AIME 2011 I · 第 7 题

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 $m \cdot (m^n)$, 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 $m^{\text{some number}} = m^n + (m-1){m^n} + (m-1)(m^{n+1})\dots$. This expression has $1 + (m-1) + (m-1) + \dots = 1 + k(m-1)$ terms, and the number of terms must be equal to $2011$ (as stated in the problem). Then $1+k(m-1) = 2011$, and we get the equation $k(m-1) = 2010 = 2\cdot3\cdot5\cdot67$. $2010$ has $16$ factors, and setting $m-1$ equal to each of these factors makes $\boxed{016}$ values of $m$.

Solution 7 (Quick Modular Arithmetic)

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 $m$.

Solution 8 (Polynomial Construction)

Let $f(a)$ be a polynomial in terms of $a$, listed below, which has roots at the values $m$ that satisfy the problem condition:

$$f(a) = a^{x_0} - \sum_{k=1}^{2011} a^{x_k}$$ It is evident to see that the polynomial has integer coefficients, which will be used later in the solution.

$f(1) = 1 - 2011 = -2010$. We use the fact that polynomials $P(x)$ with integer coefficients satisfy the condition that $(a-b)|f(a)-f(b)$ for some integers $a$ and $b$ to conclude that

$$(m-1)|f(m)-f(1)$$ Since $m$ is a solution of the polynomial, we have that $f(m) = 0$. Therefore, we have that

$$(m-1)|2010$$ Therefore, we arrive at the condition that $m-1$ must be a factor of $2010$.

Since $2010=2\cdot 3\cdot 5\cdot 67$, $2010$ has $16$ factors, so there are $\boxed{016}$ possible values of $m$.

-Vivdax

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