Difference between revisions of "1965 AHSME Problems/Problem 33"
(Created page with "==Solution== 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...") |
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<cmath>\lfloor \frac{15}{2} \rfloor + \lfloor \frac{15}{4} \rfloor + \lfloor \frac{15}{8} \rfloor = 7 + 3 + 1 = 11</cmath> | <cmath>\lfloor \frac{15}{2} \rfloor + \lfloor \frac{15}{4} \rfloor + \lfloor \frac{15}{8} \rfloor = 7 + 3 + 1 = 11</cmath> | ||
<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> <math>\boxed{D}</math> | ||
~JustinLee2017 | ~JustinLee2017 | ||
Latest revision as of 20:13, 13 February 2021
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
