Answer: 19.0

Let $A_1A_2A_3\ldots A_{12}$ be a dodecagon ( $12$ -gon). Three frogs initially sit at $A_4,A_8,$ and $A_{12}$ . At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
Define the distance between two frogs as the number of sides between them that do not contain the third frog.
Let $E(a,b,c)$ denote the expected number of minutes until the frogs stop jumping, such that the distances between the frogs are $a,b,$ and $c$ (in either clockwise or counterclockwise order). Without the loss of generality, assume that $a\leq b\leq c.$
We wish to find $E(4,4,4).$ Note that:
We have the following system of equations: \begin{align*} E(4,4,4)&=1+\frac{2}{8}E(4,4,4)+\frac{6}{8}E(2,4,6), \\ E(2,4,6)&=1+\frac{4}{8}E(2,4,6)+\frac{1}{8}E(4,4,4)+\frac{1}{8}E(2,2,8), \\ E(2,2,8)&=1+\frac{2}{8}E(2,2,8)+\frac{2}{8}E(2,4,6). \end{align*} Rearranging and simplifying each equation, we get \begin{align*} E(4,4,4)&=\frac{4}{3}+E(2,4,6), &(1) \\ E(2,4,6)&=2+\frac{1}{4}E(4,4,4)+\frac{1}{4}E(2,2,8), &\hspace{12.75mm}(2) \\ E(2,2,8)&=\frac{4}{3}+\frac{1}{3}E(2,4,6). &(3) \end{align*} Substituting $(1)$ and $(3)$ into $(2),$ we obtain \[E(2,4,6)=2+\frac{1}{4}\left[\frac{4}{3}+E(2,4,6)\right]+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(2,4,6)\right],\] from which $E(2,4,6)=4.$ Substituting this into $(1)$ gives $E(4,4,4)=\frac{16}{3}.$
Therefore, the answer is $16+3=\boxed{019}.$

Solution:
We can solve the problem by removing $1$ frog, and calculate the expected time for the remaining $2$ frogs. In the original problem, when the movement stops, $2$ of the $3$ frogs meet. Because the $3$ frogs cannot meet at one vertex, the probability that those two specific frogs meet is $\tfrac13$ . If the expected time for the two frog problem is $E'$ , then the expected time for the original problem is $\tfrac 13 E'$
The distance between the two frogs can only be $0$ $2$ $4$ $6$ . We use the distances as the states to draw the following Markov Chain . This Markov Chain is much simpler than that of Solution 1 Supplement in the Remark section.
\begin{align*} E(2) &= 1 + \frac12 \cdot E(2) + \frac14 \cdot E(4)\\ E(4) &= 1 + \frac14 \cdot E(2) + \frac12 \cdot E(4) + \frac14 \cdot E(6)\\ E(6) &= 1 + \frac12 \cdot E(4) + \frac12 \cdot E(6) \end{align*}
By solving the above system of equations, $E(4) = 16$ . The answer for the original problem is $\frac{16}{3}$ $16 + 3 = \boxed{019}$


Answer: 19.0

Let $B$ be a right rectangular prism (box) with edges lengths $1,$ $3,$ and $4$ , together with its interior. For real $r\geq0$ , let $S(r)$ be the set of points in $3$ -dimensional space that lie within a distance $r$ of some point in $B$ . The volume of $S(r)$ can be expressed as $ar^{3} + br^{2} + cr +d$ , where $a,$ $b,$ $c,$ and $d$ are positive real numbers. What is $\frac{bc}{ad}?$
$\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38$

Solution:
Split $S(r)$ into 4 regions:
1. The rectangular prism itself
2. The extensions of the faces of $B$
3. The quarter cylinders at each edge of $B$
4. The one-eighth spheres at each corner of $B$
Region 1: The volume of $B$ is $1 \cdot 3 \cdot 4 = 12$ , so $d=12$
Region 2: This volume is equal to the surface area of $B$ times $r$ (these "extensions" are just more boxes). The volume is then $\text{SA} \cdot r=2(1 \cdot 3 + 1 \cdot 4 + 3 \cdot 4)r=38r$ to get $c=38$
Region 3: We see that there are 12 quarter-cylinders, 4 of each type. We have 4 quarter-cylinders of height 1, 4 quarter-cylinders of height 3, 4 quarter-cylinders of height 4. Since 4 quarter-cylinders make a full cylinder, the total volume is $4 \cdot \dfrac{1\pi r^2}{4} + 4 \cdot \dfrac{3\pi r^2}{4} + 4 \cdot \dfrac{4 \pi r^2}{4}=8 \pi r^2$ . Therefore, $b=8\pi$
Region 4: There is an eighth-sphere of radius $r$ at each corner of $B$ . Since there are 8 corners, these eighth-spheres add up to 1 full sphere of radius $r$ . The volume of this sphere is then $\frac{4}{3}\pi \cdot r^3$ , so $a=\frac{4\pi}{3}$
Using these values, $\dfrac{bc}{ad}=\frac{(8\pi)(38)}{(4\pi/3)(12)} = \boxed{19}$


Answer: 330.0

Let $B$ be the set of all binary integers that can be written using exactly $5$ zeros and $8$ ones where leading zeros are allowed. If all possible subtractions are performed in which one element of $B$ is subtracted from another, find the number of times the answer $1$ is obtained.

Solution: When $1$ is subtracted from a binary number, the number of digits will remain constant if and only if the original number ended in $10.$ Therefore, every subtraction involving two numbers from $B$ will necessarily involve exactly one number ending in $10.$ To solve the problem, then, we can simply count the instances of such numbers. With the $10$ in place, the seven remaining $1$ 's can be distributed in any of the remaining $11$ spaces, so the answer is ${11 \choose 7} = \boxed{330}$


Answer: 588.0

Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$

Solution:
Let our polynomial be $P(x)$
It is clear that the coefficient of $x$ in $P(x)$ is $-1 + 2 - 3 + \ldots + 14 - 15 = -8$ , so $P(x) = 1 -8x + Cx^2 + Q(x)$ , where $Q(x)$ is some polynomial divisible by $x^3$
Then $P(-x) = 1 + 8x + Cx^2 + Q(-x)$ and so $P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)$ , where $R(x)$ is some polynomial divisible by $x^3$
However, we also know $P(x)\cdot P(-x) = (1 - x)(1 + x)(1 +2x)(1 - 2x) \cdots (1 - 15x)(1 + 15x)$ $= (1 - x^2)(1 - 4x^2)\cdots(1 - 225x^2)$ $= 1 - (1 + 4 + \ldots + 225)x^2 + R(x)$
Equating coefficients, we have $2C - 64 = -(1 + 4 + \ldots + 225) = -1240$ , so $-2C = 1176$ and $|C| = \boxed{588}$

Solution:
Let $S$ be the set of integers $\{-1,2,-3,\ldots,14,-15\}$ . The coefficient of $x^2$ in the expansion is equal to the sum of the product of each pair of distinct terms, or $C = \sum_{1 \le i \neq j}^{15} S_iS_j$ . Also, we know that \begin{align*}\left(\sum_{i=1}^{n} S_i\right)^2 &= \left(\sum_{i=1}^{n} S_i^2\right) + 2\left(\sum_{1 \le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*} where the left-hand sum can be computed from:
and the right-hand sum comes from the formula for the sum of the first $n$ perfect squares. Therefore, $|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}$

Solution: Let set $N$ be $\{-1, -3, \ldots -15\}$ and set $P$ be $\{2, 4, \ldots 14\}$ . The sum of the negative $x^2$ coefficients is the sum of the products of the elements in all two element sets such that one element is from $N$ and the other is from $P$ . Each summand is a term in the expansion of \[(-1 - 3 - \ldots - 15)(2 + 4 + \ldots + 14)\] which equals $-56 * 64 = -(60^2 - 4^2) = -3584$ . The sum of the positive $x^2$ coefficients is the sum of the products of all two element sets such that the two elements are either both in $N$ or both in $P$ . By counting, the sum is $2992$ , so the sum of all $x^2$ coefficients is $-588$ . Thus, the answer is $\boxed{588}$

Solution: This is just another way of summing the subsets of 2 from $[-1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15]$ . Start from the right and multiply -15 to everything on its left. Use the distributive property and add all the 14 integers together to get 7. This gives us $-15 * 7$ . Doing this for 14 gives us $14 * -7$ , and for -13 we get $-13 * 6$ . This pattern repeats where every two integers will multiple 7, 6,... to 0. Combining and simplifying the pattern give us this: $-(29 * 7 + 25 * 6 + 21 * 5 + 17 * 4 + 13 * 3 + 9 * 2 + 5*1)$ . The expression gives us -588, or $C = \boxed{588}$ . This is a good solution because it guarantees we never add a product twice, and the pattern is simple to add by hand.

Solution: We expand and obtain $\left(x-1\right)\left(2x-1\right)\left(3x-1\right)\cdots\left(15x-1\right) =  1307674368000x^{15} - 948550176000x^{14} - 689324826240x^{13} + 2733483288464x^{12} + 82808260416x^{11} - 23038684088x^{10} - 3811851848x^9 + 828730833x^8 + 81228128x^7 - 14661124x^6 - 853104x^5 + 132902x^4 + 4256x^3 - \boxed{588}x^2 - 8x + 1.$


Answer: 84.0

Let $C$ be the graph of $xy = 1$ , and denote by $C^*$ the reflection of $C$ in the line $y = 2x$ .  Let the equation of $C^*$ be written in the form
\[12x^2 + bxy + cy^2 + d = 0.\]
Find the product $bc$

Solution:
Given a point $P (x,y)$ on $C$ , we look to find a formula for $P' (x', y')$ on $C^*$ . Both points lie on a line that is perpendicular to $y=2x$ , so the slope of $\overline{PP'}$ is $\frac{-1}{2}$ . Thus $\frac{y' - y}{x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y$ . Also, the midpoint of $\overline{PP'}$ $\left(\frac{x + x'}{2}, \frac{y + y'}{2}\right)$ , lies on the line $y = 2x$ . Therefore $\frac{y + y'}{2} = x + x' \Longrightarrow 2x' - y' = y - 2x$
Solving these two equations, we find $x = \frac{-3x' + 4y'}{5}$ and $y = \frac{4x' + 3y'}{5}$ . Substituting these points into the equation of $C$ , we get $\frac{(-3x'+4y')(4x'+3y')}{25}=1$ , which when expanded becomes $12x'^2-7x'y'-12y'^2+25=0$
Thus, $bc=(-7)(-12)=\boxed{084}$

Solution:
The asymptotes of $C$ are given by $x=0$ and $y=0$ . Now if we represent the line $y=2x$ by the complex number $1+2i$ , then we find the direction of the reflection of the asymptote $x=0$ by multiplying this by $2-i$ , getting $4+3i$ . Therefore, the asymptotes of $C^*$ are given by $4y-3x=0$ and $3y+4x=0$
Now to find the equation of the hyperbola, we multiply the two expressions together to get one side of the equation: $(3x-4y)(4x+3y)=12x^2-7xy-12y^2$ . At this point, the right hand side of the equation will be determined by plugging the point $(\frac{\sqrt{2}}{2},\sqrt{2})$ , which is unchanged by the reflection, into the expression. But this is not necessary. We see that $b=-7$ $c=-12$ , so $bc=\boxed{084}$

Solution: Find some simple points on the graph of C, reflect them and note down the new coordinates, and plug 'em into the given equation. After some plugging and chugging and solving the system of equations that follow, we get that $b = -7$ and $c = -12$ , so $bc = 84$ . The answer is $\boxed{084}$


Answer: 20.0

Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$ respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$
$\text{(A) }15 \qquad \text{(B) }18 \qquad \text{(C) }20 \qquad \text{(D) }21 \qquad \text{(E) }24$

Solution:
First examine the formula $(x-10)^2+y^2=36$ , for the circle $C_1$ . Its center, $D_1$ , is located at (10,0) and it has a radius of $\sqrt{36}$ = 6. The next circle, using the same pattern, has its center, $D_2$ , at (-15,0) and has a radius of $\sqrt{81}$ = 9. So we can construct this diagram:  Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles $D_2QO$ and $D_1PO$ similar by AA, with a scale factor of $6:9$ , or $2:3$ . Next, we must subdivide the line $D_2D_1$ in a 2:3 ratio to get the length of the segments $D_2O$ and $D_1O$ . The total length is $10 - (-15)$ , or $25$ , so applying the ratio, $D_2O$ 15 and $D_1O$ 10 . These are the hypotenuses of the triangles. We already know the length of $D_2Q$ and $D_1P$ and (they're radii). So in order to find $PQ$ , we must find the length of the longer legs of the two triangles and add them.
$15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144$
$\sqrt{144} = 12$
$10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64$
$\sqrt{64} = 8$
Finally, the length of PQ is $12+8=\boxed{20}$ , or (C)

Solution:
Using the above diagram, imagine that segment $\overline{QD_2}$ is shifted to the right to match up with $\overline{PD_1}$ . Then shift $\overline{QP}$ downwards to make a right triangle. We know $\overline{D_2D_1} = 25$ from the given information and the newly created leg has length $\overline{QD_2} + \overline{PD_1} = 9 + 6 = 15$ . Hence by Pythagorean theorem $15^2 + {\overline{QP}}^2 = 25^2$
$\overline{QP} = \boxed{20}$ , or C.


Answer: 112.0

Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$

Solution:
Note that $96 = 2^5 \cdot 3$ . Since there are at most six not necessarily distinct factors $>1$ multiplying to $96$ , we have six cases: $k=1, 2, ..., 6.$ Now we look at each of the six cases.
$k=1$ : We see that there is $1$ way, merely $96$
$k=2$ : This way, we have the $3$ in one slot and $2$ in another, and symmetry. The four other $2$ 's leave us with $5$ ways and symmetry doubles us so we have $10$
$k=3$ : We have $3, 2, 2$ as our baseline. We need to multiply by $2$ in $3$ places, and see that we can split the remaining three powers of $2$ in a manner that is $3-0-0$ $2-1-0$ or $1-1-1$ . A $3-0-0$ split has $6 + 3 = 9$ ways of happening ( $24-2-2$ and symmetry; $2-3-16$ and symmetry), a $2-1-0$ split has $6 \cdot 3 = 18$ ways of happening (due to all being distinct) and a $1-1-1$ split has $3$ ways of happening ( $6-4-4$ and symmetry) so in this case we have $9+18+3=30$ ways.
$k=4$ : We have $3, 2, 2, 2$ as our baseline, and for the two other $2$ 's, we have a $2-0-0-0$ or $1-1-0-0$ split. The former grants us $4+12=16$ ways ( $12-2-2-2$ and symmetry and $3-8-2-2$ and symmetry) and the latter grants us also $12+12=24$ ways ( $6-4-2-2$ and symmetry and $3-4-4-2$ and symmetry) for a total of $16+24=40$ ways.
$k=5$ : We have $3, 2, 2, 2, 2$ as our baseline and one place to put the last two: on another two or on the three. On the three gives us $5$ ways due to symmetry and on another two gives us $5 \cdot 4 = 20$ ways due to symmetry. Thus, we have $5+20=25$ ways.
$k=6$ : We have $3, 2, 2, 2, 2, 2$ and symmetry and no more twos to multiply, so by symmetry, we have $6$ ways.
Thus, adding, we have $1+10+30+40+25+6=\boxed{112}$

Solution:
As before, note that $96=2^5\cdot3$ , and we need to consider 6 different cases, one for each possible value of $k$ , the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with $k$ factors. First, the factorization needs to contain one factor that is itself a multiple of $3$ . There are $k$ to choose from. That leaves $k-1$ slots left to fill, each of which must contain at least one factor of $2$ . Once we have filled in a $2$ to each of the remaining slots, we're left with $5-(k-1)=6-k$ twos.
Consider the remaining $6-k$ factors of $2$ left to assign to the $k$ factors. Using stars and bars, the number of ways to do this is: \[{{(6-k)+k-1}\choose{6-k}}={5\choose{6-k}}\] This makes $k{5\choose{6-k}}$ possibilities for each k.
To obtain the total number of factorizations, add all possible values for k: \[\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{112}.\]

Solution:
Begin by examining $f_1$ $f_1$ can take on any value that is a factor of $96$ except $1$ . For each choice of $f_1$ , the resulting $f_2...f_k$ must have a product of $96/f_1$ . This means the number of ways the rest $f_a$ $1<a<=k$ can be written by the scheme stated in the problem for each $f_1$ is equal to $D(96/f_1)$ , since the product of $f_2 \cdot f_3... \cdot f_k=x$ is counted as one valid product if and only if $f_1 \cdot x=96$ , the product $x$ has the properties that factors are greater than $1$ , and differently ordered products are counted separately.
For example, say the first factor is $2$ . Then, the remaining numbers must multiply to $48$ , so the number of ways the product can be written beginning with $2$ is $D(48)$ . To add up all the number of solutions for every possible starting factor, $D(96/f_1)$ must be calculated and summed for all possible $f_1$ , except $96$ and $1$ , since a single $1$ is not counted according to the problem statement. The $96$ however, is counted, but only results in $1$ possibility, the first and only factor being $96$ . This means
$D(96)=D(48)+D(32)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$
Instead of calculating D for the larger factors first, reduce $D(48)$ $D(32)$ , and $D(24)$ into sums of $D(m)$ where $m<=16$ to ease calculation. Following the recursive definition $D(n)=($ sums of $D(c))+1$ where c takes on every divisor of n except for 1 and itself, the sum simplifies to
$D(96)=(D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1)$ $(D(16)+D(8)+D(4)+D(2)+1)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1.$
$D(24)=D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$ , so the sum further simplifies to
$D(96)=3D(16)+4D(12)+5D(8)+4D(6)+5D(4)+4D(3)+5D(2)+5$ , after combining terms. From quick casework,
$D(16)=8, D(12)=8, D(8)=4, D(6)=3, D(4)=2, D(3)=1$ and $D(2)=1$ . Substituting these values into the expression above,
$D(96)=3 \cdot 8+4 \cdot 8+5 \cdot 4+4 \cdot 3+5 \cdot 2+4 \cdot 1+5 \cdot 1+5=\boxed{112}$

Solution:
Note that $96 = 3 \cdot 2^5$ , and that $D$ of a perfect power of a prime is relatively easy to calculate. Also note that you can find $D(96)$ from $D(32)$ by simply totaling the number of ways there are to insert a $3$ into a set of numbers that multiply to $32$
First, calculate $D(32)$ . Since $32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ , all you have to do was find the number of ways to divide the $2$ 's into groups, such that each group has at least one $2$ . By stars and bars, this results in $1$ way with five terms, $4$ ways with four terms, $6$ ways with three terms, $4$ ways with two terms, and $1$ way with one term. (The total, $16$ , is not needed for the remaining calculations.)
Then, to get $D(96)$ , in each possible $D(32)$ sequence, insert a $3$ somewhere in it, either by placing it somewhere next to the original numbers (in one of $n+1$ ways, where $n$ is the number of terms in the $D(32)$ sequence), or by multiplying one of the numbers by $3$ (in one of $n$ ways). There are $2+1=3$ ways to do this with one term, $3+2=5$ with two, $7$ with three, $9$ with four, and $11$ with five.
The resulting number of possible sequences is $3 \cdot 1 + 5 \cdot 4 + 7 \cdot 6 + 9 \cdot 4 + 11 \cdot 1 = \boxed{112}$ . ~ emerald_block

Solution:
Consider the arrangement of the prime factors of 96 in a line $(2,2, 2, 2, 2, 3)$ .  An arrangement of factors can be created by placing "dividers" to group primes.  For example, $(2, 2, |, 2, 2, 2, |, 3)$ is equivalent to the arrangement $4 \cdot 8 \cdot 3$ .  Because there are $6$ ways to order the prime factors, and $2^5$ ways to place dividers, this gives us an initial $6 \cdot  2^5$ ways to arrange divisors.
However, through this method, we overcount cases where $3$ is combined with another factor.  For example, the arrangement $4 \cdot 6 \cdot 4$ can be written as $(2, 2, |, 2, 3, |, 2, 2)$ or $(2, 2, |, 3, 2, |, 2, 2)$ .  Precisely, we double count any case with $6$ as a factor, triple count any case with $12$ , quadruple count any case with $24$ , etc.
Now, consider all cases where $3$ must be grouped with at least one $2$ .  This can be expressed in the same "line" format as $(2, 2, 2, 2, 6)$ , where dividers can again be placed to group divisors.  In this case, there are $5$ ways to order divisors, and $2^4$ ways to place dividers, so we have an $5 \cdot 2^4$ possible sequences for this case.  Notice that in this format, we double count cases where $12$ is a factor, we triple count cases where $24$ is a factor, etc.  Precisely, for any case counted $n$ times in the first step, it is counted $n - 1$ times in this step.  Thus, if we subtract, we count each case exactly once.
So, we get:
$6 \cdot  2^5 - 5 \cdot 2^4 = \boxed{112}$ . ~ hdai1122

Solution:
First we factor $32$ into $m$ numbers $g_1, \cdots, g_m$ where $g_i>1,i=1,\ldots,m$ . By applying stars and bars there are $\binom{5-1}{m-1}$ ways. Then we can either insert $3$ into each of the $m+1$ spaces between (or beyond) $g_i$ 's, or multiply it to one of the $g_i$ 's, a total of $2m+1$ ways. Hence the answer to the problem is
\[\sum_{m=1}^5 (2m+1)\binom{5-1}{m-1} = \sum_{n=0}^4 (2n+3) \binom{4}{n} = 8\sum_{n=0}^4 \frac{n}{4} \binom{4}{n} + 3 \sum_{n=0}^4 \binom{4}{n} = 8 \sum_{n=0}^4 \binom{3}{n-1} + 3\sum_{n=0}^4 \binom{4}{n} = 8 \cdot 2^3 + 3\cdot 2^4 = \boxed{112}.\]

Solution:
Note that $96 = 2^5 \cdot 3$ $D(n)$ depends on dividing $2^5$ into different terms, which is the integer partition of $5$
Divide $96$ into $1$ term:
There is only one way. $\underline{\textbf{1}}$
Divide $96$ into $2$ terms:
$2 + 8 = \underline{\textbf{10}}$
Divide $96$ into $3$ terms:
$12 + 18 = \underline{\textbf{30}}$
Divide $96$ into $4$ terms:
$24 + 16 = \underline{\textbf{40}}$
Divide $96$ into $5$ terms:
When dividing $5$ into $5$ parts there are $2$ cases.
$20 + 5 = \underline{\textbf{25}}$
Divide $96$ into $6$ terms:
$5 = 1 + 1 + 1 + 1 + 1$ . The number of arrangements of $(2, 2, 2, 2, 2, 3)$ is $\frac{6!}{5!} = \underline{\textbf{6}}$
$1 + 10 + 30 + 40 + 25 + 6 = \boxed{112}$

Solution:
Consider how we partition the factors of $96 = 2^5\cdot 3$ . For each $k$ , there are two cases. Either we can put the $2$ s into $k$ nonzero parts, so that the $3$ shares a partition with some $2$ s, which can be done in $k\binom{5-(k-1)+(k-1)-1}{k-2} = k\binom{4}{k-2}$ ways, or we can put the $2$ s into $k-1$ nonzero parts and put the $3$ in its own partition, which can be done in $k\binom{5-k+k-1}{k-1} = k\binom{4}{k-1}$ ways. Summing over all $k$ , we have \[\sum_{k=1}^6 \binom{4}{k-2} + \sum_{k=1}^6 \binom{4}{k-1}.\]
But $\sum_{k=1}^6 \binom{4}{k-2} = 2\binom{4}{0}+3\binom{4}{1} + \dots + 6\binom{4}{4} = 4 \cdot \sum_{i=0}^4 \binom{4}{i} = 4 \cdot 2^4 = 64$ . Similarly, $\sum_{k=1}^6 \binom{4}{k-1} = 3\cdot 2^4 = 48$ . So our answer is $64+48 = \boxed{112}$


Answer: 112.0

Solution: Ignore the $3$ first and first count $2^{x_1+x_2+...+x^n}=32$ which $x_1+x_2+...+x_n=5$ . This implies that $n$ is less than or equal to $5$ . Now, we can see that $3$ can lie between the two $x_i, x_{i+1}$ , or contribute to one of them. This gives $2k+1$ if $x_1+...+x_k=5$ . Now, just sum up gives $\binom{5-1}{5-1}\cdot 11+\binom{5-1}{4-1}\cdot 9+\binom{5-1}{3-1}\cdot 7+\binom{5-1}{2-1}\cdot 5+\binom{5-1}{1-1}\cdot 3=\boxed{112}$


Answer: 112.0

Solution:
Note that $96 = 2^5 \cdot 3$ . Since there are at most six not necessarily distinct factors $>1$ multiplying to $96$ , we have six cases: $k=1, 2, ..., 6.$ Now we look at each of the six cases.
$k=1$ : We see that there is $1$ way, merely $96$
$k=2$ : This way, we have the $3$ in one slot and $2$ in another, and symmetry. The four other $2$ 's leave us with $5$ ways and symmetry doubles us so we have $10$
$k=3$ : We have $3, 2, 2$ as our baseline. We need to multiply by $2$ in $3$ places, and see that we can split the remaining three powers of $2$ in a manner that is $3-0-0$ $2-1-0$ or $1-1-1$ . A $3-0-0$ split has $6 + 3 = 9$ ways of happening ( $24-2-2$ and symmetry; $2-3-16$ and symmetry), a $2-1-0$ split has $6 \cdot 3 = 18$ ways of happening (due to all being distinct) and a $1-1-1$ split has $3$ ways of happening ( $6-4-4$ and symmetry) so in this case we have $9+18+3=30$ ways.
$k=4$ : We have $3, 2, 2, 2$ as our baseline, and for the two other $2$ 's, we have a $2-0-0-0$ or $1-1-0-0$ split. The former grants us $4+12=16$ ways ( $12-2-2-2$ and symmetry and $3-8-2-2$ and symmetry) and the latter grants us also $12+12=24$ ways ( $6-4-2-2$ and symmetry and $3-4-4-2$ and symmetry) for a total of $16+24=40$ ways.
$k=5$ : We have $3, 2, 2, 2, 2$ as our baseline and one place to put the last two: on another two or on the three. On the three gives us $5$ ways due to symmetry and on another two gives us $5 \cdot 4 = 20$ ways due to symmetry. Thus, we have $5+20=25$ ways.
$k=6$ : We have $3, 2, 2, 2, 2, 2$ and symmetry and no more twos to multiply, so by symmetry, we have $6$ ways.
Thus, adding, we have $1+10+30+40+25+6=\boxed{112}$

Solution:
As before, note that $96=2^5\cdot3$ , and we need to consider 6 different cases, one for each possible value of $k$ , the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with $k$ factors. First, the factorization needs to contain one factor that is itself a multiple of $3$ . There are $k$ to choose from. That leaves $k-1$ slots left to fill, each of which must contain at least one factor of $2$ . Once we have filled in a $2$ to each of the remaining slots, we're left with $5-(k-1)=6-k$ twos.
Consider the remaining $6-k$ factors of $2$ left to assign to the $k$ factors. Using stars and bars, the number of ways to do this is: \[{{(6-k)+k-1}\choose{6-k}}={5\choose{6-k}}\] This makes $k{5\choose{6-k}}$ possibilities for each k.
To obtain the total number of factorizations, add all possible values for k: \[\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{112}.\]

Solution:
Begin by examining $f_1$ $f_1$ can take on any value that is a factor of $96$ except $1$ . For each choice of $f_1$ , the resulting $f_2...f_k$ must have a product of $96/f_1$ . This means the number of ways the rest $f_a$ $1<a<=k$ can be written by the scheme stated in the problem for each $f_1$ is equal to $D(96/f_1)$ , since the product of $f_2 \cdot f_3... \cdot f_k=x$ is counted as one valid product if and only if $f_1 \cdot x=96$ , the product $x$ has the properties that factors are greater than $1$ , and differently ordered products are counted separately.
For example, say the first factor is $2$ . Then, the remaining numbers must multiply to $48$ , so the number of ways the product can be written beginning with $2$ is $D(48)$ . To add up all the number of solutions for every possible starting factor, $D(96/f_1)$ must be calculated and summed for all possible $f_1$ , except $96$ and $1$ , since a single $1$ is not counted according to the problem statement. The $96$ however, is counted, but only results in $1$ possibility, the first and only factor being $96$ . This means
$D(96)=D(48)+D(32)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$
Instead of calculating D for the larger factors first, reduce $D(48)$ $D(32)$ , and $D(24)$ into sums of $D(m)$ where $m<=16$ to ease calculation. Following the recursive definition $D(n)=($ sums of $D(c))+1$ where c takes on every divisor of n except for 1 and itself, the sum simplifies to
$D(96)=(D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1)$ $(D(16)+D(8)+D(4)+D(2)+1)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1.$
$D(24)=D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$ , so the sum further simplifies to
$D(96)=3D(16)+4D(12)+5D(8)+4D(6)+5D(4)+4D(3)+5D(2)+5$ , after combining terms. From quick casework,
$D(16)=8, D(12)=8, D(8)=4, D(6)=3, D(4)=2, D(3)=1$ and $D(2)=1$ . Substituting these values into the expression above,
$D(96)=3 \cdot 8+4 \cdot 8+5 \cdot 4+4 \cdot 3+5 \cdot 2+4 \cdot 1+5 \cdot 1+5=\boxed{112}$

Solution:
Note that $96 = 3 \cdot 2^5$ , and that $D$ of a perfect power of a prime is relatively easy to calculate. Also note that you can find $D(96)$ from $D(32)$ by simply totaling the number of ways there are to insert a $3$ into a set of numbers that multiply to $32$
First, calculate $D(32)$ . Since $32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ , all you have to do was find the number of ways to divide the $2$ 's into groups, such that each group has at least one $2$ . By stars and bars, this results in $1$ way with five terms, $4$ ways with four terms, $6$ ways with three terms, $4$ ways with two terms, and $1$ way with one term. (The total, $16$ , is not needed for the remaining calculations.)
Then, to get $D(96)$ , in each possible $D(32)$ sequence, insert a $3$ somewhere in it, either by placing it somewhere next to the original numbers (in one of $n+1$ ways, where $n$ is the number of terms in the $D(32)$ sequence), or by multiplying one of the numbers by $3$ (in one of $n$ ways). There are $2+1=3$ ways to do this with one term, $3+2=5$ with two, $7$ with three, $9$ with four, and $11$ with five.
The resulting number of possible sequences is $3 \cdot 1 + 5 \cdot 4 + 7 \cdot 6 + 9 \cdot 4 + 11 \cdot 1 = \boxed{112}$ . ~ emerald_block

Solution:
Consider the arrangement of the prime factors of 96 in a line $(2,2, 2, 2, 2, 3)$ .  An arrangement of factors can be created by placing "dividers" to group primes.  For example, $(2, 2, |, 2, 2, 2, |, 3)$ is equivalent to the arrangement $4 \cdot 8 \cdot 3$ .  Because there are $6$ ways to order the prime factors, and $2^5$ ways to place dividers, this gives us an initial $6 \cdot  2^5$ ways to arrange divisors.
However, through this method, we overcount cases where $3$ is combined with another factor.  For example, the arrangement $4 \cdot 6 \cdot 4$ can be written as $(2, 2, |, 2, 3, |, 2, 2)$ or $(2, 2, |, 3, 2, |, 2, 2)$ .  Precisely, we double count any case with $6$ as a factor, triple count any case with $12$ , quadruple count any case with $24$ , etc.
Now, consider all cases where $3$ must be grouped with at least one $2$ .  This can be expressed in the same "line" format as $(2, 2, 2, 2, 6)$ , where dividers can again be placed to group divisors.  In this case, there are $5$ ways to order divisors, and $2^4$ ways to place dividers, so we have an $5 \cdot 2^4$ possible sequences for this case.  Notice that in this format, we double count cases where $12$ is a factor, we triple count cases where $24$ is a factor, etc.  Precisely, for any case counted $n$ times in the first step, it is counted $n - 1$ times in this step.  Thus, if we subtract, we count each case exactly once.
So, we get:
$6 \cdot  2^5 - 5 \cdot 2^4 = \boxed{112}$ . ~ hdai1122

Solution:
First we factor $32$ into $m$ numbers $g_1, \cdots, g_m$ where $g_i>1,i=1,\ldots,m$ . By applying stars and bars there are $\binom{5-1}{m-1}$ ways. Then we can either insert $3$ into each of the $m+1$ spaces between (or beyond) $g_i$ 's, or multiply it to one of the $g_i$ 's, a total of $2m+1$ ways. Hence the answer to the problem is
\[\sum_{m=1}^5 (2m+1)\binom{5-1}{m-1} = \sum_{n=0}^4 (2n+3) \binom{4}{n} = 8\sum_{n=0}^4 \frac{n}{4} \binom{4}{n} + 3 \sum_{n=0}^4 \binom{4}{n} = 8 \sum_{n=0}^4 \binom{3}{n-1} + 3\sum_{n=0}^4 \binom{4}{n} = 8 \cdot 2^3 + 3\cdot 2^4 = \boxed{112}.\]

Solution:
Note that $96 = 2^5 \cdot 3$ $D(n)$ depends on dividing $2^5$ into different terms, which is the integer partition of $5$
Divide $96$ into $1$ term:
There is only one way. $\underline{\textbf{1}}$
Divide $96$ into $2$ terms:
$2 + 8 = \underline{\textbf{10}}$
Divide $96$ into $3$ terms:
$12 + 18 = \underline{\textbf{30}}$
Divide $96$ into $4$ terms:
$24 + 16 = \underline{\textbf{40}}$
Divide $96$ into $5$ terms:
When dividing $5$ into $5$ parts there are $2$ cases.
$20 + 5 = \underline{\textbf{25}}$
Divide $96$ into $6$ terms:
$5 = 1 + 1 + 1 + 1 + 1$ . The number of arrangements of $(2, 2, 2, 2, 2, 3)$ is $\frac{6!}{5!} = \underline{\textbf{6}}$
$1 + 10 + 30 + 40 + 25 + 6 = \boxed{112}$

Solution:
Consider how we partition the factors of $96 = 2^5\cdot 3$ . For each $k$ , there are two cases. Either we can put the $2$ s into $k$ nonzero parts, so that the $3$ shares a partition with some $2$ s, which can be done in $k\binom{5-(k-1)+(k-1)-1}{k-2} = k\binom{4}{k-2}$ ways, or we can put the $2$ s into $k-1$ nonzero parts and put the $3$ in its own partition, which can be done in $k\binom{5-k+k-1}{k-1} = k\binom{4}{k-1}$ ways. Summing over all $k$ , we have \[\sum_{k=1}^6 \binom{4}{k-2} + \sum_{k=1}^6 \binom{4}{k-1}.\]
But $\sum_{k=1}^6 \binom{4}{k-2} = 2\binom{4}{0}+3\binom{4}{1} + \dots + 6\binom{4}{4} = 4 \cdot \sum_{i=0}^4 \binom{4}{i} = 4 \cdot 2^4 = 64$ . Similarly, $\sum_{k=1}^6 \binom{4}{k-1} = 3\cdot 2^4 = 48$ . So our answer is $64+48 = \boxed{112}$


Answer: 112.0

Solution: Ignore the $3$ first and first count $2^{x_1+x_2+...+x^n}=32$ which $x_1+x_2+...+x_n=5$ . This implies that $n$ is less than or equal to $5$ . Now, we can see that $3$ can lie between the two $x_i, x_{i+1}$ , or contribute to one of them. This gives $2k+1$ if $x_1+...+x_k=5$ . Now, just sum up gives $\binom{5-1}{5-1}\cdot 11+\binom{5-1}{4-1}\cdot 9+\binom{5-1}{3-1}\cdot 7+\binom{5-1}{2-1}\cdot 5+\binom{5-1}{1-1}\cdot 3=\boxed{112}$


Answer: 400.0

Let $E(n)$ denote the sum of the even digits of $n$ . For example, $E(5681) = 6+8 = 14$ .  Find $E(1)+E(2)+E(3)+\cdots+E(100)$
$\text{(A)}\ 200\qquad\text{(B)}\ 360\qquad\text{(C)}\ 400\qquad\text{(D)}\ 900\qquad\text{(E)}\ 2250$

Solution:
The problem is asking for the sum of all the even digits in the numbers $1$ to $100$ .  We can remove $100$ from the list, add $00$ to the list, and tack on some leading zeros to the single digit numbers without changing the sum of the even digits.  This gives the list:
$00, 01, 02, 03, ..., 10, 11, ..., 98, 99$
There are $2\cdot 100 = 200$ digits on that list, and each digit appears $\frac{200}{10} = 20$ times.
Thus, each even digit appears $20$ times, and the sum of all the even digits is $0 \cdot 20 + 2\cdot 20 + 4\cdot 20 + 6\cdot 20 + 8\cdot 20 = (0 + 2 + 4 + 6 + 8)\cdot 20 = 400$ , and the correct answer is $\boxed{400}$


Answer: 417.0

Let $EFGH$ $EFDC$ , and $EHBC$ be three adjacent square faces of a cube , for which $EC = 8$ , and let $A$ be the eighth vertex of the cube. Let $I$ $J$ , and $K$ , be the points on $\overline{EF}$ $\overline{EH}$ , and $\overline{EC}$ , respectively, so that $EI = EJ = EK = 2$ . A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\overline{AE}$ , and containing the edges, $\overline{IJ}$ $\overline{JK}$ , and $\overline{KI}$ . The surface area of $S$ , including the walls of the tunnel, is $m + n\sqrt {p}$ , where $m$ $n$ , and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$

Solution:
Set the coordinate system so that vertex $E$ , where the drilling starts, is at $(8,8,8)$ . Using a little visualization (involving some similar triangles , because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining $(1,0,0)$ to $(2,2,0)$ , and $(0,1,0)$ to $(2,2,0)$ , and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), $S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)$ , and the other two faces of the tunnel are congruent to this shape.
Observe that this shape is made up of two congruent trapezoids each with height $\sqrt {2}$ and bases $7\sqrt {3}$ and $6\sqrt {3}$ . Together they make up an area of $\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}$ . The total area of the tunnel is then $3\cdot13\sqrt {6} = 39\sqrt {6}$ . Around the corner $E$ we're missing an area of $6$ , the same goes for the corner opposite $E$ . So the outside area is $6\cdot 64 - 2\cdot 6 = 372$ . Thus the the total surface area is $372 + 39\sqrt {6}$ , and the answer is $372 + 39 + 6 = \boxed{417}$


Answer: 275.0

Let $F(z)=\dfrac{z+i}{z-i}$ for all complex numbers $z\neq i$ , and let $z_n=F(z_{n-1})$ for all positive integers $n$ . Given that $z_0=\dfrac{1}{137}+i$ and $z_{2002}=a+bi$ , where $a$ and $b$ are real numbers, find $a+b$

Solution:
Iterating $F$ we get:
\begin{align*} F(z) &= \frac{z+i}{z-i}\\ F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\ &= \frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \frac{(z+1)(2i)}{(z-1)(2)}= \frac{z+1}{z-1}i\\ F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. \end{align*}
From this, it follows that $z_{k+3} = z_k$ , for all $k$ . Thus $z_{2002} = z_{3\cdot 667+1} = z_1 = \frac{z_0+i}{z_0-i} = \frac{(\frac{1}{137}+i)+i}{(\frac{1}{137}+i)-i}= \frac{\frac{1}{137}+2i}{\frac{1}{137}}= 1+274i.$
Thus $a+b = 1+274 = \boxed{275}$


Answer: 528.0

Let $G$ be the set of polynomials of the form \[P(z)=z^n+c_{n-1}z^{n-1}+\cdots+c_2z^2+c_1z+50,\] where $c_1,c_2,\cdots, c_{n-1}$ are integers and $P(z)$ has distinct roots of the form $a+ib$ with $a$ and $b$ integers. How many polynomials are in $G$
$\textbf{(A)}\ 288\qquad\textbf{(B)}\ 528\qquad\textbf{(C)}\ 576\qquad\textbf{(D)}\ 992\qquad\textbf{(E)}\ 1056$

Solution:
If we factor into irreducible polynomials (in $\mathbb{Q}[x]$ ), each factor $f_i$ has exponent $1$ in the factorization and degree at most $2$ (since the $a+bi$ with $b\ne0$ come in conjugate pairs with product $a^2+b^2$ ). Clearly we want the product of constant terms of these polynomials to equal $50$ ; for $d\mid 50$ , let $f(d)$ be the number of permitted $f_i$ with constant term $d$ . It's easy to compute $f(1)=2$ $f(2)=3$ $f(5)=5$ $f(10)=5$ $f(25)=6$ $f(50)=7$ , and obviously $f(d) = 1$ for negative $d\mid 50$
Note that by the distinctness condition, the only constant terms $d$ that can be repeated are those with $d^2\mid 50$ and $f(d)>1$ , i.e. $+1$ and $+5$ . Also, the $+1$ s don't affect the product, so we can simply count the number of polynomials with no constant terms of $+1$ and multiply by $2^{f(1)} = 4$ at the end.
We do casework on the (unique) even constant term $d\in\{\pm2,\pm10,\pm50\}$ in our product. For convenience, let $F(d)$ be the number of ways to get a product of $50/d$ without using $\pm 1$ (so only using $\pm5,\pm25$ ) and recall $f(-1) = 1$ ; then our final answer will be $2^{f(1)}\sum_{d\in\{2,10,50\}}(f(-d)+f(d))(F(-d)+F(d))$ . It's easy to compute $F(-50)=0$ $F(50)=1$ $F(-10)=f(5)=5$ $F(10)=f(-5)=1$ $F(-2)=f(-25)+f(-5)f(5)=6$ $F(2)=f(25)+\binom{f(5)}{2}=16$ , so we get \[4 [ (1+3)(6+16) + (1+5)(1+5) + (1+7)(0+1) ] = 4[132] = \boxed{528}\]

Solution:
By Vieta's formula $50$ is the product of all $n$ roots. As the roots are all in the form $a + bi$ , there must exist a conjugate $a-bi$ for each root.
$(a+bi)(a-bi) = a^2 + b^2$
$50 = 2 \cdot 5^2$
If $a \neq b \neq 0$ , the roots can be $a \pm bi$ $-a \pm bi$ $b \pm ai$ $-b \pm ai$ , totaling $4$ pairs of roots.
If $a = b$ , the roots can be $a \pm ai$ $-a \pm ai$ , totaling $2$ pairs of roots.
If $a \neq b$ $b = 0$ , the roots can be $\pm a$ $\pm ai$ , totaling $2$ pairs of roots.
\begin{align*} 2 \cdot 25 &= (1^2+1^2)5^2 &: 2 \cdot 2 = 4\\ 2 \cdot 25 &= 2 \cdot 5^2 &: 2 \cdot 2 = 4\\ 2 \cdot 25 &= (1^2+1^2) \cdot (3^2+4^2) &: 2 \cdot 4 = 8\\ 2 \cdot 25 &= 2 \cdot (3^2+4^2) &: 2 \cdot 4 = 8 \end{align*}
\begin{align*} 10 \cdot 5 &= (1^2+3^2)(1^2+2^2) &&: 4 \cdot 4 = 16\\ 10 \cdot 5 &= 10 \cdot (1^2+2^2) &&: 2 \cdot 4 = 8\\ 10 \cdot 5 &= (1^2+3^2) \cdot 5 &&: 4 \cdot 2 = 8\\ 10 \cdot 5 &= 10 \cdot 5 &&: 2 \cdot 2 = 4\\ \end{align*}
\begin{align*} 2 \cdot 5  \cdot 5&= (1^2+1^2)(1^2+2^2)(1^2+2^2) &&: 2 \cdot 4 \cdot 4 = 32\\ 2 \cdot 5  \cdot 5&= 2 \cdot (1^2+2^2)(1^2+2^2) &&: 2 \cdot 4 \cdot 4 = 32\\ 2 \cdot 5  \cdot 5&= 2 \cdot 5 \cdot (1^2+2^2) &&: 2 \cdot 2 \cdot 4 = 16\\ 2 \cdot 5  \cdot 5&= 2 \cdot 5 \cdot 5 &&: 2 \cdot 2 \cdot 2 = 8\\ 2 \cdot 5  \cdot 5&= (1^2+1^2) \cdot 5 \cdot (1^2+2^2) &&: 2 \cdot 2 \cdot 4 = 16\\ 2 \cdot 5  \cdot 5&= (1^2+1^2) \cdot 5 \cdot 5 &&: 2 \cdot 2 \cdot 2 = 8\\ \end{align*}
\begin{align*} (1^2+7^2) &: 4\\ (5^2+5^2) &: 2\\ 50 &: 2 \end{align*}
$4+4+8+8+16+8+8+4+32+32+16+8+16+8+4+2+2 = 176$
For each case $1^2$ can be added, yielding 2 more cases $(\pm 1, \pm i)$ $176 \cdot 3 = \boxed{528}$


Answer: 528.0

Solution:
Disregard sign; we can tack on $x-1$ if the product ends up being negative.
$1: \pm i,-1$ (2) (1 is not included)
$2: \pm 2, \pm 1\pm i$ (4)
$5: \pm 2\pm i, \pm 1\pm 2i, \pm 5$ (6)
$10: \pm 3\pm i, \pm 1\pm 3i, \pm 10$ (6)
$25: \pm 25, \pm 3\pm 4i, \pm 4\pm 3i, \pm 5i$ (7)
$50: \pm 50, \pm 1\pm 7i, \pm7\pm i, \pm 5\pm 5i$ (8)
Our answer is $2^2\left(4\cdot\binom{6}{2}+6\cdot 6+4\cdot 7+8\right)=\boxed{528.}$


Answer: 36.5

Let $K$ be the measure of the area bounded by the $x$ -axis, the line $x=8$ , and the curve defined by
\[f={(x,y)\quad |\quad y=x \text{  when  } 0 \le x \le 5, y=2x-5 \text{  when  } 5 \le x \le 8}.\]
Then $K$ is:
$\text{(A) } 21.5\quad \text{(B) } 36.4\quad \text{(C) } 36.5\quad \text{(D) } 44\quad \text{(E) less than 44 but arbitrarily close to it}$

Solution: The shape can be divided into a triangle and a trapezoid.  For the triangle, the base is $5$ and the height is $5$ , so the area is $\frac{5 \cdot 5}{2} = \frac{25}{2}$ .  For the trapezoid, the two bases are $5$ and $11$ and the height is $3$ , so the area is $\frac{3(5+11)}{2} = 24$ .  Thus, the total area is $\frac{25}{2} + 24 = \frac{73}{2} = \boxed{36.5}$


Answer: 5.0

Let $K$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_n$ such that $n$ is a positive integer less than or equal to $10$ , each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$ , and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$ , inclusive. For example, $\{\}$ $\{5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 6, 7, 9\}$ is one such sequence, with $n = 5$ .What is the remainder when $K$ is divided by $10$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution:
Consider any sequence with $n$ terms.  Every 10 number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot… and appear the first time in the $n$ th spot, which means every number has $(n+1)$ choices to show up in the sequence. Consequently, for each sequence with length $n$ , there are $(n+1)^{10}$ possible ways.
Thus, the desired value is $\sum_{i=1}^{10}(i+1)^{10}\equiv \boxed{5}$

Solution:
Let $f(x,\ell)$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_\ell$ such that each $A_i$ is a subset of $\{1, 2,\dots,x\}$ , and $A_i$ is a subset of $A_{i+1}$ for $i=1$ $2\dots$ $\ell-1$ . Then $f(x,1)=2^x$ and $f(0,\ell)=1$
If $\ell\ge2$ and $x\ge1$ , we need to get a recursive formula for $f(x,\ell)$ : If $|A_1|=i$ , then $A_1$ has $\text C_x^i$ possibilities, and the subsequence $\{A_i\}_{2\le i\le\ell}$ has $f(x-i,\ell-1)$ possibilities. Hence \[f(x,\ell)=\sum_{i=0}^x\text C_x^if(x-i,\ell-1).\] By applying this formula and only considering  modulo $10$ , we get $f(1,2)=3$ $f(1,3)=4$ $f(1,4)=5$ $f(1,5)=6$ $f(1,6)=7$ $f(1,7)=8$ $f(1,8)=9$ $f(1,9)=0$ $f(1,10)=1$ $f(2,2)=9$ $f(2,3)=6$ $f(2,4)=5$ $f(2,5)=6$ $f(2,6)=9$ $f(2,7)=4$ $f(2,8)=1$ $f(2,9)=0$ $f(2,10)=1$ $f(3,2)=7$ $f(3,3)=4$ $f(3,4)=5$ $f(3,5)=6$ $f(3,6)=3$ $f(3,7)=2$ $f(3,8)=9$ $f(3,9)=0$ $f(3,10)=1$ $f(4,2)=1$ $f(4,3)=6$ $f(4,4)=5$ $f(4,5)=6$ $f(4,6)=1$ $f(4,7)=6$ $f(4,8)=1$ $f(4,9)=0$ $f(4,10)=1$ $f(5,2)=3$ $f(5,3)=4$ $f(5,4)=5$ $f(5,5)=6$ $f(5,6)=7$ $f(5,7)=8$ $f(5,8)=9$ $f(5,9)=0$ $f(5,10)=1$ $f(6,2)=9$ $f(6,3)=6$ $f(6,4)=5$ $f(6,5)=6$ $f(6,6)=9$ $f(6,7)=4$ $f(6,8)=1$ $f(6,9)=0$ $f(6,10)=1$ $f(7,2)=7$ $f(7,3)=4$ $f(7,4)=5$ $f(7,5)=6$ $f(7,6)=3$ $f(7,7)=2$ $f(7,8)=9$ $f(7,9)=0$ $f(7,10)=1$ $f(8,2)=1$ $f(8,3)=6$ $f(8,4)=5$ $f(8,5)=6$ $f(8,6)=1$ $f(8,7)=6$ $f(8,8)=1$ $f(8,9)=0$ $f(8,10)=1$ $f(9,2)=3$ $f(9,3)=4$ $f(9,4)=5$ $f(9,5)=6$ $f(9,6)=7$ $f(9,7)=8$ $f(9,8)=9$ $f(9,9)=0$ $f(9,10)=1$ $f(10,2)=9$ $f(10,3)=6$ $f(10,4)=5$ $f(10,5)=6$ $f(10,6)=9$ $f(10,7)=4$ $f(10,8)=1$ $f(10,9)=0$ $f(10,10)=1$
Lastly, we get $K\equiv\textstyle\sum\limits_{i=1}^{10}f(10,i)\equiv\boxed{5}$ .
~Quantum-Phantom

Solution:
Seeing that all the answers are different modulus 5, and that 10 is divisible by 5, we cheese this problem.
Let $A_1, A_2, \cdots, A_n$ be one sequence satisfying the constraints of the problem. Let $b_1, b_2, \cdots, b_n, b_{n+1}$ be the sequence of nonnegative integers such that $A_k$ has $b_k$ elements for all $1\leq{}k\leq{}n$ , and $b_{n+1}=10$ . Note that we can generate the number of valid sequences of $A$ by first generating all sequences of $b$ such that $b_i\leq{}b_{i+1}$ for all $1\leq{}i<n$ , then choosing the elements from $A_{k+1}$ that we keep in $A_k$ , given the sequence of $b$ as the restraint for the number of elements. For each sequence $b_1, b_2, \cdots{}, b_{n+1}$ , there are $\prod_{i=2}^{n+1}\binom{b_i}{b_{i-1}}$ corresponding sequences for $A$ . Now, consider two cases - either all terms in $b$ are either 0, 5, or 10, or there is at least one term in $b$ that is neither 0, 5 nor 10. In the second case, consider the last term in $b$ that is not 10 or 5, say $b_m$ . However, that implies $b_{m+1}=10$ , and so the number of corresponding sequences of $A$ is $\binom{10}{b_m}\cdot{}$ something or $\binom{5}{b_m}\cdot$ something, which is always a multiple of $5$ . Therefore, we only need to consider sequences of $b$ where each term is $0$ $5$ or $10$ . If all terms in $b$ are 0 or 10, then for each $1\leq{}n\leq{}10$ there are $n+1$ sequences of $b$ (since there are $n+1$ places to turn from $0$ to $10$ ), for a total of $2+3+\cdots+11=65\equiv0$ (mod 5). If there exists at least one term $b_k=5$ , then we use stars and bars to count the number of sequences of $b$ , and each sequence of $b$ corresponds to $\binom{10}{5}$ sequences of $A$ . For each $n$ , we must have at least one term of $5$ . After that, there are $n-1$ stars and $2$ bars (separating $0$ to $5$ and $5$ to $10$ ), so that is $\binom{n+1}{2}$ sequences of $b$ . So the sum is $\binom{2}{2}+\binom{3}{2}+\cdots+\binom{11}{2}=\binom{12}{3}\equiv0$ (mod 5). Therefore, the answer is 0 mod 5, and it must be $\boxed{5}$

Solution:
We observe that in each sequence, if element $e \in A_i$ , then $e \in A_j$ for all $j \geq i$ .
Therefore, to determine a sequence with a fixed length $n$ , we only need to determine the first set $A_i$ that each element in $\left\{ 1, 2, \cdots , 10 \right\}$ is inserted into, or an element is never inserted into any subset.
We have \begin{align*} K & = \sum_{n = 1}^{10} \left( n + 1 \right)^{10} \\ & = \sum_{m = 2}^{11} m^{10} . \end{align*}
Recalling or noticing that $x^n \equiv x^{n \mod 4} \pmod {10}$ , then,
Modulo 10, we have \begin{align*} K & \equiv \sum_{m = 2}^{11} m^2 \\ & \equiv \sum_{m = 1}^{11} m^2 - 1^2 \\ & \equiv \frac{11 \cdot \left( 11 + 1 \right) \left( 2 \cdot 11 + 1 \right)}{6} - 1\\ & \equiv 505 \\ & \equiv \boxed{5}


Answer: 150.0

Let $K$ be the product of all factors $(b-a)$ (not necessarily distinct) where $a$ and $b$ are integers satisfying $1\le a < b \le 20$ . Find the greatest positive integer $n$ such that $2^n$ divides $K$

Solution:
In general, there are $20-n$ pairs of integers $(a, b)$ that differ by $n$ because we can let $b$ be any integer from $n+1$ to $20$ and set $a$ equal to $b-n$ . Thus, the product is $(1^{19})(2^{18})\cdots(19^1)$ (or alternatively, $19! \cdot 18! \cdots 1!$ .)
When we count the number of factors of $2$ , we have 4 groups, factors that are divisible by $2$ at least once, twice, three times and four times.
Summing these give an answer of $\boxed{150}$


Answer: 31.0

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 $\alpha+\beta=-\frac{123}{13}+\frac{526}{13}=\frac{403}{13}=\boxed{031}$

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: 674.0

Solution: We calculate more terms: \[1,3,4,7,11,18,\ldots.\] We find a pattern: if $n$ is a multiple of $3$ , then the term is even, or else it is odd. There are $\left\lfloor \frac{2023}{3} \right\rfloor =\boxed{674}$ multiples of $3$ from $1$ to $2023$

Solution:
Like in the other solution, we find a pattern, except in a more rigorous way.
Since we start with $1$ and $3$ , the next term is $4$
We start with odd, then odd, then (the sum of odd and odd) even, (the sum of odd and even) odd, and so on. Basically the pattern goes: odd, odd, even, odd odd, even, odd, odd even…
When we take $\frac{2023}{3}$ we get $674$ with a remainder of one. So we have $674$ full cycles, and an extra odd at the end.
Therefore, there are $\boxed{674}$ evens.


Answer: 74.0

Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40.$ What is the value of $\frac{N}{M}?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 37 \qquad\textbf{(D)}\ 74 \qquad\textbf{(E)}\ 2886$

Solution: By the definition of least common mutiple, we take the greatest powers of the prime numbers of the prime factorization of all the numbers, that we are taking the $\text{lcm}$ of. In this case, \[M = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29.\] Now, using the same logic, we find that \[N = M \cdot 2 \cdot 37,\] because we have an extra power of $2$ and an extra power of $37.$ Thus, $\frac{N}{M} = 2\cdot 37 = \boxed{74}.$


Answer: 13.0

Let $M$ be the midpoint of $\overline{AB}$ in regular tetrahedron $ABCD$ . What is $\cos(\angle CMD)$
$\textbf{(A) } \frac14 \qquad \textbf{(B) } \frac13 \qquad \textbf{(C) } \frac25 \qquad \textbf{(D) } \frac12 \qquad \textbf{(E) } \frac{\sqrt{3}}{2}$

Solution:
Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $MC=MD=\sqrt3.$
Let $O$ be the center of $\triangle ABD,$ so $\overline{CO}\perp\overline{MOD}.$ Note that $MO=\frac13 MD=\frac{\sqrt{3}}{3}.$
In right $\triangle CMO,$ we have \[\cos(\angle CMD)=\frac{MO}{MC}=\boxed{13}.\] ~MRENTHUSIASM

Solution:
Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $CM=DM=\sqrt3.$
By the Law of Cosines, \[\cos(\angle CMD) = \frac{CM^2 + DM^2 - CD^2}{2(CM)(DM)} = \boxed{13}.\] ~jamesl123456

Solution: As done above, let the edge-length equal $2$ (usually better than $1$ because we can avoid fractions when dropping altitudes). Notice that the triangle stated in the question has two side-lengths that are the altitudes of two equilateral triangles. By dropping the equilateral triangles' altitude and using $30^{\circ}$ $60^{\circ}$ $90^{\circ}$ properties, we find that the other two sides are equal to $\sqrt{3}$ . Now by dropping the main triangle's altitude, we see it equals $\sqrt{2}$ from the Pythagorean Theorem. we can use the Double Angle Identities for Cosine. Doing this, we obtain \[\cos(\angle CMD) = \frac{2}{3} - \frac13 = \boxed{13}.\] ~Misclicked


Answer: 73.0

Let $M_n$ be the $n \times n$ matrix with entries as follows: for $1 \le i \le n$ $m_{i,i} = 10$ ; for $1 \le i \le n - 1$ $m_{i+1,i} = m_{i,i+1} = 3$ ; all other entries in $M_n$ are zero. Let $D_n$ be the determinant of matrix $M_n$ . Then $\sum_{n=1}^{\infty} \frac{1}{8D_n+1}$ can be represented as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$
Note: The determinant of the $1 \times 1$ matrix $[a]$ is $a$ , and the determinant of the $2 \times 2$ matrix $\left[ {\begin{array}{cc}  a & b  \\  c & d  \\  \end{array} } \right] = ad - bc$ ; for $n \ge 2$ , the determinant of an $n \times n$ matrix with first row or first column $a_1$ $a_2$ $a_3$ $\dots$ $a_n$ is equal to $a_1C_1 - a_2C_2 + a_3C_3 - \dots + (-1)^{n+1}a_nC_n$ , where $C_i$ is the determinant of the $(n - 1) \times (n - 1)$ matrix formed by eliminating the row and column containing $a_i$

Solution:
\[D_{1}=\begin{vmatrix}  10    \end{vmatrix} = 10, \quad D_{2}=\begin{vmatrix} 10 & 3  \\  3 & 10  \\ \end{vmatrix} =(10)(10) - (3)(3) = 91, \quad D_{3}=\begin{vmatrix}  10 & 3 & 0  \\  3 & 10 & 3  \\  0 & 3 & 10  \\  \end{vmatrix}.\] Using the expansionary/recursive definition of determinants (also stated in the problem):
$D_{3}=\left| {\begin{array}{ccc}  10 & 3 & 0  \\  3 & 10 & 3  \\  0 & 3 & 10  \\  \end{array} } \right|=10\left| {\begin{array}{cc}  10 & 3  \\  3 & 10  \\  \end{array} } \right| - 3\left| {\begin{array}{cc}  3 & 3  \\  0 & 10  \\  \end{array} } \right| + 0\left| {\begin{array}{cc}  3 & 10  \\  0 & 3  \\  \end{array} } \right| = 10D_{2} - 9D_{1} = 820$
This pattern repeats because the first element in the first row of $M_{n}$ is always 10, the second element is always 3, and the rest are always 0. The ten element directly expands to $10D_{n-1}$ . The three element expands to 3 times the determinant of the the matrix formed from omitting the second column and first row from the original matrix. Call this matrix $X_{n}$ $X_{n}$ has a first column entirely of zeros except for the first element, which is a three. A property of matrices is that the determinant can be expanded over the rows instead of the columns (still using the recursive definition as given in the problem), and the determinant found will still be the same. Thus, expanding over this first column yields $3D_{n-2} + 0(\text{other things})=3D_{n-2}$ . Thus, the $3 \det(X_{n})$ expression turns into $9D_{n-2}$ . Thus, the equation $D_{n}=10D_{n-1}-9D_{n-2}$ holds for all n > 2.
This equation can be rewritten as $D_{n}=10(D_{n-1}-D_{n-2}) + D_{n-2}$ . This version of the equation involves the difference of successive terms of a recursive sequence. Calculating $D_{0}$ backwards from the recursive formula and $D_{4}$ from the formula yields $D_{0}=1, D_{4}=7381$ . Examining the differences between successive terms, a pattern emerges. $D_{0}=1=9^{0}$ $D_{1}-D_{0}=10-1=9=9^{1}$ $D_{2}-D_{1}=91-10=81=9^{2}$ $D_{3}-D_{2}=820-91=729=9^{3}$ $D_{4}-D_{3}=7381-820=6561=9^{4}$ .
Thus, $D_{n}=D_{0} + 9^{1}+9^{2}+ . . . +9^{n}=\sum_{i=0}^{n}9^{i}=\frac{(1)(9^{n+1}-1)}{9-1}=\frac{9^{n+1}-1}{8}$
Thus, the desired sum is $\sum_{n=1}^{\infty}\frac{1}{8\frac{9^{n+1}-1}{8}+1}=\sum_{n=1}^{\infty}\frac{1}{9^{n+1}-1+1} = \sum_{n=1}^{\infty}\frac{1}{9^{n+1}}$
This is an infinite geometric series with first term $\frac{1}{81}$ and common ratio $\frac{1}{9}$ . Thus, the sum is $\frac{\frac{1}{81}}{1-\frac{1}{9}}=\frac{\frac{1}{81}}{\frac{8}{9}}=\frac{9}{(81)(8)}=\frac{1}{(9)(8)}=\frac{1}{72}$
Thus, $p + q = 1 + 72 = \boxed{073}$


Answer: 100.0

Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$

Solution:
Rewriting this sequence with more terms, we have
Factoring this expression yields
Next, we get
Then,
Dividing $10100$ by $1000$ yields a remainder of $\boxed{100}$

Solution:
By observation, we realize that the sequence \[(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2\] alternates every 4 terms. Simplifying, we get \[(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2 = 8a + 12\] , turning $N$ into a arithmetic sequence with 25 terms, them being $1, 5, 9, \dots ,97$ , as the series $8a + 12$ alternates every 4 terms.
Applying the sum of arithmetic sequence formula, we get
So the answer would be \[\frac{10100}{1000} = \boxed{100}\]


Answer: 477.0

Let $N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil  - \lfloor \log_{\sqrt{2}} k \rfloor )$
Find the remainder when $N$ is divided by 1000.  ( $\lfloor{k}\rfloor$ is the greatest integer less than or equal to $k$ , and $\lceil{k}\rceil$ is the least integer greater than or equal to $k$ .)

Solution:
The ceiling of a number minus the floor of a number is either equal to zero (if the number is an integer ); otherwise, it is equal to 1. Thus, we need to find when or not $\log_{\sqrt{2}} k$ is an integer.
The change of base formula shows that $\frac{\log k}{\log \sqrt{2}} = \frac{2 \log k}{\log 2}$ . For the $\log 2$ term to cancel out, $k$ is a power of $2$ . Thus, $N$ is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from $2^0 = 1$ to $2^9 = 512$
The formula for the sum of an arithmetic sequence and the sum of a geometric sequence yields that our answer is $\left[\frac{(1000 + 1)(1000)}{2} - (1 + 2 + 2^2 + \ldots + 2^9)\right] \mod{1000}$
Simplifying, we get $\left[1000\left(\frac{1000+1}{2}\right) -1023\right] \mod{1000} \equiv [500-23] \mod{1000} \equiv 477 \mod{1000}.$ The answer is $\boxed{477}$


Answer: 12.0

Let $N$ be a positive multiple of $5$ . One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\tfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\tfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \tfrac{321}{400}$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$

Solution:
Let $n = \frac{N}{5}$ . Then, consider $5$ blocks of $n$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $N + 1$ positions between the green balls to insert the red ball. Less than $\frac{3}{5}$ of the green balls will be on the same side of the red ball if the red ball is inserted inside the middle block of $n$ balls, and there are $n - 1$ positions where this happens. Thus, $P(N) = 1 - \frac{n - 1}{N + 1} = \frac{4n + 2}{5n + 1}$ , so
\[P(N) = \frac{4n + 2}{5n + 1} < \frac{321}{400}.\]
Multiplying both sides of the inequality by $400(5n+1)$ , we have
\[400(4n+2)<321(5n+1),\]
and by the distributive property,
\[1600n+800<1605n+321.\]
Subtracting $1600n+321$ on both sides of the inequality gives us
\[479<5n.\]
Therefore, $N=5n>479$ , so the least possible value of $N$ is $480$ . The sum of the digits of $480$ is $\boxed{12}$

Solution:
Let $N=5$ $P(N)=1$ (Given)
Let $N=10$ $P(N)=\frac{10}{11}$
Let $N=15$ $P(N)=\frac{14}{16}$
Notice that the fraction can be written as $1-\frac{\frac{N}{5}-1}{N+1}$
Now it's quite simple to write the inequality as $1-\frac{\frac{N}{5}-1}{N+1}<\frac{321}{400}$
We can subtract $1$ on both sides to obtain $-\frac{\frac{N}{5}-1}{N+1}<-\frac{79}{400}$
Dividing both sides by $-1$ , we derive $\frac{\frac{N}{5}-1}{N+1}>\frac{79}{400}$ . (Switch the inequality sign when dividing by $-1$
We then cross multiply to get $80N - 400 > 79N + 79$
Finally we get $N > 479$
To achieve $N = 480$
So the sum of the digits of $N$ $\boxed{12}$

Solution:
We are trying to find the number of places to put the red ball, such that $\frac{3}{5}$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $N$ : Trying a few values, we see that the ball "works" in places $1$ to $\frac{2}{5}N + 1$ and spaces $\frac{3}{5}N+1$ to $N+1$ . This is a total of $\frac{4}{5}N + 2$ spaces, over a total possible $N + 1$ places to put the ball. So:
$\frac{\frac{4}{5}N + 2}{N+1} = \frac{321}{400} \rightarrow N = 479.$ And we know that the next value is what we are looking for, so $N+1 = 480$ , and the sum of its digits is $\boxed{12}$


Answer: 12.0

Solution:
Let $n = \frac{N}{5}$ . Then, consider $5$ blocks of $n$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $N + 1$ positions between the green balls to insert the red ball. Less than $\frac{3}{5}$ of the green balls will be on the same side of the red ball if the red ball is inserted inside the middle block of $n$ balls, and there are $n - 1$ positions where this happens. Thus, $P(N) = 1 - \frac{n - 1}{N + 1} = \frac{4n + 2}{5n + 1}$ , so
\[P(N) = \frac{4n + 2}{5n + 1} < \frac{321}{400}.\]
Multiplying both sides of the inequality by $400(5n+1)$ , we have
\[400(4n+2)<321(5n+1),\]
and by the distributive property,
\[1600n+800<1605n+321.\]
Subtracting $1600n+321$ on both sides of the inequality gives us
\[479<5n.\]
Therefore, $N=5n>479$ , so the least possible value of $N$ is $480$ . The sum of the digits of $480$ is $\boxed{12}$

Solution:
Let $N=5$ $P(N)=1$ (Given)
Let $N=10$ $P(N)=\frac{10}{11}$
Let $N=15$ $P(N)=\frac{14}{16}$
Notice that the fraction can be written as $1-\frac{\frac{N}{5}-1}{N+1}$
Now it's quite simple to write the inequality as $1-\frac{\frac{N}{5}-1}{N+1}<\frac{321}{400}$
We can subtract $1$ on both sides to obtain $-\frac{\frac{N}{5}-1}{N+1}<-\frac{79}{400}$
Dividing both sides by $-1$ , we derive $\frac{\frac{N}{5}-1}{N+1}>\frac{79}{400}$ . (Switch the inequality sign when dividing by $-1$
We then cross multiply to get $80N - 400 > 79N + 79$
Finally we get $N > 479$
To achieve $N = 480$
So the sum of the digits of $N$ $\boxed{12}$

Solution:
We are trying to find the number of places to put the red ball, such that $\frac{3}{5}$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $N$ : Trying a few values, we see that the ball "works" in places $1$ to $\frac{2}{5}N + 1$ and spaces $\frac{3}{5}N+1$ to $N+1$ . This is a total of $\frac{4}{5}N + 2$ spaces, over a total possible $N + 1$ places to put the ball. So:
$\frac{\frac{4}{5}N + 2}{N+1} = \frac{321}{400} \rightarrow N = 479.$ And we know that the next value is what we are looking for, so $N+1 = 480$ , and the sum of its digits is $\boxed{12}$


Answer: 18.0

Let $N$ be the greatest five-digit number whose digits have a product of $120$ . What is the sum of the digits of $N$
$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C) }17\qquad\textbf{(D) }18\qquad\textbf{(E) }20$

Solution: If we start off with the first digit, we know that it can't be $9$ since $9$ is not a factor of $120$ . We go down to the digit $8$ , which does work since it is a factor of $120$ . Now, we have to know what digits will take up the remaining four spots. To find this result, just divide $\frac{120}{8}=15$ . The next place can be $5$ , as it is the largest factor, aside from $15$ . Consequently, our next three values will be $3,1$ and $1$ if we use the same logic. Therefore, our five-digit number is $85311$ , so the sum is $8+5+3+1+1=18\implies \boxed{18}$

Solution:
120 is 5!, so we have 5,4,3,2,1. Now look for the largest digit which you multiple numbers.
$(5)(4)(3)(2)(1) = 120$ Making the greatest integer,
$(5)(4 \cdot 2)(3)\left(\frac{2}{2}\right)(1)$ $= (5)(8)(3)(1)(1) =120$
8 is the largest value and will go in the front.
We can express the number as $85311$
$8+5+3+1+1=\boxed{18}$


Answer: 699.0

Let $N$ be the greatest four-digit positive integer with the property that whenever one of its digits is changed to $1$ , the resulting number is divisible by $7$ . Let $Q$ and $R$ be the quotient and remainder, respectively, when $N$ is divided by $1000$ . Find $Q+R$

Solution:
We note that by changing a digit to $1$ for the number $\overline{abcd}$ , we are subtracting the number by either $1000(a-1)$ $100(b-1)$ $10(c-1)$ , or $d-1$ . Thus, $1000a + 100b + 10c + d \equiv 1000(a-1) \equiv 100(b-1) \equiv 10(c-1) \equiv d-1 \pmod{7}$ . We can casework on $a$ backwards, finding the maximum value.
(Note that computing $1000 \equiv 6 \pmod{7}, 100 \equiv 2 \pmod{7}, 10 \equiv 3 \pmod{7}$ greatly simplifies computation).
Applying casework on $a$ , we can eventually obtain a working value of $\overline{abcd} = 5694 \implies \boxed{699}$ . ~akliu

Solution:
Let our four digit number be $abcd$ . Replacing digits with 1, we get the following equations:
$1000+100b+10c+d \equiv 0 \pmod{7}$
$1000a+100+10c+d \equiv 0 \pmod{7}$
$1000a+100b+10+d \equiv 0 \pmod{7}$
$1000a+100b+10c+1 \equiv 0 \pmod{7}$
Reducing, we get
$6+2b+3c+d \equiv 0 \pmod{7}$ $(1)$
$6a+2+3c+d \equiv 0 \pmod{7}$ $(2)$
$6a+2b+3+d \equiv 0 \pmod{7}$ $(3)$
$6a+2b+3c+1 \equiv 0 \pmod{7}$ $(4)$
Subtracting $(2)-(1), (3)-(2), (4)-(3), (4)-(1)$ , we get:
$3a-b \equiv 2 \pmod{7}$
$2b-3c \equiv 6 \pmod{7}$
$3c-d \equiv 2 \pmod{7}$
$6a-d \equiv 5 \pmod{7}$
For the largest 4 digit number, we test values for a starting with 9. When a is 9, b is 4, c is 3, and d is 7.  However, when switching the digits with 1, we quickly notice this doesnt work. Once we get to a=5, we get b=6,c=9,and d=4. Adding 694 with 5, we get $\boxed{699}$ -westwoodmonster

Solution:
Let our four digit number be $abcd$ . Replacing digits with 1, we get the following equations:
$1000+100b+10c+d \equiv 0 \pmod{7}$
$1000a+100+10c+d \equiv 0 \pmod{7}$
$1000a+100b+10+d \equiv 0 \pmod{7}$
$1000a+100b+10c+1 \equiv 0 \pmod{7}$
Add the equations together, we get:
$3000a+300b+30c+3d+1111 \equiv 0 \pmod{7}$
And since the remainder of 1111 divided by 7 is 5, we get:
$3abcd \equiv 2 \pmod{7}$
Which gives us:
$abcd \equiv 3 \pmod{7}$
And since we know that changing each digit into 1 will make abcd divisible by 7, we get that $d-1$ $10c-10$ $100b-100$ , and $1000a-1000$ all have a remainder of 3 when divided by 7. Thus, we get $a=5$ $b=6$ $c=9$ , and $d=4$ . Thus, we get 5694 as abcd, and the answer is $694+5=\boxed{699}$

Solution:
Let our four digit number be $abcd$ . Replacing digits with 1, we get the following equations:
$1000+100b+10c+d \equiv 0 \pmod{7}$
$1000a+100+10c+d \equiv 0 \pmod{7}$
$1000a+100b+10+d \equiv 0 \pmod{7}$
$1000a+100b+10c+1 \equiv 0 \pmod{7}$
Then, we let x, y, z, t be the smallest whole number satisfying the following equations:
$1000a \equiv x \pmod{7}$
$100b \equiv y \pmod{7}$
$10a \equiv z \pmod{7}$
$d \equiv t \pmod{7}$
Since 1000, 100, 10, and 1 have a remainder of 6, 2, 3, and 1 when divided by 7, we can get the equations of:
(1): $6+y+z+t \equiv 0 \pmod{7}$
(2): $x+2+z+t \equiv 0 \pmod{7}$
(3): $x+y+3+t \equiv 0 \pmod{7}$
(4): $x+y+z+1 \equiv 0 \pmod{7}$
Add (1), (2), (3) together, we get:
$2x+2y+2z+3t+11 \equiv 0 \pmod{7}$
We can transform this equation to:
$2(x+y+z+1)+3t+9 \equiv 0 \pmod{7}$
Since, according to (4), $x+y+z+1$ has a remainder of 0 when divided by 7, we get:
$3t+9 \equiv 0 \pmod{7}$
And because t is 0 to 6 due to it being a remainder when divided by 7, we use casework and determine that t is 4.
Using the same methods of simplification, we get that x=2, y=5, and z=6, which means that 1000a, 100b, 10c, and d has a remainder of 2, 5, 6, and 4, respectively. Since a, b, c, and d is the largest possible number between 0 to 9, we use casework to determine the answer is a=5, b=6, c=9, and d=4, which gives us an answer of $5+694=\boxed{699}$


Answer: 120.0

Let $N$ be the greatest integer multiple of 8, no two of whose digits are the same. What is the remainder when $N$ is divided by 1000?

Solution:
We want a number with no digits repeating, so we can only use the digits $0-9$ once in constructing our number. To make the greatest number, we want the greatest digit to occupy the leftmost side and the least digit to occupy the rightmost side. Therefore, the last three digits of the greatest number should be an arrangement of the digits $0,1,2$ . Since the number has to be divisible by 8, the integer formed by the arrangement of $0,1,2$ is also divisible by 8. The only arrangement that works is $120$
Therefore, the remainder when the number is divided by $1000$ is $\boxed{120}$


Answer: 816.0

Let $N$ be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of $N$ forms a perfect square. What are the leftmost three digits of $N$

Solution:
The two-digit perfect squares are $16, 25, 36, 49, 64, 81$ . We try making a sequence starting with each one:
The largest is $81649$ , so our answer is $\boxed{816}$


Answer: 131.0

Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$

Solution:
If $N$ is $22$ percent less than one integer $k$ , then $N=\frac{78}{100}k=\frac{39}{50}k$ . In addition, $N$ is $16$ percent greater than another integer $m$ , so $N=\frac{116}{100}m=\frac{29}{25}m$ . Therefore, $k$ is divisible by $50$ and $m$ is divisible by $25$ . Setting these two equal, we have $\frac{39}{50}k=\frac{29}{25}m$ . Multiplying by $50$ on both sides, we get $39k=58m$
The smallest integers $k$ and $m$ that satisfy this are $k=1450$ and $m=975$ , so $N=1131$ . The answer is $\boxed{131}$

Solution: Continuing from Solution 1, we have $N=\frac{39}{50}k$ and $N=\frac{29}{25}m$ . It follows that $k=\frac{50}{39}N$ and $m=\frac{25}{29}N$ . Both $m$ and $k$ have to be integers, so, in order for that to be true, $N$ has to cancel the denominators of both $\frac{50}{39}$ and $\frac{25}{29}$ . In other words, $N$ is a multiple of both $29$ and $39$ . That makes $N=\operatorname{lcm}(29,39)=29\cdot39=1131$ . The answer is $\boxed{131}$

Solution:
We can express $N$ as $0.78a$ and $1.22b$ , where $a$ and $b$ are some positive integers. $N=0.78a=1.22b\implies100N=78a=122b.$ Let us try to find the smallest possible value of $100N$ , first ignoring the integral constraint.
Obviously, we are just trying to find the LCM of $78$ and $116.$ They have no common factors but $2$ , so we multiply $78$ and $116$ and divide by $2$ to get $4524.$ This is obviously not divisible by $100$ , so this is not possible, as it would imply that $N=\dfrac{4524}{100},$ which is not an integer. This can be simplified to $\dfrac{1131}{25}$
We know that any possible value of $N$ will be an integral multiple of this value; the smallest such $N$ is achieved by multiplying this value by $25.$ We arrive at $1131$ , which is $\boxed{131}\mod1000.$


Answer: 440.0

Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$

Solution:
Let $a=z^{120}$ . This simplifies the problem constraint to $a^6-a \in \mathbb{R}$ . This is true if $\text{Im}(a^6)=\text{Im}(a)$ . Let $\theta$ be the angle $a$ makes with the positive x-axis. Note that there is exactly one $a$ for each angle $0\le\theta<2\pi$ . We are given $\sin\theta = \sin{6\theta}$ . Note that $\sin \theta = \sin (\pi - \theta)$ and $\sin \theta = \sin (\theta + 2\pi)$ . We can use these facts to create two types of solutions:
\[\sin \theta = \sin ((2m + 1)\pi - \theta)\]
which implies that $(2m+1)\pi-\theta = 6\theta$ and reduces to $\frac{(2m + 1)\pi}{7} = \theta$ . There are 7 solutions for this.
\[\sin \theta = \sin (2n\pi + \theta)\]
which implies that $2n\pi+\theta=6\theta$ and reduces to $\frac{2n\pi}{5} = \theta$ . There are 5 solutions for this, totaling 12 values of $a$
For each of these solutions for $a$ , there are necessarily $120$ solutions for $z$ . Thus, there are $12\cdot 120=1440$ solutions for $z$ , yielding an answer of $\boxed{440}$

Solution: The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to $0$ . Since $|z|=1$ , let $z=\cos \theta + i\sin \theta$ , then we can write the imaginary part of $\Im(z^{6!}-z^{5!})=\Im(z^{720}-z^{120})=\sin\left(720\theta\right)-\sin\left(120\theta\right)=0$ . Using the sum-to-product formula, we get $\sin\left(720\theta\right)-\sin\left(120\theta\right)=2\cos\left(\frac{720\theta+120\theta}{2}\right)\sin\left(\frac{720\theta-120\theta}{2}\right)=2\cos\left(\frac{840\theta}{2}\right)\sin\left(\frac{600\theta}{2}\right)\implies \cos\left(\frac{840\theta}{2}\right)=0$ or $\sin\left(\frac{600\theta}{2}\right)=0$ . The former yields $840$ solutions, and the latter yields $600$ solutions, giving a total of $840+600=1440$ solution, so our answer is $\boxed{440}$

Solution:
As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use exponential form of complex numbers. Let $z = e^{i \theta}$ . We have two cases to consider. Either $z^{6!} = z^{5!}$ , or $z^{6!}$ and $z^{5!}$ are reflections across the imaginary axis.
If $z^{6!} = z^{5!}$ , then $e^{6! \theta i} = e^{5! \theta i}$ . Thus, $720 \theta = 120 \theta$ or $600\theta = 0$ , giving us 600 solutions. (Equalities are taken modulo $2 \pi$ )
For the second case, $e^{6! \theta i} = e^{(\pi - 5!\theta)i}$ . This means $840 \theta = \pi$ , giving us 840 solutions.
Our total count is thus $1440$ , yielding a final answer of $\boxed{440}$

Solution: Because $|z| = 1,$ we know that $z\overline{z} = 1^2 = 1.$ Hence $\overline{z} = \frac 1 {z}.$ Because $z^{6!}-z^{5!}$ is real, it is equal to its complex conjugate. Hence $z^{6!}-z^{5!} = \overline{z^{6!}}-\overline{z^{5!}}.$ Substituting the expression we that we derived earlier, we get $z^{720}-z^{120} = \frac 1{z^{720}} - \frac 1{z^{120}}.$ This leaves us with a polynomial whose leading term is $z^{1440}.$ Hence our answer is $\boxed{440}$


Answer: 124.0

Let $N$ be the number of consecutive $0$ 's at the right end of the decimal representation of the product $1!2!3!4!\cdots99!100!.$ Find the remainder when $N$ is divided by $1000$

Solution:
A number in decimal notation ends in a zero for each power of ten which divides it.  Thus, we need to count both the number of 5s and the number of 2s dividing into our given expression.  Since there are clearly more 2s than 5s, it is sufficient to count the number of 5s.
One way to do this is as follows: $96$ of the numbers $1!,\ 2!,\ 3!,\ 100!$ have a factor of $5$ $91$ have a factor of $10$ $86$ have a factor of $15$ . And so on. This gives us an initial count of $96 + 91 + 86 + \ldots + 1$ .  Summing this arithmetic series of $20$ terms, we get $970$ .  However, we have neglected some powers of $5$ - every $n!$ term for $n\geq25$ has an additional power of $5$ dividing it, for $76$ extra; every n! for $n\geq 50$ has one more in addition to that, for a total of $51$ extra; and similarly there are $26$ extra from those larger than $75$ and $1$ extra from $100$ .  Thus, our final total is $970 + 76 + 51 + 26 + 1 = 1124$ , and the answer is $\boxed{124}$


Answer: 772.0

Let $N$ be the number of ordered pairs of nonempty sets $\mathcal{A}$ and $\mathcal{B}$ that have the following properties:
Find $N$

Solution:
Let us partition the set $\{1,2,\cdots,12\}$ into $n$ numbers in $A$ and $12-n$ numbers in $B$
Since $n$ must be in $B$ and $12-n$ must be in $A$ $n\ne6$ , we cannot partition into two sets of 6 because $6$ needs to end up somewhere, $n\ne 0$ or $12$ either).
We have $\dbinom{10}{n-1}$ ways of picking the numbers to be in $A$
So the answer is $\left(\sum_{n=1}^{11} \dbinom{10}{n-1}\right) - \dbinom{10}{5}=2^{10}-252= \boxed{772}$

Solution:
Regardless of the size $n$ of $A$ (ignoring the case when $n = 6$ ), $n$ must not be in $A$ and $12 - n$ must be in $A$
There are $10$ remaining elements whose placements have yet to be determined. Note that the actual value of $n$ does not matter; there is always $1$ necessary element, $1$ forbidden element, and $10$ other elements that need to be distributed. There are $2$ places to put each of these elements, for $2^{10}$ possibilities.
However, there is the edge case of $n = 6; 6$ is forced not the be in either set, so we must subtract the $\dbinom{10}{5}$ cases where $A$ and $B$ have size $6$
Thus, our answer is $2^{10} - \dbinom{10}{5} = 1024 - 252 = \boxed{772}$

Solution:
The total number of possible subsets is $\sum_{i=1}^{11}\dbinom{12}{i}$ , which is $2^{12}-2$ . Note that picking a subset from the set leaves the rest of the set to be in the other subset. We exclude $i=0$ and $i=12$ since they leave a set empty. We proceed with complementary counting and casework:
We apply the Principle of Inclusion and Exclusion for casework on the complementary cases. We find the ways where $|A|$ is in $A$ , which violates the first condition. Then we find the ways where the elements $|B|$ and $12-|B|$ are in set $B$ , which violates only the second condition, and not the first. This ensures we do not overcount.
Case 1: $|A|$ is an element in $A$
There are $\sum_{i=1}^{11}\dbinom{11}{i-1}$ $2^{11}-1$ ways in this case.
Case 2: $|B|$ and $12-|B|$ are in $B$
We introduce a subcase where $|B|$ is not 6, since the other element would also be 6.
There are $B-2$ elements to choose from 10 elements. Therefore, $|B|$ can be from 2 to 11.
There are $\sum_{i=2}^{11}\dbinom{10}{i-2}-\dbinom{10}{4} = 2^{10}-211$ ways.
The other subcase is when $|B|$ is equal to 6. There are $\dbinom{11}{5}=462$ ways.
Adding the subcases gives us $2^{10}+251$
Adding this with case one gives us $2^{11}+2^{10}+250$ . Subtracting this from $2^{12}-2$ gives $1024-252=\boxed{772}$


Answer: 272.0

Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions
(a) $0\le A<B<C\le99$ ,
(b) there exist integers $a$ $b$ , and $c$ , and prime $p$ where $0\le b<a<c<p$ ,
(c) $p$ divides $A-a$ $B-b$ , and $C-c$ , and
(d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$

Solution: From condition (d), we have $(A,B,C)=(B-D,B,B+D)$ and $(b,a,c)=(a-d,a,a+d)$ . Condition $\text{(c)}$ states that $p\mid B-D-a$ $p | B-a+d$ , and $p\mid B+D-a-d$ . We subtract the first two to get $p\mid-d-D$ , and we do the same for the last two to get $p\mid 2d-D$ . We subtract these two to get $p\mid 3d$ . So $p\mid 3$ or $p\mid d$ . The second case is clearly impossible, because that would make $c=a+d>p$ , violating condition $\text{(b)}$ . So we have $p\mid 3$ , meaning $p=3$ . Condition $\text{(b)}$ implies that $(b,a,c)=(0,1,2)$ or $(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})$ . Now we return to condition $\text{(c)}$ , which now implies that $(A,B,C)\equiv(-2,0,2)\pmod{3}$ . Now, we set $B=3k$ for increasing positive integer values of $k$ $B=0$ yields no solutions. $B=3$ gives $(A,B,C)=(1,3,5)$ , giving us $1$ solution. If $B=6$ , we get $2$ solutions, $(4,6,8)$ and $(1,6,11)$ . Proceeding in the manner, we see that if $B=48$ , we get 16 solutions. However, $B=51$ still gives $16$ solutions because $C_\text{max}=2B-1=101>100$ . Likewise, $B=54$ gives $15$ solutions. This continues until $B=96$ gives one solution. $B=99$ gives no solution. Thus, $N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}$

Solution:
Let $(A, B, C)$ $(B-x, B, B+x)$ and $(b, a, c) = (a-y, a, a+y)$ . Now the 3 differences would be \begin{align} \label{1} &A-a = B-x-a \\ \label{2} &B - b = B-a+y \\ \label{3} &C - c = B+x-a-y  \end{align}
Adding equations $(1)$ and $(3)$ would give $2B - 2a - y$ . Then doubling equation $(2)$ would give $2B - 2a + 2y$ . The difference between them would be $3y$ . Since $p|\{(1), (2), (3)\}$ , then $p|3y$ . Since $p$ is prime, $p|3$ or $p|y$ . However, since $p > y$ , we must have $p|3$ , which means $p=3$
If $p=3$ , the only possible values of $(b, a, c)$ are $(0, 1, 2)$ . Plugging this into our differences, we get \begin{align*}  &A-a = B-x-1 \hspace{4cm}(4)\\  &B - b = B \hspace{5.35cm}(5)\\  &C - c = B+x-2 \hspace{4cm}(6) \end{align*} The difference between $(4)$ and $(5)$ is $x+1$ , which should be divisible by 3. So $x \equiv 2 \mod 3$ . Also note that since $3|(5)$ $3|B$ . Now we can try different values of $x$ and $B$
When $x=2$ $B=3, 6, ..., 96 \Rightarrow 17$ triples.
When $x=5$ $B=6, 9, ..., 93\Rightarrow 15$ triples..
... and so on until
When $x=44$ $B=45\Rightarrow 1$ triple.
So the answer is $17 + 15 + \cdots + 1 = \boxed{272}$


Answer: 155.0

Let $N$ be the number of positive integers that are less than or equal to $2003$ and whose base- $2$ representation has more $1$ 's than $0$ 's. Find the remainder when $N$ is divided by $1000$

Solution:
In base- $2$ representation, all positive numbers have a leftmost digit of $1$ . Thus there are ${n \choose k}$ numbers that have $n+1$ digits in base $2$ notation, with $k+1$ of the digits being $1$ 's.
In order for there to be more $1$ 's than $0$ 's, we must have $k+1 > \frac{n+1}{2} \implies k > \frac{n-1}{2} \implies k \ge \frac{n}{2}$ . Therefore, the number of such numbers corresponds to the sum of all numbers on or to the right of the vertical line of symmetry in Pascal's Triangle , from rows $0$ to $10$ (as $2003 < 2^{11}-1$ ). Since the sum of the elements of the $r$ th row is $2^r$ , it follows that the sum of all elements in rows $0$ through $10$ is $2^0 + 2^1 + \cdots + 2^{10} = 2^{11}-1 = 2047$ . The center elements are in the form ${2i \choose i}$ , so the sum of these elements is $\sum_{i=0}^{5} {2i \choose i} = 1 + 2 +6 + 20 + 70 + 252 = 351$
The sum of the elements on or to the right of the line of symmetry is thus $\frac{2047 + 351}{2} = 1199$ . However, we also counted the $44$ numbers from $2004$ to $2^{11}-1 = 2047$ . Indeed, all of these numbers have at least $6$ $1$ 's in their base- $2$ representation, as all of them are greater than $1984 = 11111000000_2$ , which has $5$ $1$ 's. Therefore, our answer is $1199 - 44 = 1155$ , and the remainder is $\boxed{155}$

Solution: We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$ . If there are even digits, $2n$ , then the leftmost digit is $1$ , the rest, $2n-1$ , has odd number of digits. In order for the base-2 representation to have more $1$ 's, we will need more $1$ in the remaining $2n-1$ than $0$ 's. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$ 's at least as the number of $0$ 's. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 199$ . There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $199-44=\boxed{155}$


Answer: 144.0

Let $N$ be the number of ways to place the integers $1$ through $12$ in the $12$ cells of a $2 \times 6$ grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by $3.$ One way to do this is shown below. Find the number of positive integer divisors of $N.$ \[\begin{array}{|c|c|c|c|c|c|} \hline \,1\, & \,3\, & \,5\, & \,7\, & \,9\, & 11 \\ \hline \,2\, & \,4\, & \,6\, & \,8\, & 10 & 12 \\ \hline \end{array}\]

Solution:
We replace the numbers which are 0 mod(3) to 0, 1 mod(3) to 1, and 2 mod(3) to 2.
Then, the problem is equivalent to arranging 4 0's,4 1's, and 4 2's into the grid (and then multiplying by $4!^3$ to account for replacing the remainder numbers with actual numbers) such that no 2 of the same numbers are adjacent.
Then, the numbers which are vertically connected must be different. 2 1s must be connected with 2 2s, and 2 1s must be connected with 2 2s (vertically), as if there are less 1s connected with either, then there will be 2s or 3s which must be connected within the same number.
For instance if we did did:
- 12 x 3
- 13 x 1
Then we would be left with 23 x1, and 2 remaining 3s which aren't supposed to be connected, so it is impossible. Similar logic works for why all 1s can't be connected with all 2s.
Thus, we are left with the problem of re-arranging 2x 12 pairs, 2x 13 pairs, 2x 23 pairs. Notice that the pairs can be re-arranged horizontally in any configuration, but when 2 pairs are placed adjacent there is only 1 configuration for the rightmost pair to be set after the leftmost one has been placed.
We have $\frac{6!}{2!2!2!}=90$ ways to horizontally re-arrange the pairs, with 2 ways to set the initial leftmost column. Thus, there are 180 ways to arrange the pairs. Accounting for the initial simplification of the problem with 1-12 -> 0-3, we obtain the answer is:
\[2488320=2^{11}\cdot3^5\cdot5^1\]
The number of divisors is: $12\cdot6\cdot2=\boxed{144}.$ ~SAHANWIJETUNGA

Solution:
Let's carry out an archaeological study, that is, we will find the bones (the base) and "build up the meat."
1. Let "bones" of number $X$ be $X \pmod 3.$ Then the “skeleton” of the original table is
\[1 0 2 1 0 2\] \[2 1 0 2 1 0\] By condition, the table cannot have a column of two identical numbers (the difference of such numbers is a multiple of $3).$
There are $4$ zeros, $4$ ones and $4$ twos in the table (the order of the numbers in the columns is not important).
Therefore, there cannot be more than two identical columns (otherwise there will be four identical numbers in the remaining three, that is, the numbers in one column are the same).
Any such table has exactly $2$ columns $(0,1), 2$ columns $(0,2)$ and $2$ columns $(1,2).$
The number of possible tables of six elements of three types is \[m = \frac {6!}{2! \cdot 2! \cdot 2!} = 2 \cdot 3^2 \cdot 5.\] The number of possible tables of six elements, if the order of the digits is important, is \[M = 2^6 \cdot m = 2^7 \cdot 3^2 \cdot 5.\] 2. We are looking for the total number of options.
The column $(0,1)$ can contain the following $4^2 = 16$ pairs of numbers (the order is not important, it is already taken into account) \[(1,3), (1,6), (1,9), (1, 12), (4.3), (4.6), (4.9) (4.12), (7.3), (7.6), (7.9), (7.12), (11,3), (11,6), (11,9), (11,12).\] The second column $(0,1)$ can contain $3^2 = 9$ pairs of numbers (excluding the two that are in the first column).
Similarly with the other two columns, i.e. the total number of choices \[M \cdot 16 \cdot 9 \cdot 3 = 2^{11} \cdot 3^5 \cdot 5.\] Number of divisors \[(11+1) \cdot (5+1) \cdot (1 + 1) = \boxed{144}.\] vladimir.shelomovskii@gmail.com, vvsss


Answer: 202.0

Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$

Solution: Note that $a_2\cdot 10^2 + a_0$ is the base $100$ representation of any number from $0$ to $9999$ , and similarly $10(a_3\cdot 10^2 + a_1)$ is ten times the base $100$ representation of any number from $0$ to $9999$ . Thus, the number of solutions is just the number of solutions to $2010 = 10a+b$ where $0\le a, b\le 9999$ , which is equal to $\boxed{202}$ as $a$ can range from $0$ to $201$

Solution:
Note that $a_0 \equiv 2010\ (\textrm{mod}\ 10)$ and $a_1 \equiv 2010 - a_0\  (\textrm{mod}\ 100)$ . It's easy to see that exactly 10 values in $0 \leq a_0 \leq 99$ that satisfy our first congruence. Similarly, there are 10 possible values of $a_1$ for each choice of $a_0$ . Thus, there are $10 \times 10 = 100$ possible choices for $a_0$ and $a_1$ . We next note that if $a_0$ and $a_1$ are chosen, then a valid value of $a_3$ determines $a_2$ , so we dive into some simple casework:
Our answer is thus $6 + 0 + 196 = \boxed{202}$

Solution:
We immediately see that $a_3$ can only be $0$ $1$ or $2$ . We also note that the maximum possible value for $10a_1 + a_0$ is $990 + 99 = 1089$ . We then split into cases:
Case 1: $a_3 = 0$ .
We try to find possible values of $a_2$ . We plug in $a_3 = 0$ and $10a_1 + a_0 = 1089$ to our initial equation, which gives us $2010 = 0 + 100a_2 + 1089$ . Thus $a_2 \geq 10$ . We also see that $a_2 \leq 20$ . We now take these values of $a_2$ and find the number of pairs $(a_1, a_0)$ that work. If $a_2 = 10$ $10a_1 + a_0 = 1010$ . We see that there are $8$ possible pairs in this case. Using the same logic, there are $10$ ways for $a_2 = 11, 12 \ldots 19$ . For $a_2 = 20$ , we get the equation $10a_1 + a_0 = 10$ , for 2 ways. Thus, for $a_3 = 0$ , there are $8 + 10 \cdot 9 + 2 = 100$ ways.
Case 2: $a_3 = 1$ .
This case is almost identical to the one above, except $0 \leq a_2 \leq 10$ . We also get 100 ways.
Case 3: $a_3 = 2$ .
If $a_3 = 2$ , our initial equation becomes $100a_2 + 10a_1 + a_0 = 10$ . It is obvious that $a_2 = 0$ , and we are left with $10a_1 + a_0 = 10$ . We saw above that there are $2$ ways.
Totaling everything, we get that there are $100 + 100 + 2 = \boxed{202}$ ways.

Solution: We will represent the problem using generating functions. Consider the generating function \[f(x) = (1+x^{1000}+x^{2000}+\cdots+x^{99000})(1+x^{100}+x^{200}+\cdots+x^{9900})(1+x^{10}+x^{20}+\cdots+x^{990})(1+x+x^2+\cdots+x^{99})\] where the first factor represents $a_3$ , the second factor $a_2$ , and so forth. We want to find the coefficient of $x^{2010}$ in the expansion of $f(x)$ . Now rewriting each factor using the geometric series yields \[f(x) = \frac{\cancel{x^{100}-1}}{x-1} \cdot \frac{\cancel{x^{1000}-1}}{x^{10}-1} \cdot \frac{x^{10000}-1}{\cancel{x^{100}-1}} \cdot \frac{x^{100000}-1}{\cancel{x^{1000}-1}}=\frac{x^{10000}-1}{x-1} \cdot \frac{x^{100000}-1}{x^{10}-1} = (1+x+x^2+\cdots + x^{9999})(1+x^{10}+x^{20}+\cdots+x^{99990})\] The coefficient of $x^{2010}$ in this is simply $\boxed{202}$ , as we can choose any of the first 202 terms from the second factor and pair it with exactly one term in the first factor.

Solution:
First note that $a_3$ has to be a single-digit number( $0$ $1$ , or $2$ to be exact), and that $a_1$ has to be a two-digit multiple of ten.
Then, $a_3$ $a_2$ $a_1$ and $a_0$ can be represented as follows: \begin{align*} a_3 = a \\ a_2 = 10b+c \\ a_1= 10d+e \\ a_0 = 10f \end{align*} , where $a$ $b$ $c$ $d$ $e$ , and $f$ are all(not necessarily nonzero) digits.
Now, we can write our given equation as follows: \begin{align*} 2010 = 1000(a+b) + 100(c+d) + 10(e+f) \\ 201 = 100(a+b) + 10(c+d) + (e+f) \\ 201 = (100a + 10c + e) + (100b + d + f) \end{align*} Now, each integer between $0$ and $201$ inclusive can be represented in exactly one way as $100a + 10c + e$ , and this corresponds with one unique $100b + d + f$ , so it remains to count the number of integers between $0$ and $201$ inclusive. This is easily counted to be $\boxed{202}$ $\blacksquare$


Answer: 8.0

Let $N$ be the positive integer $7777\ldots777$ , a $313$ -digit number where each digit is a $7$ . Let $f(r)$ be the leading digit of the $r{ }$ th root of $N$ . What is \[f(2) + f(3) + f(4) + f(5)+ f(6)?\] $(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$

Solution:
We can rewrite $N$ as $\frac{7}{9}\cdot 9999\ldots999 = \frac{7}{9}\cdot(10^{313}-1)$ .
When approximating values, as we will shortly do, the minus one will become negligible so we can ignore it.
When we take the power of ten out of the square root, we’ll be multiplying by another power of ten, so the leading digit will not change. Thus the leading digit of $f(r)$ will be equal to the leading digit of $\sqrt[r]{\frac{7}{9} \cdot 10^{313(\mod r)}}$
Then $f(2)$ is the first digit of $\sqrt{\frac{7}{9}\cdot(10)} = \sqrt{\frac{70}{9}} = \sqrt{7.\ldots} \approx 2$
$f(3) - \sqrt[3]{\frac{7}{9} \cdot 10} = \sqrt[3]{\frac{70}{9}} = \sqrt[3]{7.\ldots} \approx 1$
$f(4) - \sqrt[4]{\frac{7}{9} \cdot 10} = \sqrt[4]{\frac{70}{9}} = \sqrt[4]{7.\ldots} \approx 1$
$f(5) - \sqrt[5]{\frac{7}{9} \cdot 1000} = \sqrt[5]{\frac{7000}{9}} = \sqrt[5]{777.\ldots} \approx 3$
$f(6) - \sqrt[6]{\frac{7}{9} \cdot 10} = \sqrt[6]{\frac{70}{9}} = \sqrt[6]{7.\ldots} \approx 1$
The final answer is therefore $2+1+1+3+1 = \boxed{8}$

Solution:
For notation purposes, let $x$ be the number $777 \ldots 777$ with $313$ digits, and let $B(n)$ be the leading digit of $n$ . As an example, $B(x) = 7$ , because $x = 777 \ldots 777$ , and the first digit of that is $7$
Notice that \[B(\sqrt{\frac{n}{100}}) = B(\sqrt{n})\] for all numbers $n \geq 100$ ; this is because $\sqrt{\frac{n}{100}} = \frac{\sqrt{n}}{10}$ , and dividing by $10$ does not affect the leading digit of a number. Similarly, \[B(\sqrt[3]{\frac{n}{1000}}) = B(\sqrt[3]{n}).\] In general, for positive integers $k$ and real numbers $n > 10^{k}$ , it is true that \[B(\sqrt[k]{\frac{n}{10^{k}}}) = B(\sqrt[k]{n}).\] Behind all this complex notation, all that we're really saying is that the first digit of something like $\sqrt[3]{123456789}$ has the same first digit as $\sqrt[3]{123456.789}$ and $\sqrt[3]{123.456789}$
The problem asks for \[B(\sqrt[2]{x}) + B(\sqrt[3]{x}) + B(\sqrt[4]{x}) + B(\sqrt[5]{x}) + B(\sqrt[6]{x}).\]
From our previous observation, we know that \[B(\sqrt[2]{x}) = B(\sqrt[2]{\frac{x}{100}} = B(\sqrt[2]{\frac{x}{10,000}} = B(\sqrt[2]{\frac{x}{1,000,000}} = \ldots .\] Therefore, $B(\sqrt[2]{x}) = B(\sqrt[2]{7.777 \dots})$ . We can evaluate $B(\sqrt[2]{7.777 \dots})$ , the leading digit of $\sqrt[2]{7.777 \dots}$ , to be $2$ . Therefore, $f(2) = 2$
Similarly, we have \[B(\sqrt[3]{x}) = B(\sqrt[3]{\frac{x}{1,000}} = B(\sqrt[3]{\frac{x}{1,000,000}} = B(\sqrt[3]{\frac{x}{1,000,000,000}} = \ldots .\] Therefore, $B(\sqrt[3]{x}) = B(\sqrt[3]{7.777 \ldots})$ . We know $B(\sqrt[3]{7.777 \ldots}) = 1$ , so $f(3) = 1$
Next, \[B(\sqrt[4]{x}) = B(\sqrt[4]{7.777 \ldots})\] and $B(\sqrt[4]{7.777 \ldots}) = 1$ , so $f(4) = 1$
We also have \[B(\sqrt[5]{x}) = B(\sqrt[5]{777.777 \ldots})\] and $B(\sqrt[5]{777.777 \ldots}) = 3$ , so $f(5) = 3$
Finally, \[B(\sqrt[6]{x}) = B(\sqrt[6]{7.777 \ldots})\] and $B(\sqrt[4]{7.777 \ldots}) = 1$ , so $f(6) = 1$
We have that $f(2)+f(3)+f(4)+f(5)+f(6) = 2+1+1+3+1 = \boxed{8}$

Solution: Since $7777..7$ is a $313$ digit number and $\sqrt {7}$ is around $2.5$ , we have $f(2)$ is $2$ $f(3)$ is the same story, so $f(3)$ is $1$ . It is the same as $f(4)$ as well, so $f(4)$ is also $1$ . However, $313$ is $3$ mod $5$ , so we need to take the 5th root of $777$ , which is between $3$ and $4$ , and therefore, $f(5)$ is $3$ $f(6)$ is the same as $f(4)$ , since it is $1$ more than a multiple of $6$ . Therefore, we have $2+1+1+3+1$ which is $\boxed{8}$

Solution:
First, we compute $f \left( 2 \right)$
Because $N > 4 \cdot 10^{312}$ $\sqrt{N} > 2 \cdot 10^{156}$ .
Because $N < 9 \cdot 10^{312}$ $\sqrt{N} < 3 \cdot 10^{156}$
Therefore, $f \left( 2 \right) = 2$
Second, we compute $f \left( 3 \right)$
Because $N > 1 \cdot 10^{312}$ $\sqrt[3]{N} > 1 \cdot 10^{104}$ .
Because $N < 8 \cdot 10^{312}$ $\sqrt[3]{N} < 2 \cdot 10^{104}$
Therefore, $f \left( 3 \right) = 1$
Third, we compute $f \left( 4 \right)$
Because $N > 1 \cdot 10^{312}$ $\sqrt[4]{N} > 1 \cdot 10^{78}$ .
Because $N < 16 \cdot 10^{312}$ $\sqrt[4]{N} < 2 \cdot 10^{78}$
Therefore, $f \left( 4 \right) = 1$
Fourth, we compute $f \left( 5 \right)$
Because $N > 3^5 \cdot 10^{310}$ $\sqrt[5]{N} > 3 \cdot 10^{62}$ .
Because $N < 4^5 \cdot 10^{310}$ $\sqrt[5]{N} < 4 \cdot 10^{62}$
Therefore, $f \left( 5 \right) = 3$
Fifth, we compute $f \left( 6 \right)$
Because $N > 1 \cdot 10^{312}$ $\sqrt[6]{N} > 1 \cdot 10^{52}$ .
Because $N < 2^6 \cdot 10^{312}$ $\sqrt[6]{N} < 2 \cdot 10^{52}$
Therefore, $f \left( 6 \right) = 1$
Therefore, \begin{align*} f \left( 2 \right) + f \left( 3 \right) + f \left( 4 \right) + f \left( 5 \right) + f \left( 6 \right) & = 2 + 1 + 1 + 3 + 1 \\ & = 8 . \end{align*}
Therefore, the answer is $\boxed{8}$


Answer: 3.0

Let $N$ be the second smallest positive integer that is divisible by every positive integer less than $7$ . What is the sum of the digits of $N$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$

Solution: $N$ must be divisible by every positive integer less than $7$ , or $1, 2, 3, 4, 5,$ and $6$ . Each number that is divisible by each of these is a multiple of their least common multiple. $LCM(1,2,3,4,5,6)=60$ , so each number divisible by these is a multiple of $60$ . The smallest multiple of $60$ is $60$ , so the second smallest multiple of $60$ is $2\times60=120$ . Therefore, the sum of the digits of $N$ is $1+2+0=\boxed{3}$