Difference between revisions of "2014 AIME II Problems/Problem 7"
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== Solution 1 == | == Solution 1 == | ||
− | First, let's | + | First, let's split it into two cases to get rid of the absolute value sign |
<math>\left |\sum_{k=1}^n\log_{10}f(k)\right|=1 \iff \sum_{k=1}^n\log_{10}f(k)=1,-1 </math> | <math>\left |\sum_{k=1}^n\log_{10}f(k)\right|=1 \iff \sum_{k=1}^n\log_{10}f(k)=1,-1 </math> | ||
− | Now we | + | Now we simplify using product-sum logarithmic identites: |
− | <math>\log_{10}{f(1)}+\log_{10}{f(2)}+...+\ | + | <math>\log_{10}{f(1)}+\log_{10}{f(2)}+...+\log_{10}{f(n)}=\log_{10}{f(1)\cdot f(2) \cdot ... \cdot f(n)}=1,-1</math> |
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− | + | Note that the exponent <math>\cos{\pi(x)}</math> is either <math>-1</math> if <math>x</math> is odd or <math>1</math> if <math>x</math> is even. | |
− | + | Writing out the first terms we have | |
− | + | <math>\frac{1}{(2)(3)}(3)(4)\frac{1}{(4)(5)} \ldots</math> | |
− | <math> | + | This product clearly telescopes (i.e. most terms cancel) and equals either <math>10</math> or <math>\frac{1}{10}</math>. But the resulting term after telescoping depends on parity (odd/evenness), so we split it two cases, one where <math>n</math> is odd and another where <math>n</math> is even. |
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− | + | <math>\textbf{Case 1: Odd n}</math> | |
− | + | For odd <math>n</math>, it telescopes to <math>\frac{1}{2(n+2)}</math> where <math>n</math> is clearly <math>3</math>. | |
− | <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{021}</math> | |
− | + | ==Solution 2== | |
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Note that <math>\cos(\pi x)</math> is <math>-1</math> when <math>x</math> is odd and <math>1</math> when <math>x</math> is even. Also note that <math>x^2+3x+2=(x+1)(x+2)</math> for all <math>x</math>. Therefore | Note that <math>\cos(\pi x)</math> is <math>-1</math> when <math>x</math> is odd and <math>1</math> when <math>x</math> is even. Also note that <math>x^2+3x+2=(x+1)(x+2)</math> for all <math>x</math>. Therefore | ||
− | <cmath>\log_{10}f(x)=\log_{10}(x+1)+\log_{10}(x+2)\ \ \ \text{if x is even}</cmath> | + | <cmath>\log_{10}f(x)=\log_{10}(x+1)+\log_{10}(x+2)\ \ \ \text{if }x \text{ is even}</cmath> |
− | <cmath>\log_{10}f(x)=-\log_{10}(x+1)-\log_{10}(x+2)\ \ \ \text{if x is odd}</cmath> | + | <cmath>\log_{10}f(x)=-\log_{10}(x+1)-\log_{10}(x+2)\ \ \ \text{if }x \text{ is odd}</cmath> |
Because of this, <math>\sum_{k=1}^n\log_{10}f(k)</math> is a telescoping series of logs, and we have | Because of this, <math>\sum_{k=1}^n\log_{10}f(k)</math> is a telescoping series of logs, and we have | ||
− | <cmath>\sum_{k=1}^n\log_{10}f(k)= \log_{10}(n+2)-\log_{10}2=\log_{10}\frac{n+2}{2}\ \ \ \text{if n is even}</cmath> | + | <cmath>\sum_{k=1}^n\log_{10}f(k)= \log_{10}(n+2)-\log_{10}2=\log_{10}\frac{n+2}{2}\ \ \ \text{if }n \text{ is even}</cmath> |
− | <cmath>\sum_{k=1}^n\log_{10}f(k)= -\log_{10}(n+2)-\log_{10}2=-\log_{10}2(n+2)\ \ \ \text{if n is odd}</cmath> | + | <cmath>\sum_{k=1}^n\log_{10}f(k)= -\log_{10}(n+2)-\log_{10}2=-\log_{10}2(n+2)\ \ \ \text{if }n \text{ is odd}</cmath> |
Setting each of the above quantities to <math>1</math> and <math>-1</math> and solving for <math>n</math>, | Setting each of the above quantities to <math>1</math> and <math>-1</math> and solving for <math>n</math>, | ||
we get possible values of <math>n=3</math> and <math>n=18</math> so our desired answer is <math>3+18=\boxed{021}</math> | we get possible values of <math>n=3</math> and <math>n=18</math> so our desired answer is <math>3+18=\boxed{021}</math> | ||
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+ | == See also == | ||
+ | {{AIME box|year=2014|n=II|num-b=6|num-a=8}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:13, 2 December 2017
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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