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Difference between revisions of "1965 AHSME Problems/Problem 33"

m (Put a {} around 11)
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==Solution==
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== Problem ==
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If the number <math>15!</math>, that is, <math>15 \cdot 14 \cdot 13 \dots 1</math>, ends with <math>k</math> zeros when given to the base <math>12</math> and ends with <math>h</math> zeros
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when given to the base <math>10</math>, then <math>k + h</math> equals:
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<math>\textbf{(A)}\ 5 \qquad
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\textbf{(B) }\ 6 \qquad
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\textbf{(C) }\ 7 \qquad
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\textbf{(D) }\ 8 \qquad
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\textbf{(E) }\ 9  </math> 
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== Solution ==
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We can use Legendre's to find the number of <math>0</math>s in base <math>10</math>
 
We can use Legendre's to find the number of <math>0</math>s in base <math>10</math>
 
<cmath>\lfloor \frac{15}{5} \rfloor + \lfloor \frac{15}{25} \rfloor = 3</cmath>
 
<cmath>\lfloor \frac{15}{5} \rfloor + \lfloor \frac{15}{25} \rfloor = 3</cmath>

Latest revision as of 13:54, 16 July 2024

Problem

If the number $15!$, that is, $15 \cdot 14 \cdot 13 \dots 1$, ends with $k$ zeros when given to the base $12$ and ends with $h$ zeros when given to the base $10$, then $k + h$ equals:

$\textbf{(A)}\ 5 \qquad \textbf{(B) }\ 6 \qquad \textbf{(C) }\ 7 \qquad \textbf{(D) }\ 8 \qquad \textbf{(E) }\ 9$

Solution

We can use Legendre's to find the number of $0$s in base $10$ \[\lfloor \frac{15}{5} \rfloor + \lfloor \frac{15}{25} \rfloor = 3\] So $h = 3$. Likewise, we are looking for the number of $2^2$s and $3$s that divide $15!$, so we use Legendre's again. \[\lfloor \frac{15}{2} \rfloor + \lfloor \frac{15}{4} \rfloor + \lfloor \frac{15}{8} \rfloor = 7 + 3 + 1 = 11\] \[\lfloor \frac{15}{3} \rfloor + \lfloor \frac{15}{9} \rfloor = 5 + 1 = 6\] Thus, $3^6 \vert 15!$ and $2^{11} \vert 15! \Rrightarrow (2^2)^5 \vert 15!$ So $k = 5$, and $5+3 = 8$ $\boxed{D}$

~JustinLee2017

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