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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>  
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<cmath>\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.</cmath>
  
 
== Solution 1 ==
 
== Solution 1 ==
−
First, let's simplify that big ugly sigma notation:
+
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 write out the notation and simplify:
+
Now we simplify using product-sum logarithmic identites:
  
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<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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<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:
 
  
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<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:
+
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.  
  
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<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>
+
Writing out the first terms we have
  
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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:
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<math>\frac{1}{(2)(3)}(3)(4)\frac{1}{(4)(5)} \ldots</math>
  
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<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>
+
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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−
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:
 
  
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'''Case 1: n is odd'''
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<math>\textbf{Case 1: Odd n}</math>
  
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Okk so if n is odd, then the exponent of f(n) is -1 and we have
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For odd <math>n</math>, it telescopes to <math>\frac{1}{2(n+2)}</math> where <math>n</math> is clearly <math>3</math>.
  
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<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>
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<math>\textbf{Case 2: Even n}</math>
  
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Now we simply solve for n in both situations and see which one gives us an integer n:
+
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>
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<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>
 
−
 
 
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<math>\frac{1}{2(n+2)}=1/10 \iff n=3 \text{      Yay! One value of n down, 2 more to check!}</math>
 
−
 
 
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'''Case 2 : n is even'''
 
−
 
 
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Okk so if n is even, then the exponent of f(n) is 1 and we have:
 
−
 
 
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<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:
 
−
 
 
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<math>\frac{n+2}{2}=10 \iff n=18 \text{      Yay!}</math>
 
−
 
 
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<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>
 
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OKKK FINALLY BACK TO THE SOLUTION:
 
−
 
 
−
We've got n=18,3. So the sum is clearly <math>\boxed{021}</math>
 
  
 
==Solution 2==
 
==Solution 2==
  
 
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>
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<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>
 +
 +
== See also ==
 +
{{AIME box|year=2014|n=II|num-b=6|num-a=8}}
 +
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 01:10, 12 December 2023

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 split it into two cases to get rid of the absolute value sign

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

Now we simplify using product-sum logarithmic identites:

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


Note that the exponent $\cos{\pi(x)}$ is either $-1$ if $x$ is odd or $1$ if $x$ is even.

Writing out the first terms we have

$\frac{1}{(2)(3)}(3)(4)\frac{1}{(4)(5)} \ldots$

This product clearly telescopes (i.e. most terms cancel) and equals either $10$ or $\frac{1}{10}$. But the resulting term after telescoping depends on parity (odd/evenness), so we split it two cases, one where $n$ is odd and another where $n$ is even.

$\textbf{Case 1: Odd n}$

For odd $n$, it telescopes to $\frac{1}{2(n+2)}$ where $n$ is clearly $3$.

$\textbf{Case 2: Even n}$

For even $n$, it telescopes to $\frac{n+2}{2}$ where $18$ is the only possible $n$ value. Thus the answer is $\boxed{021}$

Solution 2

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 \[\log_{10}f(x)=\log_{10}(x+1)+\log_{10}(x+2)\ \ \ \text{if }x \text{ is even}\] \[\log_{10}f(x)=-\log_{10}(x+1)-\log_{10}(x+2)\ \ \ \text{if }x \text{ is odd}\] Because of this, $\sum_{k=1}^n\log_{10}f(k)$ is a telescoping series of logs, and we have \[\sum_{k=1}^n\log_{10}f(k)= \log_{10}(n+2)-\log_{10}2=\log_{10}\frac{n+2}{2}\ \ \ \text{if }n \text{ is even}\] \[\sum_{k=1}^n\log_{10}f(k)= -\log_{10}(n+2)-\log_{10}2=-\log_{10}2(n+2)\ \ \ \text{if }n \text{ is odd}\] 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 $3+18=\boxed{021}$

See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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