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

(Created page with "== Problem == The number <math>10!</math> (<math>10</math> is written in base <math>10</math>), when written in the base <math>12</math> system, ends with exactly <math>k</math>...")
 
 
(One intermediate revision by one other user not shown)
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== Solution ==
 
== Solution ==
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<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>.
 +
 
 +
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 ==
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{{AHSME box|year=1970|num-b=22|num-a=24}}   
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{{AHSME 35p box|year=1970|num-b=22|num-a=24}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

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 (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

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