2014 AIME II Problems/Problem 7

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Problem

Let $f(x)=(x^2+3x+2)^{\cos(\pi x)}$. Find the sum of all positive integers $n$ for which \[\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.\]

Solution 1

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

$\left |\sum_{k=1}^n\log_{10}f(k)\right|=1 \iff \sum_{k=1}^n\log_{10}f(k)=1,-1$

Now we write out the notation and simplify:

$\log_{10}{f(1)}+\log_{10}{f(2)}+...+\log{10}{f(n)}=\log_{10}{f(1)\cdot f(2) \cdot ... \cdot f(n)}=1,-1$

Converting to exponential form we have the much nicer equation:

$f(1) \cdot f(2) \cdot... \cdot f(n)=10,\frac{1}{10}$

OKAY. Now let's look at the function f. Well we have the base which factors nicely into $(x+2)(x+1)$. 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:

$f(1) \cdot f(2) \cdot... \cdot f(n)=\left (\frac{1}{(1+1)(1+2)}\right ) ((2+1)(2+2)) \left (\frac{1}{(3+1)(3+2)}\right )...$

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:

$\left (\frac{1}{(x+1)(x+2)}\right )(((x+1)+1)((x+1)+2)) \left (\frac{1}{((x+2)+1)((x+2)+2)} \right )...$

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

$\left (\frac{1}{(1+1)\cancel{(1+2)}}\right )(\cancel{(2+1)}\cancel{(2+2)})...\left (\frac{1}{\cancel{(n+1)}(n+2)}\right )=\frac{1}{2(n+2)}=10,1/10$

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

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

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

Case 2 : n is even

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

$\left (\frac{1}{(1+1)\cancel{(1+2)}} \right )(\cancel{(2+1)}\cancel{(2+2)})...(\cancel{(n+1)}(n+2))=\frac{n+2}{2}=10,1/10$

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

$\frac{n+2}{2}=10 \iff n=18 \text{      Yay!}$

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

OKKK FINALLY BACK TO THE SOLUTION:

We've got n=18,3. So the sum is clearly $\boxed{021}$

Solution2

Note that $\cos(\pi x)$ is $-1$ when $x$ is odd and $1$ when $x$ is even. Also note that $x^2+3x+2=(x+1)(x+2)$ for all $x$. Therefore \[\log_{10}f(x)=\log_{10}(x+1)+\log_{10}(x+2)\ \ \ \text{if x is even}\] \[\log_{10}f(x)=-\log_{10}(x+1)-\log_{10}(x+2)\ \ \ \text{if x is odd}\] Because of this, $\sum_{k=1}^n\log_{10}f(k)$ is a telescoping series of logs, and we have \[\sum_{k=1}^n\log_{10}f(k)= \log_{10}(n+2)-\log_{10}2=\log_{10}\frac{n+2}{2}\ \ \ \text{if n is even}\] \[\sum_{k=1}^n\log_{10}f(k)= -\log_{10}(n+2)-\log_{10}2=-\log_{10}2(n+2)\ \ \ \text{if n is odd}\] Setting each of the above quantities to $1$ and $-1$ and solving for $n$, we get possible values of $n=3$ and $n=18$ so our desired answer is $3+18=\boxed{021}$

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