# Difference between revisions of "2014 AIME II Problems/Problem 7"

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<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== | + | == Solution 1 == |

+ | First, let's simplify that big ugly sigma notation: | ||

+ | |||

+ | <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 write out the notation and simplify: | ||

+ | |||

+ | <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> | ||

+ | |||

+ | Converting to exponential form we have the much nicer equation: | ||

+ | |||

+ | <math>f(1) \cdot f(2) \cdot... \cdot f(n)=10,\frac{1}{10}</math> | ||

+ | |||

+ | OKAY. Now let's look at the function f. Well we have the base which factors nicely into <math>(x+2)(x+1)</math>. And then there's the exponent. Hmm well there's a pi inside. That must count for something. Well, if x is odd, then the exponent will be -1 because the cosine of an odd multiple of pi is always -1. However, if it's an even multiple of pi, the cosine is 1. Remember raising to an exponent of -1 just gives the reciprocal. So we have fractions and then anti-fractions and we're multiplying them? Let's plug in the values without simplifying: | ||

+ | |||

+ | <math>f(1) \cdot f(2) \cdot... \cdot f(n)=\left (\frac{1}{(1+1)(1+2)}\right ) ((2+1)(2+2)) \left (\frac{1}{(3+1)(3+2)}\right )...</math> | ||

+ | |||

+ | Aha! MASS CANCELATION...however, notice we can't really end because we don't know if the value of n is going to be odd or even. We can prove this mass cancelation happens by simply looking at consecutive functions of f: | ||

+ | |||

+ | <math>\left (\frac{1}{(x+1)(x+2)}\right )(((x+1)+1)((x+1)+2)) \left (\frac{1}{((x+2)+1)((x+2)+2)} \right )...</math> | ||

+ | |||

+ | Therefore this does indeed cancel and was not a clever trap set by AIME committee. However, we still don't know where to end. So we branch off into 2 cases here: | ||

+ | |||

+ | '''Case 1: n is odd''' | ||

+ | |||

+ | Okk so if n is odd, then the exponent of f(n) is -1 and we have | ||

+ | |||

+ | <math>\left (\frac{1}{(1+1)\cancel{(1+2)}}\right )(\cancel{(2+1)}\cancel{(2+2)})...\left (\frac{1}{\cancel{(n+1)}(n+2)}\right )=\frac{1}{2(n+2)}=10,1/10</math> | ||

+ | |||

+ | Now we simply solve for n in both situations and see which one gives us an integer n: | ||

+ | |||

+ | <math>\frac{1}{2(n+2)}=10 \iff n=-1.95 \text{ Err...not only is it not an integer, it's negative too.}</math> | ||

+ | |||

+ | <math>\frac{1}{2(n+2)}=1/10 \iff n=3 \text{ Yay! One value of n down, 2 more to check!}</math> | ||

+ | |||

+ | '''Case 2 : n is even''' | ||

+ | |||

+ | Okk so if n is even, then the exponent of f(n) is 1 and we have: | ||

+ | |||

+ | <math>\left (\frac{1}{(1+1)\cancel{(1+2)}} \right )(\cancel{(2+1)}\cancel{(2+2)})...(\cancel{(n+1)}(n+2))=\frac{n+2}{2}=10,1/10</math> | ||

+ | |||

+ | Now we simply solve for n in both situations and see which one gives us an integer n: | ||

+ | |||

+ | <math>\frac{n+2}{2}=10 \iff n=18 \text{ Yay!}</math> | ||

+ | |||

+ | <math>\frac{n+2}{2}=1/10 \iff n=-1.8 \text{ Err...not only is it not an integer, it's negative too.}</math> | ||

+ | |||

+ | OKKK FINALLY BACK TO THE SOLUTION: | ||

+ | |||

+ | We've got n=18,3. So the sum is clearly <math>\boxed{021}</math> | ||

+ | |||

+ | ==Solution2== | ||

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 | ||

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<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 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 | + | 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> |

## Revision as of 16:01, 27 March 2014

## Problem

Let . Find the sum of all positive integers for which

## Solution 1

First, let's simplify that big ugly sigma notation:

Now we write out the notation and simplify:

Converting to exponential form we have the much nicer equation:

OKAY. Now let's look at the function f. Well we have the base which factors nicely into . And then there's the exponent. Hmm well there's a pi inside. That must count for something. Well, if x is odd, then the exponent will be -1 because the cosine of an odd multiple of pi is always -1. However, if it's an even multiple of pi, the cosine is 1. Remember raising to an exponent of -1 just gives the reciprocal. So we have fractions and then anti-fractions and we're multiplying them? Let's plug in the values without simplifying:

Aha! MASS CANCELATION...however, notice we can't really end because we don't know if the value of n is going to be odd or even. We can prove this mass cancelation happens by simply looking at consecutive functions of f:

Therefore this does indeed cancel and was not a clever trap set by AIME committee. However, we still don't know where to end. So we branch off into 2 cases here:

**Case 1: n is odd**

Okk so if n is odd, then the exponent of f(n) is -1 and we have

Now we simply solve for n in both situations and see which one gives us an integer n:

**Case 2 : n is even**

Okk so if n is even, then the exponent of f(n) is 1 and we have:

Now we simply solve for n in both situations and see which one gives us an integer n:

OKKK FINALLY BACK TO THE SOLUTION:

We've got n=18,3. So the sum is clearly

## Solution2

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