Difference between revisions of "2014 AMC 12B Problems/Problem 25"
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<cmath>2c - \frac{2014}{c} \equiv 0 \mod 2</cmath> | <cmath>2c - \frac{2014}{c} \equiv 0 \mod 2</cmath> | ||
− | In order for this to be satisfied, <math>\frac{2014}{c}</math> must be an even integer. Factoring <math>2014 = 2 \cdot 19 \cdot 53</math>, we see that | + | In order for this to be satisfied, <math>\frac{2014}{c}</math> must be an even integer. Factoring <math>2014 = 2 \cdot 19 \cdot 53</math>, we see that <math>c</math> must be a positive odd integer that divides <math>2014</math>. Our only positive valid <math>c</math> are <math>c = 1, 19, 53, 1007</math>. Our answer is just <math>\pi(1+19+53+1007) = 1080\pi \implies \textbf{(D)}</math>. |
Revision as of 16:26, 26 September 2022
Contents
[hide]Problem
Find the sum of all the positive solutions of
Solution 1
Rewrite as
. Now let
, and let
. We have:
Therefore, .
Notice that either and
or
and
. For the first case,
only when
and
is an integer.
when
is an even multiple of
, and since
,
only when
is an odd divisor of
. This gives us these possible values for
:
For the case where
,
, so
, where m is odd.
must also be an odd multiple of
in order for
to equal
, so
must be odd. We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for
, and therefore no cases where
and
. Therefore, the sum of all our possible values for
is
Solution 2
Very similar to the solution above, re-write the expression using :
Now, expand the LHS and cancel terms:
Now we use product-to-sum identities to get:
Notice that for any ,
. This is achieved when
, or equivalently
We can cleverly assume for some real
. Then, we must have
In order for this to be satisfied, must be an even integer. Factoring
, we see that
must be a positive odd integer that divides
. Our only positive valid
are
. Our answer is just
.
-FIREDRAGONMATH16
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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 12 Problems and Solutions |
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