Difference between revisions of "2005 AMC 10A Problems/Problem 21"
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==Problem== | ==Problem== | ||
− | For how many positive integers <math>n</math> does <math> 1+2+...+n </math> evenly divide <math>6n</math>? | + | For how many positive integers <math>n</math> does <math> 1+2+...+n </math> evenly divide from <math>6n</math>? |
− | <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]]. | If <math> 1+2+...+n </math> evenly [[divide]]s <math>6n</math>, then <math>\frac{6n}{1+2+...+n}</math> is an [[integer]]. | ||
− | Since <math> 1+2+...+n = \frac{n(n+1)}{2} </math> we may substitute the [[RHS]] in the above [[fraction]]. | + | 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>. | ||
− | The factors of <math>12</math> are <math>1 | + | 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>. |
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− | + | 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>. | |
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− | + | ==Video Solution== | |
+ | CHECK OUT Video Solution: https://youtu.be/WRv86DHa3zY | ||
− | + | ==See also== | |
+ | {{AMC10 box|year=2005|ab=A|num-b=20|num-a=22}} | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:54, 14 December 2021
Contents
Problem
For how many positive integers does
evenly divide from
?
Solution
If evenly divides
, then
is an integer.
Since we may substitute the RHS in the above fraction. So the problem asks us for how many positive integers
is
an integer, or equivalently when
for a positive integer
.
is an integer when
is a factor of
.
The factors of are
and
, so the possible values of
are
and
.
But since isn't a positive integer, only
and
are the possible values of
. Therefore the number of possible values of
is
.
Video Solution
CHECK OUT 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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