1965 AHSME Problems/Problem 33: Difference between revisions
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==Solution== | == Problem == | ||
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 | |||
when given to the base <math>10</math>, then <math>k + h</math> equals: | |||
<math>\textbf{(A)}\ 5 \qquad | |||
\textbf{(B) }\ 6 \qquad | |||
\textbf{(C) }\ 7 \qquad | |||
\textbf{(D) }\ 8 \qquad | |||
\textbf{(E) }\ 9 </math> | |||
== Solution == | |||
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> | ||
Revision as of 12:54, 16 July 2024
Problem
If the number 
, that is, 
, ends with 
zeros when given to the base 
and ends with 
zeros
when given to the base 
, then 
equals:
Solution
We can use Legendre's to find the number of 
s in base
![\[\lfloor \frac{15}{5} \rfloor + \lfloor \frac{15}{25} \rfloor = 3\]](http://latex.artofproblemsolving.com/3/e/0/3e0d7ca65804a10128b7f54b77fec8e41729fcfa.png)
So 
.
Likewise, we are looking for the number of 
s and 
s that divide , so we use Legendre's again.
![\[\lfloor \frac{15}{2} \rfloor + \lfloor \frac{15}{4} \rfloor + \lfloor \frac{15}{8} \rfloor = 7 + 3 + 1 = 11\]](http://latex.artofproblemsolving.com/3/7/8/378195d015eae6f05632ab620b62d19dd528df06.png)
![\[\lfloor \frac{15}{3} \rfloor + \lfloor \frac{15}{9} \rfloor = 5 + 1 = 6\]](http://latex.artofproblemsolving.com/1/0/a/10a0f48b6948282f2c4c840101c47a6aca96d1ba.png)
Thus, 
and 
So 
, and 
~JustinLee2017
