Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A=(24,-1)$ , and let $M$ be the line perpendicular to line $L$ that contains the point $B=(5,6)$ .  The original coordinate axes are erased, and line $L$ is made the $x$ -axis and line $M$ the $y$ -axis.  In the new coordinate system, point $A$ is on the positive $x$ -axis, and point $B$ is on the positive $y$ -axis.  The point $P$ with coordinates $(-14,27)$ in the original system has coordinates $(\alpha,\beta)$ in the new coordinate system.  Find $\alpha+\beta$

Solution: Given that $L$ has slope $\frac{5}{12}$ and contains the point $A=(24,-1)$ , we may write the point-slope equation for $L$ as $y+1=\frac{5}{12}(x-24)$ .
Since $M$ is perpendicular to $L$ and  contains the point $B=(5,6)$ , we have that the slope of $M$ is $-\frac{12}{5}$ , and consequently that the point-slope equation for $M$ is $y-6=-\frac{12}{5}(x-5)$
Converting both equations to the form $0=Ax+By+C$ , we have that $L$ has the equation $0=5x-12y-132$ and that $M$ has the equation $0=12x+5y-90$ .
Applying the point-to-line distance formula, $\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$ , to point $P$ and lines $L$ and $M$ , we find that the distance from $P$ to $L$ and $M$ are $\frac{526}{13}$ and $\frac{123}{13}$ , respectively.
Since $A$ and $B$ lie on the positive axes of the shifted coordinate plane, we may show by graphing the given system that point P will lie in the second quadrant in the new coordinate system. Therefore, the $x$ -coordinate of $P$ is negative, and is therefore $-\frac{123}{13}$ ; similarly, the $y$ -coordinate of $P$ is positive and is therefore $\frac{526}{13}$
Thus, we have that $\alpha=-\frac{123}{13}$ and that $\beta=\frac{526}{13}$ . It follows that $\alpha+\beta=-\frac{123}{13}+\frac{526}{13}=\frac{403}{13}=\boxed{031}$

Answer:  031
Problem: Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A=(24,-1)$ , and let $M$ be the line perpendicular to line $L$ that contains the point $B=(5,6)$ .  The original coordinate axes are erased, and line $L$ is made the $x$ -axis and line $M$ the $y$ -axis.  In the new coordinate system, point $A$ is on the positive $x$ -axis, and point $B$ is on the positive $y$ -axis.  The point $P$ with coordinates $(-14,27)$ in the original system has coordinates $(\alpha,\beta)$ in the new coordinate system.  Find $\alpha+\beta$

Solution: The equations for the axes are $\frac{5}{12} (x-24) = y+1$ and $-\frac{12}{5}(x-5) = y - 6$ . We can solve the system to find that they intersect at the point $\left( \frac{1740}{169},\frac{-1134}{169} \right)$
The unit basis vectors of our new axes are $\begin{pmatrix} 12/13 \\ 5/13 \end{pmatrix}$ and $\begin{pmatrix} -5/13 \\ 12/13 \end{pmatrix}$ for the $x$ and $y$ axes respectively (taking into account which direction is positive).
Then, we solve the following system for $\alpha$ and $\beta$
\[\alpha \begin{pmatrix} 12/13 \\ 5/13 \end{pmatrix} + \beta \begin{pmatrix} -5/13 \\ 12/13 \end{pmatrix} + \begin{pmatrix} 1740/169 \\ -1134/169 \end{pmatrix} = \begin{pmatrix} -146 \\ 27 \end{pmatrix}\]
Painful bashing gives $\alpha = -\frac{123}{13}$ and $\beta = \frac{526}{13}$ . Adding gives $\alpha + \beta = \frac{403}{13} = \boxed{031}$

Answer:  031
Problem: In triangle $ABC$ $AB=125$ $AC=117$ and $BC=120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$

Solution: Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$ , respectively. [asy] defaultpen(fontsize(10)+0.8); size(200); pen p=fontsize(9)+linewidth(3); pair A,B,C,D,K,L,M,N,P,Q; A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L); draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5); dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p); label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10));  [/asy] Since ${BM}$ is the angle bisector of angle $B$ and ${CM}$ is perpendicular to ${BM}$ $\triangle BCP$ must be an isoceles triangle, so $BP=BC=120$ , and $M$ is the midpoint of ${CP}$ . For the same reason, $AQ=AC=117$ , and $N$ is the midpoint of ${CQ}$ .
Hence $MN=\tfrac 12 PQ$ . Since \[PQ=BP+AQ-AB=120+117-125=112,\] so $MN=\boxed{056}$

Answer:  056
Problem: In triangle $ABC$ $AB=125$ $AC=117$ and $BC=120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$

Solution: Let $I$ be the incenter of $ABC$ . Since $I$ lies on $BM$ and $AN$ $IM \perp MC$ and $IN \perp NC$ , so $\angle IMC + \angle INC = 180^\circ$ . This means that $CMIN$ is a cyclic quadrilateral. By the Law of Sines, $\frac{MN}{\sin \angle MIN} = \frac{2R}{\sin \angle CMI} = 2R = CI$ , where $R$ is the radius of the circumcircle of $CMIN$ . Since $\sin \angle MIN = \sin \angle BIA = \sin (90^\circ + \tfrac 12 \angle BCA) = \cos \tfrac 12 \angle BCA = \cos \angle BCI$ , we have that $MN = CI \cdot \sin \angle MIN = CI \cdot \cos \angle BCI$ . Letting $H$ be the point of contact of the incircle of $ABC$ with side $BC$ , we have $MN = CI \cdot \cos \angle BCI = CI \cdot \frac{CH}{CI} = CH$ . Thus, $MN = s - AB = \frac{117+120-125}{2}=\boxed{056}$

Answer:  056
Problem: In triangle $ABC$ $AB=125$ $AC=117$ and $BC=120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$

Solution: Project $I$ onto $AC$ and $BC$ as $D$ and $E$ $ID$ and $IE$ are both in-radii of $\triangle ABC$ so we get right triangles with legs $r$ (the in-radius length) and $s - c = 56$ . Since $IC$ is the hypotenuse for the 4 triangles ( $\triangle INC, \triangle IMC, \triangle IDC,$ and $\triangle IEC$ ), $C, D, M, I, N, E$ are con-cyclic on a circle we shall denote as $\omega$ which is also the circumcircle of $\triangle CMN$ and $\triangle CDE$ . To find $MN$ , we can use the Law of Cosines on $\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})$ where $O$ is the center of $\omega$ . Now, the circumradius $R$ can be found with Pythagorean Theorem with $\triangle CDI$ or $\triangle CEI$ $r^2 + 56^2 = (2R)^2$ . To find $r$ , we can use the formula $rs = [ABC]$ and by Heron's, $[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}$ . To find $\angle MCN$ , we can find $\angle MIN$ since $\angle MCN = 180 - \angle MIN$ $\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}$ . Thus, $\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}$ and since $\angle A + \angle B + \angle C = 180$ , we have $\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}$ . Plugging this into our Law of Cosines (LoC) formula gives $MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})$ . To find $\cos{\angle C}$ , we use LoC on $\triangle ABC \implies \cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}$ . Our formula now becomes $MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}$ . After simplifying, we get $MN^2 = 3136 \implies MN = \boxed{056}$

Answer:  056
Problem: In triangle $ABC$ $AB=125$ $AC=117$ and $BC=120$ . The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$ , and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$ . Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$ , respectively. Find $MN$

Solution: Because $\angle CMI = \angle CNI = 90$ $CMIN$ is cyclic.
Applying Ptolemy's theorem on CMIN:
$CN \cdot MI+CM \cdot IN=CI \cdot MN$
$CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI \cdot MN$
$MN = CI \sin \angle MCN$ by sine angle addition formula.
$\angle MCN = 180 - \angle MIN = 90 - \angle BCI$
Let $H$ be where the incircle touches $BC$ , then $CI \cos \angle BCI = CH = \frac{a+b-c}{2}$ $a=120, b=117, c=125$ , for a final answer of $\boxed{056}$

Answer:  056
Problem: The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3.  Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane.  Find the number of distinguishable acceptable arrangements.

Solution: First, we determine which possible combinations of digits $1$ through $9$ will yield sums that are multiples of $3$ . It is simplest to do this by looking at each of the digits $\bmod{3}$
We see that the numbers $1, 4,$ and $7$ are congruent to $1 \pmod{3}$ , that the numbers $2, 5,$ and $8$ are congruent to $2 \pmod{3}$ , and that the numbers $3, 6,$ and $9$ are congruent to $0 \pmod{3}$ . In order for a sum of three of these numbers to be a multiple of three, the mod $3$ sum must be congruent to $0$ . Quick inspection reveals that the only possible combinations are $0+0+0, 1+1+1, 2+2+2,$ and $0+1+2$ . However, every set of three consecutive vertices must sum to a multiple of three, so using any of $0+0+0, 1+1+1$ , or $2+2+2$ would cause an adjacent sum to include exactly $2$ digits with the same $\bmod{3}$ value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different $\bmod{3}$ values.
We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to $1 \pmod{3}$ can be located counterclockwise of a digit congruent to $0$ and clockwise of a digit congruent to $2 \pmod{3}$ , or the reverse can be true.
We set the first digit as $3$ avoid overcounting rotations, so we have one option as a choice for the first digit.  The other two $0 \pmod{3}$ numbers can be arranged in $2!=2$ ways.  The three $1 \pmod{3}$ and three $2 \pmod{3}$ can both be arranged in $3!=6$ ways.  Therefore, the desired result is $2(2 \times 6 \times 6)=\boxed{144}$

Answer:  144
Problem: The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3.  Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane.  Find the number of distinguishable acceptable arrangements.

Solution: Notice that there are three triplets of congruent integers $\mod 3$ $(1,4,7),$ $(2,5,8),$ and $(3,6,9).$ There are $3!$ ways to order each of the triplets individually and $3!$ ways to order the triplets as a group (see solution 1). Rotations are indistinguishable, so there are $(3!)^4/9=\boxed{144}$ total arrangements.

Answer:  144
Problem: Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$ , where $a > 0$ and $a + b + c$ is an integer.  The minimum possible value of $a$ can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers.  Find $p + q$

Solution: If the vertex is at $\left(\frac{1}{4}, -\frac{9}{8}\right)$ , the equation of the parabola can be expressed in the form \[y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}.\] Expanding, we find that \[y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8},\] and \[y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}.\] From the problem, we know that the parabola can be expressed in the form $y=ax^2+bx+c$ , where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$ $-\frac{a}{2}=b$ , and $\frac{a}{16}-\frac{9}{8}=c$ . Adding up all of these gives us \[\frac{9a-18}{16}=a+b+c.\] We know that $a+b+c$ is an integer, so $9a-18$ must be divisible by $16$ .  Let $9a=z$ . If ${z-18}\equiv {0} \pmod{16}$ , then ${z}\equiv {2} \pmod{16}$ . Therefore, if $9a=2$ $a=\frac{2}{9}$ . Adding up gives us $2+9=\boxed{011}$

Answer:  011
Problem: Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$ , where $a > 0$ and $a + b + c$ is an integer.  The minimum possible value of $a$ can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers.  Find $p + q$

Solution: Complete the square. Since $a>0$ , the parabola must be facing upwards. $a+b+c=\text{integer}$ means that $f(1)$ must be an integer. The function can be recasted into $a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}$ because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than $-\frac{9}{8}$ is $-1$ . So the $y$ -coordinate must change by $\frac{1}{8}$ and the $x$ -coordinate must change by $1-\frac{1}{4}=\frac{3}{4}$ . Thus, $a\left(\frac{3}{4}\right)^2=\frac{1}{8}\implies \frac{9a}{16}=\frac{1}{8}\implies a=\frac{2}{9}$ . So $2+9=\boxed{011}$

Answer:  011
Problem: Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$ , where $a > 0$ and $a + b + c$ is an integer.  The minimum possible value of $a$ can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers.  Find $p + q$

Solution: To do this, we can use the formula for the minimum (or maximum) value of the $x$ coordinate at a vertex of a parabola, $-\frac{b}{2a}$ and equate this to $\frac{1}{4}$ . Solving, we get $-\frac{a}{2}=b$ . Enter $x=\frac{1}{4}$ to get $-\frac{9}{8}=\frac{a}{16}+\frac{b}{4}+c=-\frac{a}{16}+c$ so $c=\frac{a-18}{16}$ . This means that $\frac{9a-18}{16}\in \mathbb{Z}$ so the minimum of $a>0$ is when the fraction equals -1, so $a=\frac{2}{9}$ . Therefore, $p+q=2+9=\boxed{011}$ .
-Gideontz

Answer:  011
Problem: 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: $m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}$ . 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 $2\cdot 1\frac{1}{4}$ '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.

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

Answer:  016
Problem: 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: 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 $2^4 = \boxed{016}$

Answer:  016
Problem: 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: 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 $2^4 = \boxed{016}$

Answer:  016
Problem: 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: 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.

Answer:  016
Problem: 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: 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.

Answer:  016