Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence.  Find the difference between the largest and smallest geometric numbers.

Solution: The smallest geometric number is $124$ because $123$ and any number containing a zero does not work. $964$ is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives $\boxed{840}.$

Answer:  840
Problem: There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$

Solution: Let $z = a + 164i$
Then \[\frac {a + 164i}{a + 164i + n} = 4i\] and \[a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.\]
By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,
we conclude that \[a = -656.\]
By equating the imaginary terms on each side of the equation,
we conclude that \[164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).\]
We now have an equation for $n$ \[4i \left (-656 + n \right ) = 164i,\]
and this equation shows that $n = \boxed{697}.$

Answer:  697
Problem: There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$

Solution: \[\frac {z}{z+n}=4i\]
\[1-\frac {n}{z+n}=4i\]
\[1-4i=\frac {n}{z+n}\]
\[\frac {1}{1-4i}=\frac {z+n}{n}\]
\[\frac {1+4i}{17}=\frac {z}{n}+1\]
Since their imaginary part has to be equal,
\[\frac {4i}{17}=\frac {164i}{n}\]
\[n=\frac {(164)(17)}{4}=697\]
\[n = \boxed{697}.\]

Answer:  697
Problem: There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$

Solution: Below is an image of the complex plane. Let $\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$ [asy] unitsize(1cm);   xaxis("Re",Arrows); yaxis("Im",Arrows);  real f(real x) {return 164;} pair F(real x) {return (x,f(x));}  draw(graph(f,-700,100),red,Arrows);  label("Im$(z)=164$",F(-330),N);   pair z = (-656,164); dot(Label("$z$",align=N),z); dot(Label("$z+n$",align=N),z+(697,0));  draw(Label("$4x$"),z--(0,0)); draw(Label("$x$"),(0,0)--z+(697,0));  markscalefactor=2; draw(rightanglemark(z,(0,0),z+(697,0))); [/asy] $z$ must lie on the line $\operatorname{Im}(z)=164$ $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$
Consider $z$ and $z+n$ in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have $z_1z_2 = r_1\angle\theta_1 \cdot r_2\angle\theta_2 = r_1r_2\angle(\theta_1+\theta_2)$ and $\frac{z_1}{z_2} = \frac{r_1\angle\theta_1}{r_2\angle\theta_2} = \frac{r_1}{r_2}\angle(\theta_1-\theta_2)$ , where $r$ is the magnitude and $\theta$ is the phase, and $z_n=r_n\angle\theta_n$
Since $4i$ has magnitude $4$ and phase $90^\circ$ (since the positive imaginary axis points in a direction $90^\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$ . We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$ . Additionally, $z$ , the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$
This means that $z$ , the origin, and $z+n$ form a right triangle. The hypotenuse is the length from $z$ to $z+n$ and has length $n$ , since $n$ is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as $\frac{x \cdot 4x}{2}$ , or using the hypotenuse and its corresponding altitude, as $\frac{164n}{2}$ , so $\frac{x \cdot 4x}{2} = \frac{164n}{2} \implies x^2 = 41n$ . By Pythagorean Theorem, $x^2+(4x)^2 = n^2 \implies 17x^2 = n^2$ . Substituting out $x^2$ using the earlier equation, we get $17\cdot41n = n^2 \implies n = \boxed{697}$ . ~ emerald_block

Answer:  697
Problem: There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$

Solution: Taking the reciprocal of our equation gives us $1 + \frac{n}{z} = \frac{1}{4i}.$ Therefore, \[\frac{n}{z} = \frac{1-4i}{4i} = \frac{17}{-16+4i}.\] Since $z$ has an imaginary part of $164$ , we must multiply both sides of our RHS fraction by $\frac{164}{4} = 41$ so that its denominator's imaginary part matches the LHS fraction's denominator's imaginary part. That looks like this: \[\frac{n}{z} = \frac{697}{-656 + 164i}.\] Therefore, we can conclude the the real part of $z$ is $-656$ and $n = \boxed{697}.$ (it wasn't necessary to find the real part)

Answer:  697
Problem: A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: The probability of three heads and five tails is $\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\binom {8}{3}p^5(1-p)^3$
\begin{align*} 25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\ 25(1-p)^2&=p^2 \\ 25p^2-50p+25&=p^2 \\ 24p^2-50p+25&=0 \\ p&=\frac {5}{6}\end{align*}
Therefore, the answer is $5+6=\boxed{011}$

Answer:  011
Problem: A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: We start as shown above. However, when we get to $25(1-p)^2=p^2$ , we square root both sides to get $5(1-p)=p$ . We can do this because we know that both $p$ and $1-p$ are between $0$ and $1$ , so they are both positive. Now, we have:
\begin{align*} 5(1-p)&=p \\ 5-5p&=p \\ 5&=6p \\ p&=\frac {5}{6}\end{align*}
Now, we get $5+6=\boxed{011}$

Answer:  011
Problem: A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Rewrite it as : $(P)^3$ $(1-P)^5=\frac {1}{25}$ $(P)^5$ $(1-P)^3$
This can be simplified as $24P^2 -50P + 25 = 0$
This can be factored into $(4P-5)(6P-5)$
This yields two solutions: $\frac54$ (ignored because it would result in $1-p<0$ ) or $\frac56$
Therefore, the answer is $5+6$ $\boxed{011}$

Answer:  011
Problem: In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$

Solution: One of the ways to solve this problem is to make this parallelogram a straight line.
So the whole length of the line is $APC$ $AMC$ or $ANC$ ), and $ABC$ is $1000x+2009x=3009x.$
$AP$ $AM$ or $AN$ ) is $17x.$
So the answer is $3009x/17x = \boxed{177}$

Answer:  177
Problem: In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$

Solution: Draw a diagram with all the given points and lines involved. Construct parallel lines $\overline{DF_2F_1}$ and $\overline{BB_1B_2}$ to $\overline{MN}$ , where for the lines the endpoints are on $\overline{AM}$ and $\overline{AN}$ , respectively, and each point refers to an intersection. Also, draw the median of quadrilateral $BB_2DF_1$ $\overline{E_1E_2E_3}$ where the points are in order from top to bottom. Clearly, by similar triangles, $BB_2 = \frac {1000}{17}MN$ and $DF_1 = \frac {2009}{17}MN$ . It is not difficult to see that $E_2$ is the center of quadrilateral $ABCD$ and thus the midpoint of $\overline{AC}$ as well as the midpoint of $\overline{B_1}{F_2}$ (all of this is easily proven with symmetry). From more triangle similarity, $E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP$ $= \boxed{177}AP$

Answer:  177
Problem: In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$

Solution: Assume, for the ease of computation, that $AM=AN=17$ $AB=1000$ , and $AD=2009$ .  Now, let line $MN$ intersect line $CD$ at point $X$ and let $Y$ be a point such that $XY\parallel AD$ and $AY\parallel DX$ .  As a result, $ADXY$ is a parallelogram.  By construction, $\triangle MAN\sim \triangle MYX$ so \[\frac{MY}{MA}=\frac{YX}{AN}=\frac{AD}{AN}=\frac{2009}{17}\implies MY=2009\] and $AY=DX=2009-17$ .  Also, because $AM\parallel XC$ , we have $\triangle PAM\sim \triangle PCX$ so \[\frac{PC}{PA}=\frac{CX}{AM}=\frac{DX+CD}{AM}=\frac{2009-17+1000}{17}=176.\] Hence, $\frac{AC}{AP}=\frac{PC}{PA}+1=\boxed{177}.$

Answer:  177
Problem: In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$

Solution: Assign $A = (0,0)$ . Since there are no constraints in the problem against this, assume $ABCD$ to be a rectangle with dimensions $1000 \times 2009.$ Now, we can assign \[A=(0,0)\] \[B=(1000, 0)\] \[C=(1000,-2009)\] \[D=(0,-2009).\]
Then, since $\frac{AM}{AB} = \frac{17}{1000}$ and $AB = 1000$ , we can place $M$ at $(17, 0).$ Similarly, place $AN$ at $(0, 17).$ Then, the equation of line $MN$ is $y=x-17,$ and the equation of $AC$ is $y=\frac{-2009}{1000}x.$ Solve to find point $P$ at $\left( \frac{1000}{177}, \frac{-2009}{177} \right)$
We can calculate vectors to represent the distances: \[\overrightarrow{AC}= <1000, -2009>\] \[\overrightarrow{AP}= \frac{1}{177}<1000, -2009>.\] In this way, we can see that \[AC:AP = 177:1,\] and our answer is $\boxed{177}.$

Answer:  177
Problem: Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$ . If $AM = 180$ , find $LP$

Solution: Since $K$ is the midpoint of $\overline{PM}$ and $\overline{AC}$ , quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$
Thus,
\[\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1\]
Now let's apply the angle bisector theorem.
\[\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}\]
\[\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}\]
\[\frac {180}{LP}=\frac {5}{2}\]
\[LP=\boxed{072}\]

Answer:  072
Problem: Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$ . If $AM = 180$ , find $LP$

Solution: Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: \[\frac{BL}{CB}=\frac{AL}{CA}\implies\frac{BL}{300}=\frac{AL}{450}\implies 3BL=2AL\] So, we can weight $A$ as $2$ and $B$ as $3$ and $L$ as $5$ . Since $K$ is the midpoint of $A$ and $C$ , the weight of $A$ is equal to the weight of $C$ , which equals $2$ .
Also, since the weight of $L$ is $5$ and $C$ is $2$ , we can weight $P$ as $7$
By the definition of mass points, \[\frac{LP}{CP}=\frac{2}{5}\implies LP=\frac{2}{5}CP\] By vertical angles, angle $MKA =$ angle $PKC$ . 
Also, it is given that $AK=CK$ and $PK=MK$
By the SAS congruence, $\triangle MKA$ $\triangle PKC$ . So, $MA$ $CP$ $180$ .
Since $LP=\frac{2}{5}CP$ $LP = \frac{2}{5}(180) = \boxed{072}$

Answer:  072
Problem: Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$ . If $AM = 180$ , find $LP$

Solution: Using the diagram from solution $1$ , we can also utilize the fact that $AMCP$ forms a parallelogram. Because of that, we know that $AM = CP = 180$
Applying the angle bisector theorem to $\triangle CKB$ , we get that $\frac{KP}{PB} = \frac{225}{300} = \frac{3}{4}.$ So, we can let $MK = KP = 3x$ and $BP = 4x$
Now, apply law of cosines on $\triangle CKP$ and $\triangle CPB.$
If we let $\angle KCP = \angle PCB = \alpha$ , then the law of cosines gives the following system of equations:
\[9x^2 = 225^2 + 180^2 - 2\cdot 225 \cdot 180 \cdot \cos \alpha\] \[16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos \alpha.\]
Bashing those out, we get that $x = 15 \sqrt{13}$ and $\cos \alpha = \frac{7}{10}.$
Since $\cos \alpha = \frac{7}{10}$ , we can use the double angle formula to calculate that $\cos \left(2 \alpha \right) = -\frac{1}{50}.$
Now, apply Law of Cosines on $\triangle ABC$ to find $AB$
We get: \[AB^2 = 450^2 + 300^2 - 2 \cdot 450 \cdot 300 \cdot \left(- \frac{1}{50} \right).\]
Bashing gives $AB = 30 \sqrt{331}.$
From the angle bisector theorem on $\triangle ABC$ , we know that $\frac{AL}{BL} = \frac{450}{300} = \frac{3}{2}.$ So, $AL = 18 \sqrt{331}$ and $BL = 12 \sqrt{331}.$
Now, we apply Law of Cosines on $\triangle ALC$ and $\triangle BLC$ in order to solve for the length of $LC$
We get the following system:
\[(18 \sqrt{331})^2 = 450^2 + LC^2 - 2 \cdot 450 \cdot LC \cdot \frac{7}{10}\] \[(12 \sqrt{331})^2 = LC^2 + 300^2 - 2 \cdot 300 \cdot LC \cdot \frac{7}{10}\]
The first equation gives $LC = 252$ or $378$ and the second gives $LC = 252$ or $168$
The only value that satisfies both equations is $LC = 252$ , and since $LP = LC - PC$ , we have \[LC = 252 - 180 = \boxed{072}.\]

Answer:  072
Problem: How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$

Solution: First, $x$ must be less than $5$ , since otherwise $x^{\lfloor x\rfloor}$ would be at least $3125$ which is greater than $1000$
Because ${\lfloor x\rfloor}$ must be an integer, let’s do case work based on ${\lfloor x\rfloor}$
For ${\lfloor x\rfloor}=0$ $N=1$ as long as $x \neq 0$ . This gives us $1$ value of $N$
For ${\lfloor x\rfloor}=1$ $N$ can be anything between $1^1$ to $2^1$ excluding $2^1$
Therefore, $N=1$ . However, we got $N=1$ in case 1 so it got counted twice.
For ${\lfloor x\rfloor}=2$ $N$ can be anything between $2^2$ to $3^2$ excluding $3^2$
This gives us $3^2-2^2=5$ $N$ 's
For ${\lfloor x\rfloor}=3$ $N$ can be anything between $3^3$ to $4^3$ excluding $4^3$
This gives us $4^3-3^3=37$ $N$ 's
For ${\lfloor x\rfloor}=4$ $N$ can be anything between $4^4$ to $5^4$ excluding $5^4$
This gives us $5^4-4^4=369$ $N$ 's
Since $x$ must be less than $5$ , we can stop here and the answer is $1+5+37+369= \boxed{412}$ possible values for $N$

Answer:  412
Problem: How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$

Solution: For a positive integer $k$ , we find the number of positive integers $N$ such that $x^{\lfloor x\rfloor}=N$ has a solution with ${\lfloor x\rfloor}=k$ . Then $x=\sqrt[k]{N}$ , and because $k \le x < k+1$ , we have $k^k \le x^k < (k+1)^k$ , and because $(k+1)^k$ is an integer, we get $k^k \le x^k \le (k+1)^k-1$ . The number of possible values of $x^k$ is equal to the number of integers between $k^k$ and $(k+1)^k-1$ inclusive, which is equal to the larger number minus the smaller number plus one or $((k+1)^k-1)-(k^k)+1$ , and this is equal to $(k+1)^k-k^k$ . If $k>4$ , the value of $x^k$ exceeds $1000$ , so we only need to consider $k \le 4$ . The requested number of values of $N$ is the same as the number of values of $x^k$ , which is $\sum^{4}_{k=1} [(k+1)^k-k^k]=2-1+9-4+64-27+625-256=\boxed{412}$

Answer:  412
Problem: The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$ . Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$

Solution: The best way to solve this problem is to get the iterated part out of the exponent: \[5^{(a_{n + 1} - a_n)} = \frac {1}{n + \frac {2}{3}} + 1\] \[5^{(a_{n + 1} - a_n)} = \frac {n + \frac {5}{3}}{n + \frac {2}{3}}\] \[5^{(a_{n + 1} - a_n)} = \frac {3n + 5}{3n + 2}\] \[a_{n + 1} - a_n = \log_5{\left(\frac {3n + 5}{3n + 2}\right)}\] \[a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}\] Plug in $n = 1, 2, 3, 4$ to see the first few terms of the sequence: \[\log_5{5},\log_5{8}, \log_5{11}, \log_5{14}.\] We notice that the terms $5, 8, 11, 14$ are in arithmetic progression. Since $a_1 = 1 = \log_5{5} = \log_5{(3(1) + 2)}$ , we can easily use induction to show that $a_n = \log_5{(3n + 2)}$ . So now we only need to find the next value of $n$ that makes $\log_5{(3n + 2)}$ an integer. This means that $3n + 2$ must be a power of $5$ . We test $25$ \[3n + 2 = 25\] \[3n = 23\] This has no integral solutions, so we try $125$ \[3n + 2 = 125\] \[3n = 123\] \[n = \boxed{041}\]

Answer:  041
Problem: The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$ . Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$

Solution: We notice that by multiplying the equation from an arbitrary $a_n$ all the way to $a_1$ , we get: \[5^{a_n-a_1}=\dfrac{n+\tfrac23}{1+\tfrac23}\] This simplifies to \[5^{a_n}=3n+2.\] We can now test powers of $5$
$5$ - that gives us $n=1$ , which is useless.
$25$ - that gives a non-integer $n$
$125$ - that gives $n=\boxed{41}$

Answer:  41
Problem: Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$

Solution: Find the positive differences in all $55$ pairs and you will get $\boxed{398}$ . 
(This is not recommended unless you can't find any other solutions to this problem)

Answer:  398
Problem: Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$

Solution: When computing $N$ , the number $2^x$ will be added $x$ times (for terms $2^x-2^0$ $2^x-2^1$ , ..., $2^x - 2^{x-1}$ ), and subtracted $10-x$ times. Hence $N$ can be computed as $N=10\cdot 2^{10} + 8\cdot 2^9 + 6\cdot 2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0$ . Evaluating $N \bmod {1000}$ yields:
\begin{align*} N  & = 10(2^{10}-1) + 8(2^9 - 2^1) + 6(2^8-2^2) + 4(2^7-2^3) + 2(2^6-2^4) \\ & = 10(1023) + 8(510) + 6(252) + 4(120) + 2(48) \\ & = 10(1000+23) + 8(500+10) + 6(250+2) + 480 + 96 \\ & \equiv (0 + 230) + (0 + 80) + (500 + 12) + 480 + 96 \\ & \equiv \boxed{398}

Answer:  398
Problem: Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$

Solution: This solution can be generalized to apply when $10$ is replaced by other positive integers.
Extending from Solution 2, we get that the sum $N$ of all possible differences of pairs of elements in $S$ when $S = \{2^0,2^1,2^2,\ldots,2^{n}\}$ is equal to $\sum_{x=0}^{n} (2x-n) 2^x$ . Let $A = \sum_{x=0}^{n} x2^x$ $B=\sum_{x=0}^{n} 2^x$ . Then $N=2A - nB$
For $n = 10$ $B = 2^{11}-1 = 2047 \equiv 47 \pmod{1000}$ by the geometric sequence formula.
$2A = \sum_{x=1}^{n+1} (x-1)2^x$ , so $2A - A = A = n2^{n+1} - \sum_{x=1}^{n} 2^x$ . Hence, for $n = 10$ $A = 10 \cdot 2^{11} - 2^{11} + 2 = 9 \cdot 2^{11} + 2 \equiv 48 \cdot 9 + 2 =$
$434 \pmod{1000}$ , by the geometric sequence formula and the fact that $2^{10} = 1024 \equiv 24 \pmod{1000}$
Thus, for $n = 10$ $N = 2A - 10B \equiv 2\cdot 434 - 10\cdot 47 = 868 - 470 = \boxed{398}$

Answer:  398
Problem: A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $$ 1$ to $$ 9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were $1, 1, 1, 1, 3, 3, 3$ . Find the total number of possible guesses for all three prizes consistent with the hint.

Solution: [Clarification: You are supposed to find the number of all possible tuples of prices, $(A, B, C)$ , that could have been on that day.]
Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$ , then the string is
\[1131331.\]
Since the strings have seven digits and three threes, there are $\binom{7}{3}=35$ arrangements of all such strings.
In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups.
Let's look at the example. We have to partition it into 3 groups with each group having at least 1 digit. In other words, we need to find the solution to
\[x+y+z=7, x,y,z>0.\]
This gives us
\[\binom{6}{2}=15\]
ways by balls and urns. But we have counted the one with 5 digit numbers; that is, $(5,1,1),(1,1,5),(1,5,1)$
Thus, each arrangement has \[\binom{6}{2}-3=12\] ways per arrangement, and there are $12\times35=\boxed{420}$ ways.

Answer:  420