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Difference between revisions of "2005 AMC 10A Problems/Problem 21"

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<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 11 </math>
 
<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 11 </math>
  
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==Solution 1==
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== Solution ==
 
If <math> 1+2+...+n </math> evenly [[divide]]s <math>6n</math>, then <math>\frac{6n}{1+2+...+n}</math> is an [[integer]].  
 
If <math> 1+2+...+n </math> evenly [[divide]]s <math>6n</math>, then <math>\frac{6n}{1+2+...+n}</math> is an [[integer]].  
  
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Since <math> 1+2+...+n = \frac{n(n+1)}{2} </math> we may substitute the [[RHS]] in the above [[fraction]].  
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Since <math> 1+2+...+n = \frac{n(n+1)}{2} </math> we may substitute the [[RHS]] in the above [[fraction]]. So the problem asks us for how many [[positive integer]]s <math>n</math> is <math>\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}</math> an integer, or equivalently when <math>k(n+1) = 12</math> for a positive integer <math>k</math>.  
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So the problem asks us for how many [[positive integer]]s <math>n</math> is <math>\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}</math> an integer.  
 
  
 
<math>\frac{12}{n+1}</math> is an integer when <math>n+1</math> is a [[divisor |factor]] of <math>12</math>.  
 
<math>\frac{12}{n+1}</math> is an integer when <math>n+1</math> is a [[divisor |factor]] of <math>12</math>.  
  
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The factors of <math>12</math> are <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>6</math>, and <math>12</math>.
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The factors of <math>12</math> are <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>6</math>, and <math>12</math>, so the possible values of <math>n</math> are <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, and <math>11</math>.
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So the possible values of <math>n</math> are <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, and <math>11</math>.
 
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But <math>0</math> isn't a positive integer, so only <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, and <math>11</math> are possible values of <math>n</math>.
 
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Therefore the number of possible values of <math>n</math> is <math>5\Longrightarrow \mathrm{(B)}</math>
 
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== Solution 2 ==
 
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The sum of the first <math>n</math> positive integers is <math>\frac{n(n+1)}{2}</math>. If this is to divide <math>6n</math>, then there exists a positive integer <math>k</math> such that:
 
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<math>k \cdot \frac{n(n+1)}{2} = 6n</math>
 
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<math>k(n+1) = 12</math>
 
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Therefore, <math>k</math> and <math>n+1</math> are divisors of <math>12</math>. There are <math>6</math> divisors of <math>12</math>, which are <math>1, 2, 3, 4, 6, 12</math>. The divisors which multiply to <math>12</math> can be assigned to <math>k</math> and <math>n+1</math> in either order. However, when <math>1</math> is assigned to <math>n+1</math>, then <math>n=0</math>, which is not possible, because <math>n</math> must be positive. Therefore, we have <math>6-1=5</math> values of <math>n</math> <math>\Rightarrow \boxed{B}</math>
 
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==See Also==
 
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*[[2005 AMC 10A Problems]]
 
  
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*[[2005 AMC 10A Problems/Problem 20|Previous Problem]]
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But <math>0</math> isn't a positive integer, so only <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, and <math>11</math> are possible values of <math>n</math>. Therefore the number of possible values of <math>n</math> is <math>5\Longrightarrow \boxed{\mathrm{(B)}}</math>.
  
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*[[2005 AMC 10A Problems/Problem 22|Next Problem]]
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==See also==
 +
{{AMC10 box|year=2005|ab=A|num-b=20|num-a=22}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Revision as of 12:49, 31 July 2008

Problem

For how many positive integers $n$ does $1+2+...+n$ evenly divide from $6n$?

$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 11$

Solution

If $1+2+...+n$ evenly divides $6n$, then $\frac{6n}{1+2+...+n}$ is an integer.

Since $1+2+...+n = \frac{n(n+1)}{2}$ we may substitute the RHS in the above fraction. So the problem asks us for how many positive integers $n$ is $\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}$ an integer, or equivalently when $k(n+1) = 12$ for a positive integer $k$.

$\frac{12}{n+1}$ is an integer when $n+1$ is a factor of $12$.

The factors of $12$ are $1$, $2$, $3$, $4$, $6$, and $12$, so the possible values of $n$ are $0$, $1$, $2$, $3$, $5$, and $11$.

But $0$ isn't a positive integer, so only $1$, $2$, $3$, $5$, and $11$ are possible values of $n$. Therefore the number of possible values of $n$ is $5\Longrightarrow \boxed{\mathrm{(B)}}$.

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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
All AMC 10 Problems and Solutions
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