2012 AMC 10A Problems/Problem 22
Contents
[hide]Problem
The sum of the first positive odd integers is
more than the sum of the first
positive even integers. What is the sum of all possible values of
?
Solution 1
The sum of the first odd integers is given by
. The sum of the first
even integers is given by
.
Thus, . Since we want to solve for n, rearrange as a quadratic equation:
.
Use the quadratic formula: . Since
is clearly an integer,
must be not only a perfect square, but also an odd perfect square for
to be an integer.
Let ; note that this means
. It can be rewritten as
, so
. Factoring the left side by using the difference of squares, we get
.
Our goal is to find possible values for , then use the equation above to find
. The difference between the factors is
We have three pairs of factors,
and
. The differences between these factors are
,
, and
- those are all possible values for
. Thus the possibilities for
are
,
, and
.
Now plug in these values into the equation , so
can equal
,
, or
, hence the answer is
.
~Edits by BakedPotato66
Solution 2
As above, start off by noting that the sum of the first odd integers
and the sum of the first
even integers
. Clearly
, so let
, where
is some positive integer. We have:
.
Expanding, grouping like terms and factoring, we get:
.
We know that and
are both positive integers, so we need only check values of
from
to
(
). Plugging in, the only values of
that give integral solutions are
and
. These gives
values of
and
, respectively.
. Hence, the answer is
.
Solution 3
Using the closed forms for the sums, we get , or
. We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to let us complete the square. Our equation is now
. Complete the square on the right hand side:
. Move over the
and factor to get
. The second factor is clearly greater than the first, and the only possible factor pairs are
and
,
and
,
and
. In each of these cases, solve for
and
and we find the solutions
. The sum of all possible values of
is
.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc10a/252
~dolphin7
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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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