Difference between revisions of "2013 AMC 10B Problems/Problem 24"
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This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>. We know <math>2016</math> works so it suffices to check whether <math>2012</math> works. | This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>. We know <math>2016</math> works so it suffices to check whether <math>2012</math> works. | ||
− | <math>2012=4*503</math> so we have that a factor of <math>2</math> must go to both <math>(a+1)</math> and <math>(b+1)</math>. So we have that <math>(a+1)</math> and <math>(b+1)</math> equal the numbers <math>(2+503)(2+1)</math>, but this contradicts our assumption for the case. | + | <math>2012=4*503</math> so we have that a factor of <math>2</math> must go to both <math>(a+1)</math> and <math>(b+1)</math>. So we have that <math>(a+1)</math> and <math>(b+1)</math> equal the numbers <math>(2+503)(2+1)</math>, but this contradicts our assumption for the case. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math> as <math>2016</math> is the only solution. |
− | |||
− | Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math> as <math>2016</math> is the only solution |
Revision as of 14:40, 22 February 2013
Problem
A positive integer is nice if there is a positive integer
with exactly four positive divisors (including
and
) such that the sum of the four divisors is equal to
. How many numbers in the set
are nice?
Solution
A positive integer with only four positive divisors has its prime factorization in the form of , where
and
are both prime positive integers or c^3 where c is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of
. The four factors of this number would be
,
,
, and
. The sum of these would be
, which can be factored into the form
. Easily we can see that Now we can take cases again.
Case 1: Either or
is 2.
If this is true then we have to have that one of or
is even and that one is 3. So we have that in this case the only numbers that work are odd multiples of 3 which are 2010 and 2016. So we just have to check if either
or
is a prime. We see that
is the only one that works in this case.
Case 2: Both and
are odd primes.
This implies that both and
are even which implies that in this case the number must be divisible by
. This leaves only
and
. We know
works so it suffices to check whether
works.
so we have that a factor of
must go to both
and
. So we have that
and
equal the numbers
, but this contradicts our assumption for the case. Thus the answer is
as
is the only solution.