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

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== Solution ==
 
== Solution ==
  
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We can use Legendre's to find the number of <math>0</math>s in base <math>10</math>
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We can use [[Legendre's Formula]] 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>
 
So <math>h = 3</math>.
 
So <math>h = 3</math>.
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<cmath>\lfloor \frac{15}{3} \rfloor + \lfloor \frac{15}{9} \rfloor = 5 + 1 = 6</cmath>
 
<cmath>\lfloor \frac{15}{3} \rfloor + \lfloor \frac{15}{9} \rfloor = 5 + 1 = 6</cmath>
 
Thus, <math>3^6 \vert 15!</math> and <math>2^{11} \vert 15! \Rrightarrow (2^2)^5 \vert 15!</math>
 
Thus, <math>3^6 \vert 15!</math> and <math>2^{11} \vert 15! \Rrightarrow (2^2)^5 \vert 15!</math>
−
So <math>k = 5</math>, and <math>5+3 = 8</math> <math>\boxed{D}</math>
+
So <math>k = 5</math>, and <math>5+3 = 8</math>, which corresponds to answer <math>\fbox{\textbf{(D)}}</math>.
  
 
~JustinLee2017
 
~JustinLee2017
 +
 +
== See Also ==
 +
{{AHSME 40p box|year=1965|num-b=32|num-a=34}}
 +
{{MAA Notice}}
 +
[[Category:Introductory Number Theory Problems]]

Latest revision as of 11:05, 19 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 Formula 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 $5+3 = 8$, which corresponds to answer $\fbox{\textbf{(D)}}$.

~JustinLee2017

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 32 		Followed by
Problem 34
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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