \Big\lfloor \frac
 +
c{k}\Big\rfloor
 +
\quad
 +
\geq
 +
\quad
 +
\Big\lfloor \frac {a+b+c}k\Big\rfloor - 2
 +
</cmath>
 +
 
 +
Therefore for any <math>a,b,c</math> the result is at least <math>\left\lfloor \frac n{5}\right\rfloor +
 +
\left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor +
 +
\left\lfloor \frac n{625}\right\rfloor - 8 = 401 + 80 + 16 + 3 - 8 = 500 - 8 = 492</math>.
 +
 
 +
If we now show how to pick <math>a,b,c</math> so that we'll get the result <math>492</math>, we will be done.
 +
 
 +
Consider the row with <math>625</math> in the denominator. We need to achieve sum <math>1</math> in this row,
 +
hence we need to make two of the numbers smaller than <math>625</math>. Choosing <math>a=b=624</math>
 +
does this, and it will give us the largest possible remainders for <math>a</math> and <math>b</math> in
 +
the other three rows, so this is a pretty good candidate. We can compute
 +
<math>c=2006-a-b=758</math> and verify that this triple gives the desired result <math>\boxed{492}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=21|num-a=23}}
 
{{AMC12 box|year=2006|ab=B|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Latest revision as of 15:52, 23 June 2021

Problem

Suppose $a$, $b$ and $c$ are positive integers with $a+b+c=2006$, and $a!b!c!=m\cdot 10^n$, where $m$ and $n$ are integers and $m$ is not divisible by $10$. What is the smallest possible value of $n$?

$\mathrm{(A)}\ 489 \qquad \mathrm{(B)}\ 492  \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 498 \qquad \mathrm{(E)}\ 501$

Solution 1

The power of $10$ for any factorial is given by the well-known algorithm \[\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \cdots\] It is rational to guess numbers right before powers of $5$ because we won't have any extra numbers from higher powers of $5$. As we list out the powers of 5, it is clear that $5^{4}=625$ is less than 2006 and $5^{5}=3125$ is greater. Therefore, set $a$ and $b$ to be 624. Thus, c is $2006-(624\cdot 2)=758$. Applying the algorithm, we see that our answer is $152+152+188= \boxed{492}$.

Solution 2

Clearly, the power of $2$ that divides $n!$ is larger or equal than the power of $5$ which divides it. Hence we are trying to minimize the power of $5$ that will divide $a!b!c!$.

Consider $n! = 1\cdot 2 \cdot \dots \cdot n$. Each fifth term is divisible by $5$, each $25$-th one by $25$, and so on. Hence the total power of $5$ that divides $n$ is $\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \cdots$. (For any $n$ only finitely many terms in the sum are non-zero.)

In our case we have $a<2006$, so the largest power of $5$ that will be less than $a$ is at most $5^4 = 625$. Therefore the power of $5$ that divides $a!$ is equal to $\left\lfloor \frac a{5}\right\rfloor + \left\lfloor \frac a{25}\right\rfloor + \left\lfloor \frac a{125}\right\rfloor + \left\lfloor \frac a{625}\right\rfloor$. The same is true for $b$ and $c$.

Intuition may now try to lure us to split $2006$ into $a+b+c$ as evenly as possible, giving $a=b=669$ and $c=668$. However, this solution is not optimal.

To see how we can do better, let's rearrange the terms as follows:

\begin{align*} \text{result} & = \Big\lfloor \frac a{5}\Big\rfloor + \Big\lfloor \frac b{5}\Big\rfloor + \Big\lfloor \frac c{5}\Big\rfloor \\ & + \Big\lfloor \frac a{25}\Big\rfloor + \Big\lfloor \frac b{25}\Big\rfloor + \Big\lfloor \frac c{25}\Big\rfloor \\ & + \Big\lfloor \frac a{125}\Big\rfloor + \Big\lfloor \frac b{125}\Big\rfloor + \Big\lfloor \frac c{125}\Big\rfloor \\ & + \Big\lfloor \frac a{625}\Big\rfloor + \Big\lfloor \frac b{625}\Big\rfloor + \Big\lfloor \frac c{625}\Big\rfloor \end{align*}

The idea is that the rows of the above equation are roughly equal to $\left\lfloor \frac n{5}\right\rfloor$, $\left\lfloor \frac n{25}\right\rfloor$, etc.

More precisely, we can now notice that for any positive integers $a,b,c,k$ we can write $a,b,c$ in the form $a=a_0k + a_1$, $b=b_0k+b_1$, $c=c_0k + c_1$, where all $a_i,b_i,c_i$ are integers and $0\leq a_1,b_1,c_1<k$.

It follows that \[\Big\lfloor \frac a{k}\Big\rfloor + \Big\lfloor \frac b{k}\Big\rfloor + \Big\lfloor \frac c{k}\Big\rfloor = a_0+b_0+c_0\] and \[\Big\lfloor \frac {a+b+c}k\Big\rfloor = a_0 + b_0 + c_0 + \Big\lfloor \frac {a_1+b_1+c_1}k\Big\rfloor \leq a_0 + b_0 + c_0 + 2\]

Hence we get that for any positive integers $a,b,c,k$ we have \[\Big\lfloor \frac a{k}\Big\rfloor + \Big\lfloor \frac b{k}\Big\rfloor + \Big\lfloor \frac c{k}\Big\rfloor \quad \geq \quad \Big\lfloor \frac {a+b+c}k\Big\rfloor - 2\]

Therefore for any $a,b,c$ the result is at least $\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \left\lfloor \frac n{625}\right\rfloor - 8 = 401 + 80 + 16 + 3 - 8 = 500 - 8 = 492$.

If we now show how to pick $a,b,c$ so that we'll get the result $492$, we will be done.

Consider the row with $625$ in the denominator. We need to achieve sum $1$ in this row, hence we need to make two of the numbers smaller than $625$. Choosing $a=b=624$ does this, and it will give us the largest possible remainders for $a$ and $b$ in the other three rows, so this is a pretty good candidate. We can compute $c=2006-a-b=758$ and verify that this triple gives the desired result $\boxed{492}$.

See also

2006 AMC 12B (ProblemsAnswer KeyResources)
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 12 Problems and Solutions

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