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Difference between revisions of "2011 AIME I Problems/Problem 7"

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==Solution==
 
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Let <math>P(m) = m^x_0 - m^x_1 -m^x_2 - .... - m^x_2011</math>. The problem then becomes finding the number of positive integer roots <math>m</math> for which <math>P(m) = 0</math> and <math>x_0, x_1, ..., x_2011</math> are nonnegative integers. We plug in <math>m = 1</math> and see that <math>P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010</math>. Now, we can say that <math>P(m) = (m-1)Q(m) - 2010</math> for some polynomial <math>Q(m)</math> with integer coefficients. Then if <math>P(m) = 0</math>, <math>(m-1)Q(m) = 2010</math>. Thus, it is only possible for <math>m-1 | 2010</math> if <math>P(m) = 0</math>.
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Let <math>P(m) = m^x_0 - m^x_1 -m^x_2 - .... - m^x_{2011}</math>. The problem then becomes finding the number of positive integer roots <math>m</math> for which <math>P(m) = 0</math> and <math>x_0, x_1, ..., x_2011</math> are nonnegative integers. We plug in <math>m = 1</math> and see that <math>P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010</math>. Now, we can say that <math>P(m) = (m-1)Q(m) - 2010</math> for some polynomial <math>Q(m)</math> with integer coefficients. Then if <math>P(m) = 0</math>, <math>(m-1)Q(m) = 2010</math>. Thus, it is only possible for <math>m-1 | 2010</math> if <math>P(m) = 0</math>.
 
Now, we need to show that for all <math>m-1 | 2010</math>, <math><cmath> m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}. </cmath></math>. We try with the first few <math>m</math> that satisfy this.
 
Now, we need to show that for all <math>m-1 | 2010</math>, <math><cmath> m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}. </cmath></math>. We try with the first few <math>m</math> that satisfy this.
 
For <math>m = 2</math>, we see we can satisfy this if <math>x_0 = 2010</math>, <math>x_1 = 2009</math>, <math>x_2 = 2008</math>, <math>\cdots</math> , <math>x_2008 = 2</math>, <math>x_2009 = 1</math>, <math> x_2010 = 0</math>, <math>x_2011 = 0</math>, because <math>2^2009 + 2^2008 + \cdots + 2^1 + 2^0 +2^ 0 = 2^2009 + 2^2008 + \cdots + 2^1 + 2^1 = \cdots</math> (based on the idea <math>2^n + 2^n = 2^{n+1}</math>, leading to a chain of substitutions of this kind) <math>= 2^2009 + 2^2008 + 2^2008 = 2^2009 + 2^2009 = '''2^2010'''</math>. Thus <math>2</math> is a possible value of <math>m</math>. For other values, for example <math>m = 3</math>, we can use the same strategy, with <math>x_2011 = x_2010 = x_2009 = 0</math>, <math>x_2008 = x_2007 = 1</math>, <math>x_2006 = x_2005 = 2</math>, <math>\cdots</math>, <math>x_2 = x_1 = 1004</math> and <math>x_0 = 1005</math>, because  
 
For <math>m = 2</math>, we see we can satisfy this if <math>x_0 = 2010</math>, <math>x_1 = 2009</math>, <math>x_2 = 2008</math>, <math>\cdots</math> , <math>x_2008 = 2</math>, <math>x_2009 = 1</math>, <math> x_2010 = 0</math>, <math>x_2011 = 0</math>, because <math>2^2009 + 2^2008 + \cdots + 2^1 + 2^0 +2^ 0 = 2^2009 + 2^2008 + \cdots + 2^1 + 2^1 = \cdots</math> (based on the idea <math>2^n + 2^n = 2^{n+1}</math>, leading to a chain of substitutions of this kind) <math>= 2^2009 + 2^2008 + 2^2008 = 2^2009 + 2^2009 = '''2^2010'''</math>. Thus <math>2</math> is a possible value of <math>m</math>. For other values, for example <math>m = 3</math>, we can use the same strategy, with <math>x_2011 = x_2010 = x_2009 = 0</math>, <math>x_2008 = x_2007 = 1</math>, <math>x_2006 = x_2005 = 2</math>, <math>\cdots</math>, <math>x_2 = x_1 = 1004</math> and <math>x_0 = 1005</math>, because  

Revision as of 13:24, 27 December 2011

Problem 7

Find the number of positive integers $m$ for which there exist nonnegative integers $x_0$, $x_1$ , $\dots$ , $x_{2011}$ such that \[m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.\]

Solution

Let $P(m) = m^x_0 - m^x_1 -m^x_2 - .... - m^x_{2011}$. The problem then becomes finding the number of positive integer roots $m$ for which $P(m) = 0$ and $x_0, x_1, ..., x_2011$ are nonnegative integers. We plug in $m = 1$ and see that $P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010$. Now, we can say that $P(m) = (m-1)Q(m) - 2010$ for some polynomial $Q(m)$ with integer coefficients. Then if $P(m) = 0$, $(m-1)Q(m) = 2010$. Thus, it is only possible for $m-1 | 2010$ if $P(m) = 0$. Now, we need to show that for all $m-1 | 2010$, $<cmath> m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}. </cmath>$. We try with the first few $m$ that satisfy this. For $m = 2$, we see we can satisfy this if $x_0 = 2010$, $x_1 = 2009$, $x_2 = 2008$, $\cdots$ , $x_2008 = 2$, $x_2009 = 1$, $x_2010 = 0$, $x_2011 = 0$, because $2^2009 + 2^2008 + \cdots + 2^1 + 2^0 +2^ 0 = 2^2009 + 2^2008 + \cdots + 2^1 + 2^1 = \cdots$ (based on the idea $2^n + 2^n = 2^{n+1}$, leading to a chain of substitutions of this kind) $= 2^2009 + 2^2008 + 2^2008 = 2^2009 + 2^2009 = '''2^2010'''$. Thus $2$ is a possible value of $m$. For other values, for example $m = 3$, we can use the same strategy, with $x_2011 = x_2010 = x_2009 = 0$, $x_2008 = x_2007 = 1$, $x_2006 = x_2005 = 2$, $\cdots$, $x_2 = x_1 = 1004$ and $x_0 = 1005$, because $3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\cdots+3^1004 +3^1004 = 3^1+3^1+3^1+3^2+3^2+\cdots+3^1004 +3^1004 = 3^2+3^2+3^2+\cdots+3^1004 +3^1004 = \cdots$ $=3^1004 +3^1004+3^1004 = '''3^1005'''$. It's clearly seen we can use the same strategy for all $m-1 |2010$. We count all positive $m$ satisfying $m-1 |2010$, and see there are $\boxed{16}$ One notices that $m-1 \mid 2010$ if and only if there exist non-negative integers $x_0,x_1,\ldots,x_{2011}$ such that $m^{x_0} = \sum_{k=1}^{2011}m^{x^k}$.

Solution 2

To prove the forward case, we proceed by directly finding $x_0,x_1,\ldots,x_{2011}$. Suppose $m$ is an integer such that $m^{x_0} = \sum_{k=1}^{2011}m^{x^k}$. We will count how many $x_k = 0$, how many $x_k = 1$, etc. Suppose the number of $x_k = 0$ is non-zero. Then, there must be at least $m$ such $x_k$ since $m$ divides all the remaining terms, so $m$ must also divide the sum of all the $m^0$ terms. Thus, if we let $x_k = 0$ for $k = 1,2,\ldots,m$, we have, \[m^{x_0} = m + \sum_{k=m+1}^{2011}m^{x_k}.\] Well clearly, $m^{x_0}$ is greater than $m$, so $m^2 \mid m^{x_0}$. $m^2$ will also divide every term, $m^{x_k}$, where $x_k \geq 2$. So, all the terms, $m^{x_k}$, where $x_k < 2$ must sum to a multiple of $m^2$. If there are exactly $m$ terms where $x_k = 0$, then we must have at least $m-1$ terms where $x_k = 1$. Suppose there are exactly $m-1$ such terms and $x_k = 1$ for $k = m+1,m+2,2m-1$. Now, we have, \[m^{x_0} = m^2 + \sum_{k=2m}^{2011}m^{x_k}.\] One can repeat this process for successive powers of $m$ until the number of terms reaches 2011. Since there are $m + j(m-1)$ terms after the $j$th power, we will only hit exactly 2011 terms if $m-1$ is a factor of 2010. To see this, 

\[m+j(m-1) = 2011 \Rightarrow m-1+j(m-1) &= 2010 \Rightarrow (m-1)(j+1) = 2010.\] (Error compiling LaTeX. Unknown error_msg)

Thus, when $j = 2010/(m-1) - 1$ (which is an integer since $m-1 \mid 2010$ by assumption, there are exactly 2011 terms. To see that these terms sum to a power of $m$, we realize that the sum is a geometric series: 

\[1 + (m-1) + (m-1)m+(m-1)m^2 + \cdots + (m-1)m^j &= 1+(m-1)\frac{m^{j+1}-1}{m-1} = m^{j+1}.\] (Error compiling LaTeX. Unknown error_msg)

Thus, we have found a solution for the case $m-1 \mid 2010$.

Now, for the reverse case, we use the formula \[x^k-1 = (x-1)(x^{k-1}+x^{k-2}+\cdots+1).\] Suppose $m^{x_0} = \sum_{k=1}^{2011}m^{x^k}$ has a solution. Subtract 2011 from both sides to get \[m^{x_0}-1-2010 = \sum_{k=1}^{2011}(m^{x^k}-1).\] Now apply the formula to get \[(m-1)a_0-2010 = \sum_{k=1}^{2011}[(m-1)a_k],\] where $a_k$ are some integers. Rearranging this equation, we find \[(m-1)A = 2010,\] where $A = a_0 - \sum_{k=1}^{2011}a_k$. Thus, if $m$ is a solution, then $m-1 \mid 2010$.

So, there is one positive integer solution corresponding to each factor of 2010. Since $2010 = 2\cdot 3\cdot 5\cdot 67$, the number of solutions is $2^4 = \boxed{16}$.

See also

2011 AIME I (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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