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2005 AMC 10A Problems/Problem 21: Difference between revisions

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==Problem==
==Problem==
For how many positive integers <math>n</math> does <math> 1+2+...+n </math> evenly divide from <math>6n</math>?  
For how many positive integers <math>n</math> does <math>1+2+\dotsb+n</math> evenly divide <math>6n</math>?  


<math> \textbf{(A) } 3\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 9\qquad \textbf{(E) } 11 </math>
<math>
\textbf{(A) } 3\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 9\qquad \textbf{(E) } 11
</math>


== Solution ==
== Solution ==
If <math> 1+2+...+n </math> evenly [[divide]]s <math>6n</math>, then <math>\frac{6n}{1+2+...+n}</math> is an [[integer]].  
By a standard result, <math>1+2+\dotsb+n = \frac{n(n+1)}{2}</math>, so this will evenly divide <math>6n</math> precisely if <math>\frac{6n}{\left(\frac{n(n+1)}{2}\right)} = \frac{12}{n+1}</math> is an integer, or equivalently, <math>(n+1)</math> is a factor of <math>12</math>.  


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>.
The positive 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> (i.e. <math>6</math> factors in total), but as <math>n</math> must be a positive integer, we must have <math>n+1 \geq 1+1 = 2</math>. Therefore, the factor <math>1</math> cannot be used, but all the others can, giving <math>6-1 = \boxed{\textbf{(B) } 5}</math> possible values of <math>n</math> (namely, <math>n</math> can be <math>2-1 = 1</math>, <math>3-1 = 2</math>, <math>4-1 = 3</math>, <math>6-1 = 5</math>, or <math>12-1 = 11</math>).
 
<math>\frac{12}{n+1}</math> is an integer when <math>n+1</math> is a [[divisor |factor]] of <math>12</math>.
 
The factors of <math>12</math> are <math>1, 2, 3, 4, 6,</math> and <math>12</math>, so the possible values of <math>n</math> are <math>0, 1, 2, 3, 5,</math> and <math>11</math>.
 
But since <math>0</math> isn't a positive integer, only <math>1, 2, 3, 5,</math> and <math>11</math> are the possible values of <math>n</math>. Therefore the number of possible values of <math>n</math> is <math>\boxed{\textbf{(B) }5}</math>.


==Video Solution==
==Video Solution==
CHECK OUT Video Solution: https://youtu.be/WRv86DHa3zY
https://youtu.be/WRv86DHa3zY


==See also==
==See also==

Latest revision as of 01:09, 2 July 2025

Problem

For how many positive integers $n$ does $1+2+\dotsb+n$ evenly divide $6n$?

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

Solution

By a standard result, $1+2+\dotsb+n = \frac{n(n+1)}{2}$, so this will evenly divide $6n$ precisely if $\frac{6n}{\left(\frac{n(n+1)}{2}\right)} = \frac{12}{n+1}$ is an integer, or equivalently, $(n+1)$ is a factor of $12$.

The positive factors of $12$ are $1$, $2$, $3$, $4$, $6$, and $12$ (i.e. $6$ factors in total), but as $n$ must be a positive integer, we must have $n+1 \geq 1+1 = 2$. Therefore, the factor $1$ cannot be used, but all the others can, giving $6-1 = \boxed{\textbf{(B) } 5}$ possible values of $n$ (namely, $n$ can be $2-1 = 1$, $3-1 = 2$, $4-1 = 3$, $6-1 = 5$, or $12-1 = 11$).

Video Solution

https://youtu.be/WRv86DHa3zY

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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