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> | ||
− | ==Solution== | + | ==Solution 1== |
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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Therefore the number of possible values of <math>n</math> is <math>5\Longrightarrow \mathrm{(B)}</math> | Therefore the number of possible values of <math>n</math> is <math>5\Longrightarrow \mathrm{(B)}</math> | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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: | ||
+ | |||
+ | <math>k \cdot \frac{n(n+1)}{2} = 6n</math> | ||
+ | |||
+ | <math>k(n+1) = 12</math> | ||
+ | |||
+ | 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> | ||
==See Also== | ==See Also== |
Revision as of 03:15, 31 July 2008
Contents
Problem
For how many positive integers does
evenly divide from
?
Solution 1
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.
is an integer when
is a factor of
.
The factors of are
,
,
,
,
, and
.
So the possible values of are
,
,
,
,
, and
.
But isn't a positive integer, so only
,
,
,
, and
are possible values of
.
Therefore the number of possible values of is
Solution 2
The sum of the first positive integers is
. If this is to divide
, then there exists a positive integer
such that:
Therefore, and
are divisors of
. There are
divisors of
, which are
. The divisors which multiply to
can be assigned to
and
in either order. However, when
is assigned to
, then
, which is not possible, because
must be positive. Therefore, we have
values of