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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− | ==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> | ||
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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 13:54, 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
So
.
Likewise, we are looking for the number of
s and
s that divide
, so we use Legendre's again.
Thus,
and
So
, and
~JustinLee2017