Difference between revisions of "1965 AHSME Problems/Problem 33"
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| − | ==Solution== | + | == Problem == |
| − | We can use Legendre's to find the number of <math>0</math>s in base <math>10</math> | + | |
| + | 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 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>. | ||
| Line 6: | Line 18: | ||
<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>\ | + | 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 
, 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 Formula 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 
, which corresponds to answer 
.
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
