Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
m (→‎Solution 2: 2011 should be 2010)
Line 8: Line 8:
 
==Solution 2==
 
==Solution 2==
 
Let <cmath>P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}</cmath>. 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 <cmath>P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010</cmath> Now, we can say that <cmath>P(m) = (m-1)Q(m) - 2010</cmath> 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, if <math>P(m) = 0</math>, then <math>m-1 | 2010</math> .
 
Let <cmath>P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}</cmath>. 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 <cmath>P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010</cmath> Now, we can say that <cmath>P(m) = (m-1)Q(m) - 2010</cmath> 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, if <math>P(m) = 0</math>, then <math>m-1 | 2010</math> .
−
Now, we need to show that <cmath>m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}\forall m-1|2011 </cmath> We try with the first few <math>m</math> that satisfy this.
+
Now, we need to show that <cmath>m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}\forall m-1|2010 </cmath> 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 <cmath>2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots</cmath> (based on the idea <math>2^n + 2^n = 2^{n+1}</math>, leading to a chain of substitutions of this kind) <cmath>= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}</cmath>. 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 <cmath>x_{2011} = x_{2010} = x_{2009} = 0</cmath>, <cmath>x_{2008} = x_{2007} = 1</cmath>, <cmath>x_{2006} = x_{2005} = 2</cmath>, <cmath>\cdots</cmath>, <cmath>x_2 = x_1 = 1004</cmath> and <cmath>x_0 = 1005</cmath>, 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 <cmath>2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots</cmath> (based on the idea <math>2^n + 2^n = 2^{n+1}</math>, leading to a chain of substitutions of this kind) <cmath>= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}</cmath>. 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 <cmath>x_{2011} = x_{2010} = x_{2009} = 0</cmath>, <cmath>x_{2008} = x_{2007} = 1</cmath>, <cmath>x_{2006} = x_{2005} = 2</cmath>, <cmath>\cdots</cmath>, <cmath>x_2 = x_1 = 1004</cmath> and <cmath>x_0 = 1005</cmath>, because  
 
<cmath>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}</cmath>
 
<cmath>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}</cmath>

Revision as of 19:32, 1 September 2024

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 1

$m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}$. Now, divide by $m^{x_0}$ to get $1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}$. Notice that since we can choose all nonnegative $x_0,...,x_{2011}$, we can make $x_n-x_0$ whatever we desire. WLOG, let $x_0\geq...\geq x_{2011}$ and let $a_n=x_n-x_0$. Notice that, also, $m^{a_{2011}}$ doesn't matter if we are able to make $m^{a_1} +m^{a_2} + .... + m^{a_{2010}}$ equal to $1-\left(\frac{1}{m}\right)^x$ for any power of $x$. Consider $m=2$. We can achieve a sum of $1-\left(\frac{1}{2}\right)^x$ by doing $\frac{1}{2}+\frac{1}{4}+...$ (the "simplest" sequence). If we don't have $\frac{1}{2}$, to compensate, we need $2\cdot \frac{1}{4}$'s. Now, let's try to generalize. The "simplest" sequence is having $\frac{1}{m}$ $m-1$ times, $\frac{1}{m^2}$ $m-1$ times, $\ldots$. To make other sequences, we can split $m-1$ $\frac{1}{m^i}$s into $m(m-1)$ $\text{ }\frac{1}{m^{i+1}}$s since $(m-1)\cdot\frac{1}{m^{i}} = m(m-1)\cdot\frac{1}{m^{i+1}}$. Since we want $2010$ terms, we have $\sum$ $(m-1)\cdot m^x=2010$. However, since we can set $x$ to be anything we want (including 0), all we care about is that $(m-1)\text{ }|\text{ } 2010$ which happens $\boxed{016}$ times.

Solution 2

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, if $P(m) = 0$, then $m-1 | 2010$ . Now, we need to show that \[m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}\forall m-1|2010\] 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 $\boxed{016}$

Solution 3

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 $m^{x_0} = \sum_{k=1}^{2011}m^{x_k}$.

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, \[m+j(m-1) = 2011 \Rightarrow m-1+j(m-1) = 2010 \Rightarrow (m-1)(j+1) = 2010.\] 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: \[1 + (m-1) + (m-1)m+(m-1)m^2 + \cdots + (m-1)m^j = 1+(m-1)\frac{m^{j+1}-1}{m-1} = m^{j+1}.\] 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 $2^4 = \boxed{016}$.

Solution 4 (for noobs like me)

The problem is basically asking how many integers $m$ have a power that can be expressed as the sum of 2011 other powers of $m$ (not necessarily distinct). Notice that $2+2+4+8+16=32$, $3+3+3+9+9+27+27+81+81=243$, and $4+4+4+4+16+16+16+64+64+64+256+256+256=1024$. Thus, we can safely assume that the equation $2011 = (m-1)x + m$ must have an integer solution $x$. To find the number of $m$-values that allow the aforementioned equation to have an integer solution, we can subtract 1 from the constant $m$ to make the equation equal a friendlier number, $2010$, instead of the ugly prime number $2011$: $2010 = (m-1)x+(m-1)$. Factor the equation and we get $2010 = (m-1)(x+1)$. The number of values of $m-1$ that allow $x+1$ to be an integer is quite obviously the number of factors of $2010$. Factoring $2010$, we obtain $2010 = 2 \times 3 \times 5 \times 67$, so the number of positive integers $m$ that satisfy the required condition is $2^4 = \boxed{016}$.

-fidgetboss_4000

Solution 5

First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of $m$ we can multiply to get to non-negative powers. Now we see that our problem is just a matter of m-chopping blocks. What is meant by $m$-chopping is taking an existing block of say $m^{k}$ and turning it into $m$ blocks of $m^{k-1}$. This process increases the total number of blocks by $m-1$ per chop.The problem wants us to find the number of positive integers $m$ where some number of chops will turn $1$ block into $2011$ such blocks, thus increasing the total amount by $2010= 2 \cdot 3 \cdot 5 \cdot 67$. Thus $m-1 | 2010$, and a cursory check on extreme cases will confirm that there are indeed $\boxed{016}$ possible $m$s.

Solution 6 (fast, easy)

The problem is basically saying that we want to find 2011 powers of m that add together to equal another power of m. If we had a power of m, m^n, then to get to m^(n+1) or m*(m^n), we have to add m^n, m-1 times. Then when we are at m^(n+1), to get to m^(n+2), it is similar. So we have to have m^(some number) = m^n + (m-1)(m^n) + (m-1)(m^(n+1))...<=This expression has 1 + (m-1) + (m-1) + ... = 1 + k(m-1) terms, and the number of terms must be equal to 2011 (the problem said). Then 1+k(m-1) = 2011, and we get the equation k*(m-1) = 2010 = 2*3*5*67. 2010 has 16 factors, and setting m-1 = each of these factors makes $\boxed{016}$ values of m.

Solution 7 (xooks-rbo)

Take$\pmod {m-1}$ of both sides \[m^{x_0} \equiv 1 \pmod {m-1}\] \[\sum_{k = 1}^{2011} m^{x_k} \equiv 2011 \pmod {m-1}\] Setting both of these to be congruent to eachother, we get: \[1 \equiv 2011 \pmod {m-1}\] Subtract $1$ from both sides to get: \[0 \equiv 2010 \pmod {m-1}\] $m-1$ must be a factor of $2010$.

$2010$ has $16$ factors, so there are $\boxed{16}$ values of $m$.

An Olympiad Problem that's (almost) the exact same (and it came before 2011, MAA)

https://artofproblemsolving.com/community/c6h84155p486903 2001 Austrian-Polish Math Individual Competition #1 ~MSC

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&oldid=226166"