Difference between revisions of "1970 AHSME Problems/Problem 23"

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== Solution ==
 
== Solution ==
<math>\fbox{D}</math>
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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>.
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There are <math>4</math> factors of <math>3</math>:  <math>3, 6, 9</math>, and an extra factor from <math>9</math>.
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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 05:42, 15 July 2019

Problem

The number $10!$ ($10$ is written in base $10$), when written in the base $12$ system, ends with exactly $k$ zeros. The value of $k$ is

$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$

Solution

A number in base $b$ that ends in exactly $k$ zeros will be divisible by $b^k$, but not by $b^{k+1}$. Thus, we want to find the highest $k$ for which $12^k | 10!$.

There are $4$ factors of $3$: $3, 6, 9$, and an extra factor from $9$.

There are $8$ factors of $2$: $2, 4, 6, 8, 10$, an extra factor from $4, 8$, and a third extra factor from $8$.

So, $2^8 \cdot 3^4 = 4^4 \cdot 3^4 = 12^4$ will divide $10!$. Thus, the answer is $\fbox{D}$.

See also

1970 AHSC (ProblemsAnswer KeyResources)
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

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