Difference between revisions of "2014 AIME II Problems/Problem 7"
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Let <math>f(x)=(x^2+3x+2)^{\cos(\pi x)}</math>. Find the sum of all positive integers <math>n</math> for which | Let <math>f(x)=(x^2+3x+2)^{\cos(\pi x)}</math>. Find the sum of all positive integers <math>n</math> for which | ||
− | <cmath>\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.</cmath> | + | <cmath>\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.</cmath> |
== Solution 1 == | == Solution 1 == | ||
Line 29: | Line 29: | ||
<math>\textbf{Case 2: Even n}</math> | <math>\textbf{Case 2: Even n}</math> | ||
− | For even <math>n</math>, it telescopes to <math>\frac{n+2}{2}</math> where <math>18</math> is the only possible <math>n</math> value. Thus the answer is <math>\boxed{ | + | For even <math>n</math>, it telescopes to <math>\frac{n+2}{2}</math> where <math>18</math> is the only possible <math>n</math> value. Thus the answer is <math>\boxed{021}</math> |
==Solution 2== | ==Solution 2== |
Latest revision as of 02:10, 12 December 2023
Contents
Problem
Let . Find the sum of all positive integers
for which
Solution 1
First, let's split it into two cases to get rid of the absolute value sign
Now we simplify using product-sum logarithmic identites:
Note that the exponent is either
if
is odd or
if
is even.
Writing out the first terms we have
This product clearly telescopes (i.e. most terms cancel) and equals either or
. But the resulting term after telescoping depends on parity (odd/evenness), so we split it two cases, one where
is odd and another where
is even.
For odd , it telescopes to
where
is clearly
.
For even , it telescopes to
where
is the only possible
value. Thus the answer is
Solution 2
Note that is
when
is odd and
when
is even. Also note that
for all
. Therefore
Because of this,
is a telescoping series of logs, and we have
Setting each of the above quantities to
and
and solving for
,
we get possible values of
and
so our desired answer is
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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