Difference between revisions of "1970 AHSME Problems/Problem 23"
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== Solution == | == Solution == | ||
| − | <math>\fbox{D}</math> | + | A number in base <math>b</math> that ends in exactly <math>k</math> zeros will be divisible by <math>b^k</math>, but not by <math>b^{k+1}</math>. Thus, we want to find the highest <math>k</math> for which <math>12^k | 10!</math>. |
| + | |||
| + | There are <math>4</math> factors of <math>3</math>: <math>3, 6, 9</math>, and an extra factor from <math>9</math>. | ||
| + | |||
| + | There are <math>8</math> factors of <math>2</math>: <math>2, 4, 6, 8, 10</math>, an extra factor from <math>4, 8</math>, and a third extra factor from <math>8</math>. | ||
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| + | So, <math>2^8 \cdot 3^4 = 4^4 \cdot 3^4 = 12^4</math> will divide <math>10!</math>. Thus, the answer is <math>\fbox{D}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 06:42, 15 July 2019
Problem
The number 
(
is written in base ), when written in the base 
system, ends with exactly 
zeros. The value of is
Solution
A number in base 
that ends in exactly zeros will be divisible by 
, but not by 
. Thus, we want to find the highest for which 
.
There are 
factors of 
: 
, and an extra factor from 
.
There are 
factors of 
: 
, an extra factor from 
, and a third extra factor from .
So, 
will divide . Thus, the answer is 
.
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
