Difference between revisions of "2012 AMC 10A Problems/Problem 22"
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− | Using the closed forms for the sums, we get <math>m^2=n(n+1)+212</math>, or <math>m^2=n^2+n+212</math>. 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 <math>4m^2=4n^2+4n+848</math>. Complete the square on the right hand side: <math>4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847</math>. Move over the <math>(2n+1)^2</math> and factor to get <math>(2m-2n-1)(2m+2n+1)=847=7 | + | Using the closed forms for the sums, we get <math>m^2=n(n+1)+212</math>, or <math>m^2=n^2+n+212</math>. 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 <math>4m^2=4n^2+4n+848</math>. Complete the square on the right hand side: <math>4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847</math>. Move over the <math>(2n+1)^2</math> and factor to get <math>(2m-2n-1)(2m+2n+1)=847=7\cdot11\cdot11</math>. The second factor is clearly greater than the first, and the only possible factor pairs are <math>1</math> and <math>847</math>, <math>7</math> and <math>121</math>, <math>11</math> and <math>77</math>. In each of these cases, solve for <math>m</math> and <math>n</math> and we find the solutions <math>(m,n)=(212,211), (32,28), (22,16)</math>. The sum of all possible values of <math>n</math> is <math>211+28+16=255</math>. |
== See Also == | == See Also == |
Revision as of 11:18, 14 January 2017
Problem
The sum of the first positive odd integers is 212 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: .
is clearly an integer, so
must be not only a perfect square, but also an odd perfect square. This is because the entire expression must be an integer, and for the numerator to be even (divisible by 2),
must be odd.
Let =
. (Note that this means that
.) This can be rewritten as
, which can then be rewritten to
. Factor the left side by using the difference of squares.
.
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 .
can equal
,
, or
. Add
. The answer is
.
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
.
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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