Answer: 192.0

Find the smallest positive integer whose cube ends in $888$

Solution:
$n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8$ and $n^3 \equiv 13 \pmod{125}$ $n \equiv 2 \pmod 5$ due to the last digit of $n^3$ . Let $n = 5a + 2$ . By expanding, $125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}$
By looking at the last digit again, we see $a \equiv 3 \pmod5$ , so we let $a = 5a_1 + 3$ where $a_1 \in \mathbb{Z^+}$ . Plugging this in to $5a^2 + 12a \equiv 1 \pmod{25}$ gives $10a_1 + 6 \equiv 1 \pmod{25}$ . Obviously, $a_1 \equiv 2 \pmod 5$ , so we let $a_1 = 5a_2 + 2$ where $a_2$ can be any non-negative integer.
Therefore, $n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67$ $n^3$ must also be a multiple of $8$ , so $n$ must be even. $125a_2 + 67 \equiv 0 \pmod 2 \implies a_2 \equiv 1 \pmod 2$ . Therefore, $a_2 = 2a_3 + 1$ , where $a_3$ is any non-negative integer. The number $n$ has form $125(2a_3+1)+67 = 250a_3+192$ . So the minimum $n = \boxed{192}$

Solution:
Let $x^3 = 1000a + 888$ . We factor an $8$ out of the right hand side, and we note that $x$ must be of the form $x = 2y$ , where $y$ is a positive integer. Then, this becomes $y^3 = 125a + 111$ . Taking mod $5$ $25$ , and $125$ , we get $y^3 \equiv 1\pmod 5$ $y^3 \equiv 11\pmod{25}$ , and $y^3 \equiv 111\pmod{125}$
We can work our way up, and find that $y\equiv 1\pmod 5$ $y\equiv 21\pmod{25}$ , and finally $y\equiv 96\pmod{125}$ . This gives us our smallest value, $y = 96$ , so $x = \boxed{192}$ , as desired. - Spacesam

Solution: Let this integer be $x.$ Note that \[x^3 \equiv 888 \pmod{1000} \implies x \equiv 0 \pmod {2}~~ \cap ~~ x \equiv 2 \pmod{5}.\] We wish to find the residue of $x$ mod $125.$ Note that \[x \equiv 2,7,12,17, \text{ or } 22 \pmod{25}\] using our congruence in mod $5.$ The residue that works must also satisfy $x^3 \equiv 13 \pmod{25}$ from our original congruence. Noting that $17^3 \equiv (-8)^3 \equiv -512 \equiv 13 \pmod{25}$ (and bashing out the other residues perhaps but they're not that hard), we find that \[x \equiv 17 \pmod{25}.\] Thus, \[x \equiv 17,42,67,92,117 \pmod{125}.\] The residue that works must also satisfy $x^3 \equiv 13 \pmod{125}$ from our original congruence. It is easy to memorize that \[17^3 \equiv \mathbf{4913} \equiv 38 \pmod{125}.\] Also, \[42^3 \equiv 42^2 \cdot 42 \equiv 1764 \cdot 42 \equiv 14 \cdot 42 \equiv 88 \pmod{125}.\] Finally, \[67^3 \equiv 67^2 \cdot 67 \equiv 4489 \cdot 67 \equiv (-11) \cdot 67 \equiv -737 \equiv 13 \pmod{125},\] as desired. Thus, $x$ must satisfy \[x \equiv 0 \pmod{2}~~ \cap ~~x \equiv 67 \pmod{125} \implies x \equiv 192 \pmod{250} \implies x=\boxed{192}.\] ~samrocksnature

Solution:
This number is in the form of $10k+2$ , after binomial expansion, we only want $600k^2+120k\equiv 880 \pmod{1000}$ . We realize that $600,120$ are both multiples of $8$ , we only need that $600k^2+120k \equiv 5\pmod{125}$ , so we write $600k^2+120k=125x+5; 120k^2+24k=25x+1, 24(5k^2+k)=25x+1, 5k^2+k\equiv -1\pmod{25}$
Then, we write $5k^2+k=25m-1, 5k^2+k+1=25m$ so $k+1$ must be a multiple of $5$ at least, so $k\equiv {-1, -6, -11, -16, -21} \pmod {25}$ after checking, when $k=-6, 5k^2+k+1=175=25\cdot 7$ . So $k\equiv -6 \pmod{25}$ , smallest $k=19$ , the number is $\boxed{192}$


Answer: 16.0

Find the smallest whole number that is larger than the sum
\[2\dfrac{1}{2}+3\dfrac{1}{3}+4\dfrac{1}{4}+5\dfrac{1}{5}.\]
$\text{(A)}\ 14 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 17 \qquad \text{(E)}\ 18$

Solution:
Adding the whole numbers gives $2 + 3 + 4 + 5 = 14$
Adding the fractions gives $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{30}{60} + \frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{77}{60}$ .  This will create one more whole, and a fraction that is less than $1$ .  Thus, the smallest whole number that is less than $15$ plus some fractional part is $16$ , and the answer is $\boxed{16}$

Solution: Convert the fractional parts to decimals, and approximate the answer. $2.5 + 3.33 + 4.25 + 5.2 = 15.28$ , and the answer is $16$ , which is $\boxed{16}$

Solution:
Clearly, break the mixed numbers into $2+3+4+5+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
This is the same as $14+(\frac{1}{2} + \frac{1}{3}) + (\frac{1}{4} + \frac{1}{5})$ . Notice that clearly, $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ , and $1>\frac{1}{4}+\frac{1}{5}>\frac{1}{6}$ (clearly, $\frac{1}{4}>\frac{1}{6}$ ). This means that the fractional parts is between 1 and 2, so when that is added to 14, the original expression is between 15, and 16. So the least integer that is bigger than this sum is clearly 16. So select $\boxed{16}$


Answer: 371.0

Find the sum of all positive integers $b < 1000$ such that the base- $b$ integer $36_{b}$ is a perfect square and the base- $b$ integer $27_{b}$ is a perfect cube.

Solution:
The first step is to convert $36_{b}$ and $27_{b}$ into base-10 numbers. Then, we can write \[36_{b} = 3b + 6\] and \[27_{b} = 2b + 7\] . It should also be noted that $8 \leq b < 1000$
Because there are less perfect cubes than perfect squares for the restriction we are given on $b$ , it is best to list out all the perfect cubes. Since the maximum $b$ can be is 1000 and $2$ $1000 + 7 = 2007$ , we can list all the perfect cubes less than 2007.
Now, $2b + 7$ must be one of \[3^3, 4^3, ... , 12^3\] . However, $2b + 7$ will always be odd, so we can eliminate the cubes of the even numbers and change our list of potential cubes to \[3^3, 5^3, 7^3, 9^3\text{, and }11^3\]
Because $3b + 6$ is a perfect square and is clearly divisible by 3, it must be divisible by 9, so $b + 2$ is divisible by 3. Thus the cube, which is \[2b + 7 = 2(b + 2) + 3\] , must also be divisible by 3. Therefore, the only cubes that $2b + 7$ could potentially be now are $3^3$ and $9^3$
We need to test both of these cubes to make sure $3b + 6$ is a perfect square.
$\textbf{Case 1:}$ If we set \[3^3 = 2b + 7\] so \[b = 10\] . If we plug this value of b into $3b + 6$ , the expression equals $36$ , which is indeed a perfect square.
$\textbf{Case 2:}$ If we set \[9^3 = 2b + 7\] so \[b = 361\] . If we plug this value of b into $3b + 6$ , the expression equals $1089$ , which is $33^2$
We have proven that both $b = 10$ and $b = 361$ are the only solutions, so \[10 + 361 = \boxed{371}\]

Solution:
The conditions are: \[3b+6 = n^2\] \[2b+7 = m^3\] We can see $n$ is multiple is 3, so let $n=3k$ , then $b= 3k^2-2$ . Substitute $b$ into second condition and we get $m^3=3(2k^2+1)$ . Now we know $m$ is both a multiple of 3 and odd. Also, $m$ must be smaller than 13 for $b$ to be smaller than 1000. So the only two possible values for $m$ are 3 and 9. Test and they both work. The final answer is $10 + 361 =$ $\boxed{371}$ .
-Mathdummy

Solution:
As shown above, let \[3b+6 = n^2\] \[2b+7 = m^3\] such that \[6b+12=2n^2\] \[6b+21=3m^3\]
Subtracting the equations we have \[3m^3-2n^2=9 \implies 3m^3-2n^2-9=0.\]
We know that $m$ and $n$ both have to be integers, because then the base wouldn't be an integer. Furthermore, any integer solution $m$ must divide $9$ by the Rational Root Theorem.
We can instantly know $m \neq -9,-3,-1,1$ since those will have negative solutions.
When $m=3$ we have $n=6$ , so then $b=10$
When $m=9$ we have $n=33$ , so then $b=361$
Therefore, the sum of all possible values of $b$ is \[10+361=\boxed{371}.\]


Answer: 38.0

Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square

Solution:
If $n^2-19n+99=x^2$ for some positive integer $x$ , then rearranging we get $n^2-19n+99-x^2=0$ . Now from the quadratic formula,
Because $n$ is an integer, this means $4x^2-35=q^2$ for some nonnegative integer $q$ . Rearranging gives $(2x+q)(2x-q)=35$ . Thus $(2x+q, 2x-q)=(35, 1)$ or $(7,5)$ , giving $x=3$ or $9$ . This gives $n=1, 9, 10,$ or $18$ , and the sum is $1+9+10+18=\boxed{38}$

Solution:
When $n \geq 12$ , we have \[(n-10)^2 < n^2 -19n + 99 < (n-8)^2.\]
So if $n \geq 12$ and $n^2 -19n + 99$ is a perfect square, then \[n^2 -19n + 99 = (n-9)^2\]
or $n = 18$
For $1 \leq n < 12$ , it is easy to check that $n^2 -19n + 99$ is a perfect square when $n = 1, 9$ and $10$ ( using the identity $n^2 -19n + 99 = (n-10)^2 + n - 1.)$
We conclude that the answer is $1 + 9 + 10 + 18 = \boxed{38}.$


Answer: 195.0

Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer.

Solution: Manipulating the given expression, $\sqrt{n^2+85n+2017}=\frac{1}{2}\sqrt{4n^2+340n+8068}=\frac{1}{2}\sqrt{(2n+85)^2+843}$ . The expression under the radical must be an square number for the entire expression to be an integer, so $(2n+85)^2+843=s^2$ . Rearranging, $s^2-(2n+85)^2=843$ . By difference of squares, $(s-(2n+85))(s+(2n+85))=1\times843=3\times281$ . It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, $2n+85$ is found to be $421$ and $139$ . The two values of $n$ that satisfy one of the equations are $168$ and $27$ . Summing these together gives us the answer ; $168+27=\boxed{195}$

Solution:
Clearly, the result when $n$ is plugged into the given expression is larger than $n$ itself. Let $x$ be the positive difference between that result and $n$ , so that $\sqrt{n^2+85n+2017}=n+x$ . Squaring both sides and canceling the $n^2$ terms gives $85n+2017=2xn+x^2$ . Combining like terms, $(85-2x)n=x^2-2017$ , so
\[n=\frac{x^2-2017}{85-2x}.\]
Since $n$ is positive, there are two cases, which are simple (luckily). Remembering that $x$ is a positive integer, then $x^2-2017$ and $85-2x$ are either both positive or both negative. The smallest value for which $x^2>2017$ is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that $x<45$ (from the numerator) and $85-2x<0$ , which means $x>42$ . This only gives two solutions, $x=43, 44$ . Plugging these into the expression for $n$ , we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is $168+27=\boxed{195}$

Solution:
Let the integer given by the square root be represented by $x$ . Then $0 = n^2 + 85n + 2017 - x^2$ . For this to have rational solutions for $n$ (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)
Thus, $b^2 - 4ac = 7225 + 4x^2 - 8068 = y^2$ for some integer $y$ . Then $4x^2 - 843 = y^2$ . Rearranging this equation yields that $843 = (2x+y)(2x-y)$ . Noticing that there are 2 factor pairs of $843$ , namely, $1*843$ and $3*281$ , there are 2 systems to solve for $x$ and $y$ that create rational $n$ . These yield solutions $(x,y)$ of $(211, 421)$ and $(71, 139)$
The solution to the initial quadratic in $n$ must then be $\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}$ . Noticing that for each value of $x$ that has rational solutions for $n$ , the corresponding value of the square root of the discriminant is $y$ , the formula can be rewritten as $n = \frac{-85 \pm y}{2}$ . One solution is $\frac{421 - 85}{2} = 168$ and the other solution is $\frac{139 - 85}{2} = 27$ . Thus the answer is $168 + 27 = \boxed{195}$ as both rational solutions are integers.

Solution: Notice that $(n+42)^2= n^2+84n+1764$ . Also note that $(n+45)^2= n^2+90n+2025$ . Thus, \[(n+42)^2< n^2+85n+2017<(n+45)^2\] where $n^2+85n+2017$ is a perfect square. Hence, \[n^2+85n+2017= (n+43)^2\] or \[n^2+85n+2017= (n+44)^2.\] Solving the two equations yields the two solutions $n= 168, 27$ . Therefore, our answer is $\boxed{195}$

Solution: Let the expression be equal to $a$ . This expression can be factored into $\sqrt{(n+44)^2-3n+81}$ . Then square both sides, and the expression becomes $(n+44)^2-3n+81=a^2$ . We have a difference of two squares. Rearranging, we have $(n+44+a)(n+44-a)=3(n-27)$ . By inspection, the only possible values for $(n+44-a)$ are 0 and 1. When $(n+44-a)=0$ , we must have $n-27=0$ . Therefore, $27$ is a solution. When we have $(n+44-a)=1$ , so $n=a-43$ . Plugging this back to $(n+44+a)=3(n-27)$ (since $(n+44-a)=1$ ), we find that $a=211 \implies n=168$ . Thus, the answer is $27+168= \boxed{195}$

Solution:
More intuitive, but a little bit slower considering the decimals.
Label the entire given expression as $k^2$
Instinctively we can do a crude completion of the square, resulting in $k^2$ $(n+42.5)^2+210.75$ Rearrange the equation to get a difference of squares.
$k^2-(n+42.5)^2 = 210.75$
$(k+n+42.5)(k-n-42.5) = 210.75$
Factor $21,075$ to get $3^1$ $5^2$ , and $281^1$
Now the two factors given are either divided by 10 each or one being divided by 100. Let's start with the former case.
If you try $281*3/10$ and $5*5/10$ , you quickly realize that $n$ becomes negative. Naturally, you will realize you want the number's difference to be larger. Try $281*5/10$ and $3*5/10$ . This gives an answer of $27$ for $x$ . The next largest possibility also works, giving an $n$ of $168$ . As you rise, some numbers don't work because it results in an n that is not an integer, as in the example of $\frac{281*5*5}{10}$ and $\frac{3}{10}$
Now if you continue on with the next case, where one factor is divided by $100$ , very swiftly will you realize most don't work simply because the difference is too small, or it doesn't give an integer. It helps a lot when you realize that the decimal does not end in a $5$ , the answer will not be an integer. After a few short tests, we get $168+27=\boxed{195}$


Answer: 239.0

Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$ , the remainder is $17$

Solution:
The formula for the sum of cubes, also known as Nicomachus's Theorem, is as follows: \[1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2=\left(\frac{k(k+1)}{2}\right)^2\] for any positive integer $k$
So let's apply this to this problem.
Let $m=n+5$ . Then we have \begin{align*} 1^3+2^3+3^3+\dots+(m-5)^3&\equiv 17 \mod m \\ \left(\frac{(m-5)(m-4)}{2}\right)^2&\equiv 17 \mod m \\ \left(\dfrac{m(m-9)+20}2\right)^2&\equiv 17\mod m \\ \left(\dfrac{20}2\right)^2&\equiv 17\mod m \\ \frac{400}{4}&\equiv 17 \mod m \\ 332 &\equiv 0 \mod m \\ \end{align*} So, $m\in\{83,166,332\}$ . Testing the cases, only $332$ fails. This leaves $78+161=\boxed{239}$

Solution:
Using the formula for $\sum_{k=1}^n  k^3$ \[1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}\] Since $1^3 + 2^3 + 3^3 + ... + n^3$ divided by $n + 5$ has a remainder of $17$ \[\frac{n^2(n+1)^2}{4} \equiv 17\pmod {n + 5}\] Using the rules of modular arithmetic, \[n^2(n+1)^2 \equiv 68\pmod {n + 5}\] \[n^2(n+1)^2 - 68\equiv 0\pmod {n + 5}\] Expanding the left hand side, \[n^4 + 2 n^3 + n^2 - 68\equiv 0\pmod {n + 5}\] This means that $n^4 + 2 n^3 + n^2 - 68$ is divisible by ${n + 5}$
\[(n + 5) | (n^4 + 2 n^3 + n^2 - 68)\] Dividing polynomials, \[\frac{n^4 + 2 n^3 + n^2 - 68}{n + 5}\] \[= n^3 - 3 n^2 + 16n - 80 + \frac{332}{(n + 5)}\] $(n + 5)$ $|$ $(n^4 + 2 n^3 + n^2 - 68)$ $\iff$ $\frac{332}{(n + 5)}$ $\in$ $\mathbb{Z}$ $\frac{332}{(n + 5)}$ $\in$ $\mathbb{Z}$ $\iff$ $(n + 5) = \pm 1, \pm 2, \pm 4, \pm 83, \pm 166, \pm 332$ Note that $n$ $\in$ $\mathbb{N}$ and $n + 5 > 17$ (because the remainder when dividing by $n + 5$ is $17$ , so $n + 5$ must be greater than $17$ ), so all options $\leq 17$ can be eliminated. \[(n + 5) = 83, 166, 332\] \[n = 78, 161, 327\] Checking all 3 cases, $n = 78$ and $n = 161$ work; $n = 327$ fails. Therefore, the answer is $78 + 161 = \boxed{239} ~

Solution:
As before, we note that $(5+a)^3 + (n-a)^3 \equiv (5+a)^3 - (n+5 - (n-a))^3 \equiv 0 \pmod {n+5}.$ Thus, we can pair up the terms from $5^3$ to $n^3$ and cancel them. We have to deal with two cases:
If $n$ is even, then $5^3+6^3 + \cdots + n^3 \equiv 0 \pmod {n+5},$ as there are an even number of terms and they pair and cancel. We thus get $1^2+2^3+3^3+4^3 = 100 \equiv 17 \pmod {n+5},$ or $(n+5) | 83,$ which yields $n=78.$
If $n$ is odd, then $1^3+2^3+\cdots + n^3 \equiv 1^3+2^3+3^3+4^3+\left( \frac{n+5}{2} \right)^3 \equiv 17 \pmod {n+5}.$ Letting $k = \frac{n+5}{2}$ yields $k^2 + 83 \equiv 0 \pmod {2k}.$ However, this means that $83$ is divisible by $k,$ so $k=1,83.$ Plugging this back into $n$ yields $n=2(83)-5 = 161$ in the latter case.
Thus, the sum of all possible $n$ is just $78+161 = \boxed{239}.$


Answer: 71.0

Find the sum of all positive integers $n$ such that, given an unlimited supply of stamps of denominations $5,n,$ and $n+1$ cents, $91$ cents is the greatest postage that cannot be formed.

Solution:
By the Chicken McNugget theorem, the least possible value of $n$ such that $91$ cents cannot be formed satisfies $5n - (5 + n) = 91 \implies n = 24$ , so $n$ must be at least $24$
For a value of $n$ to work, we must not only be unable to form the value $91$ , but we must also be able to form the values $92$ through $96$ , as with these five values, we can form any value greater than $96$ by using additional $5$ cent stamps.
Notice that we must form the value $96$ without forming the value $91$ . If we use any $5$ cent stamps when forming $96$ , we could simply remove one to get $91$ . This means that we must obtain the value $96$ using only stamps of denominations $n$ and $n+1$
Recalling that $n \geq 24$ , we can easily figure out the working $(n,n+1)$ pairs that can used to obtain $96$ , as we can use at most $\frac{96}{24}=4$ stamps without going over. The potential sets are $(24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96)$ , and $(96, 97)$
The last two obviously do not work, since they are too large to form the values $92$ through $95$ . By a little testing, only $(24, 25)$ and $(47, 48)$ can form the necessary values, so $n \in \{24, 47\}$ $24 + 47 = \boxed{071}$

Solution:
Notice that once we hit all residues $\bmod 5$ , we'd be able to get any number greater (since we can continually add $5$ to each residue). Furthermore, $n\not\equiv 0,1\pmod{5}$ since otherwise $91$ is obtainable (by repeatedly adding $5$ to either $n$ or $n+1$ ) Since the given numbers are $5$ $n$ , and $n+1$ , we consider two cases: when $n\equiv 4\pmod{5}$ and when $n$ is not that.
When $n\equiv 4 \pmod{5}$ , we can only hit all residues $\bmod 5$ once we get to $4n$ (since $n$ and $n+1$ only contribute $1$ more residue $\bmod 5$ ). Looking at multiples of $4$ greater than $91$ with $n\equiv 4\pmod{5}$ , we get $n=24$ . It's easy to check that this works. Furthermore, any $n$ greater than this does not work since $91$ isn't the largest unobtainable value (can be verified using Chicken McNugget Theorem).
Now, if $n\equiv 2,3\pmod{5}$ , then we'd need to go up to $2(n+1)=2n+2$ until we can hit all residues $\bmod 5$ since $n$ and $n+1$ create $2$ distinct residues $\bmod{5}$ . Checking for such $n$ gives $n=47$ and $n=48$ . It's easy to check that $n=47$ works, but $n=48$ does not (since $92$ is unobtainable). Furthermore, any $n$ greater than this does not work since $91$ isn't the largest unobtainable value in those cases (can be verified using Chicken McNugget Theorem). (Also note that in the $3 \pmod{5}$ case, the residue $2 \pmod{5}$ has will not be produced until $3(n+1)$ while the $1\pmod5$ case has already been produced, so the highest possible value that cannot be produced would not be a number equivalent to $1 \pmod5$
Since we've checked all residues $\bmod 5$ , we can be sure that these are all the possible values of $n$ . Hence, the answer is $24+47=\boxed{071}$ . - ktong

Solution:
Obviously $n\le 90$ . We see that the problem's condition is equivalent to: 96 is the smallest number that can be formed which is 1 mod 5, and 92, 93, 94 can be formed (95 can always be formed). Now divide this up into cases. If $n\equiv 0\pmod{5}$ , then 91 can be formed by using $n+1$ and some 5's, so there are no solutions for this case. If $n\equiv 1\pmod{5}$ , then 91 can be formed by using $n$ and some 5's, so there are no solutions for this case either.
For $n\equiv 2\pmod{5}$ $2n+2$ is the smallest value that can be formed which is 1 mod 5, so $2n+2=96$ and $n=47$ . We see that $92=45+47$ $93=48+45$ , and $94=47+47$ , so $n=47$ does work. If $n\equiv 3\pmod{5}$ , then the smallest value that can be formed which is 1 mod 5 is $2n$ , so $2n=96$ and $n=48$ . We see that $94=49+45$ and $93=48+45$ , but 92 cannot be formed, so there are no solutions for this case. If $n\equiv 4\pmod{5}$ , then we can just ignore $n+1$ since it is a multiple of 5, meaning that the Chicken McNuggest theorem is a both necessary and sufficient condition, and it states that $5n-n-5=91$ meaning $4n=96$ and $n=24$ .
Hence, the only two $n$ that work are $n=24$ and $n=47$ , so our answer is $24+47=\boxed{071}$ .
-Stormersyle


Answer: 400.0

Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms

Solution:
There are 8 fractions which fit the conditions between 0 and 1: $\frac{1}{30},\frac{7}{30},\frac{11}{30},\frac{13}{30},\frac{17}{30},\frac{19}{30},\frac{23}{30},\frac{29}{30}$
Their sum is 4. Note that there are also 8 terms between 1 and 2 which we can obtain by adding 1 to each of our first 8 terms. For example, $1+\frac{19}{30}=\frac{49}{30}.$ Following this pattern, our answer is $4(10)+8(1+2+3+\cdots+9)=\boxed{400}.$

Solution: Note that if $x$ is a solution, then $(300-x)$ is a solution. We know that $\phi(300) = 80.$ Therefore the answer is $\frac{80}{2} \cdot\frac{300}{30} = \boxed{400}.$


Answer: 630.0

Find the sum of all positive two-digit integers that are divisible by each of their digits.

Solution:
Let our number be $10a + b$ $a,b \neq 0$ . Then we have two conditions: $10a + b \equiv 10a \equiv 0 \pmod{b}$ and $10a + b \equiv b \pmod{a}$ , or $a$ divides into $b$ and $b$ divides into $10a$ . Thus $b = a, 2a,$ or $5a$ (note that if $b = 10a$ , then $b$ would not be a digit).
If we ignore the case $b = 0$ as we have been doing so far, then the sum is $495 + 120 + 15 = \boxed{630}$

Solution: Using casework, we can list out all of these numbers: \[11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.\]

Solution:
In this solution, we will do casework on the ones digit.
Before we start, let's make some variables. Let $a$ be the ones digit, and $b$ be the tens digit. Let $n$ equal our number. Our number can be expressed as $10b+a$ . We can easily see that $b|a$ , since $b|n$ , and $b|10b$ . Therefore, $b|(n-10b)$ .
Now, let's start with the casework.
Case 1: $a=1$ Since $b|a$ $b=1$ . From this, we get that $n=11$ satisfies the condition.
Case 2: $a=2$ We either have $b=1$ , or $b=2$ . From this, we get that $n=12$ and $n=22$ satisfy the condition.
Case 3: $a=3$ We have $b=3$ . From this, we get that $n=33$ satisfies the condition. Note that $b=1$ was not included because $3$ does not divide $13$
Case 4: $a=4$ We either have $b=2$ or $b=4$ . From this, we get that $n=24$ and $n=44$ satisfy the condition. $b=1$ was not included for similar reasons as last time.
Case 5: $a=5$ We either have $b=1$ or $b=5$ . From this, we get that $n=15$ and $n=55$ satisfy the condition.
Continuing with this process up to $a=9$ , we get that $n$ could be $11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99$ . Summing, we get that the answer is $\boxed{630}$ . A clever way to sum would be to group the multiples of $11$ together to get $11+22+\dots+99=(45)(11)=495$ , and then add the remaining $12+24+15+36+48=135$


Answer: 118.0

Find the sum of all prime numbers between $1$ and $100$ that are simultaneously $1$ greater than a multiple of $4$ and $1$ less than a multiple of $5$
$\mathrm{(A) \ } 118 \qquad \mathrm{(B) \ }137 \qquad \mathrm{(C) \ } 158 \qquad \mathrm{(D) \ } 187 \qquad \mathrm{(E) \ } 245$

Solution:
Numbers that are $1$ less than a multiple of $5$ all end in $4$ or $9$
No prime number ends in $4$ , since all numbers that end in $4$ are divisible by $2$ .  Thus, we are only looking for numbers that end in $9$
Writing down the ten numbers that so far qualify, we get $9, 19, 29, 39, 49, 59, 69, 79, 89, 99$
Crossing off multiples of $3$ gives $19, 29, 49, 59, 79, 89$
Crossing off numbers that are not $1$ more than a multiple of $4$ (in other words, numbers that are $1$ less than a multiple of $4$ , since all numbers are odd), we get:
$29, 49, 89.$
Noting that $49$ is not prime, we have only $29$ and $89$ , which give a sum of $118$ , so the answer is $\boxed{118}$


Answer: 1080.0

Find the sum of all the positive solutions of
$2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1$
$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$

Solution:
Rewrite $\cos{4x} - 1$ as $2\cos^2{2x} - 2$ .  Now let $a = \cos{2x}$ , and let $b = \cos{\left( \frac{2014\pi^2}{x} \right) }$ .  We have: \[2a(a - b) = 2a^2 - 2\]
Therefore, \[ab = 1\]
Notice that either $a = 1$ and $b = 1$ or $a = -1$ and $b = -1$ .  For the first case, $a = 1$ only when $x = k\pi$ and $k$ is an integer. $b = 1$ when $\frac{2014\pi^2}{k\pi}$ is an even multiple of $\pi$ , and since $2014 = 2*19*53$ $b =1$ only when $k$ is an odd divisor of $2014$ .  This gives us these possible values for $x$ \[x= \pi, 19\pi, 53\pi, 1007\pi\] For the case where $a = -1$ $\cos{2x} = -1$ , so $x = \frac{m\pi}{2}$ , where m is odd. $\frac{2014\pi^2}{\frac{m\pi}{2}}$ must also be an odd multiple of $\pi$ in order for $b$ to equal $-1$ , so $\frac{4028}{m}$ must be odd.  We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for $m$ , and therefore no cases where $a = -1$ and $b = -1$ . Therefore, the sum of all our possible values for $x$ is \[\pi + 19\pi + 53\pi + 1007\pi = \boxed{1080}\]

Solution:
Rewriting $\cos{4x} - 1$ as $2\cos^2{2x} - 2$ and transposing $2\cos{2x}$ from the LHS to the RHS, we get,
\[\cos{2x} - \cos{\left(\frac{2014\pi^2}{x}\right)} = 1 - \frac{1}{\cos{2x}}\] \[\implies \underbrace{\cos{2x} + \frac{1}{\cos{2x}}}_{\text{LHS}} = \underbrace{1 + \cos{\left(\frac{2014\pi^2}{x}\right)}}_{\text{RHS}}\]
By the AM-GM Inequality
\[\cos{2x} + \frac{1}{\cos{2x}} \in (-\infty, -2] \cup [2, \infty)\]
Also, because of the range of $\cos$
\[1 + \cos{\left(\frac{2014\pi^2}{x}\right)} \in [0, 2]\]
Hence, $\text{LHS} = \text{RHS} = 2$ , and we get ( $m, n \in \mathbb{Z}$ ),
From $(1)$ and $(2)$ \[m = \frac{1007}{n}\] \[\implies m \in \{1, 19, 53, 1007\}\] \[\implies x \in \{1\pi, 19\pi, 53\pi, 1007\pi\}\]
Therefore, sum of values of $x$ is \[\pi + 19\pi + 53\pi + 1007\pi = \boxed{1080}\]


Answer: 180001.0

Find the sum of digits of all the numbers in the sequence $1,2,3,4,\cdots ,10000$
$\text{(A) } 180001\quad \text{(B) } 154756\quad \text{(C) } 45001\quad \text{(D) } 154755\quad \text{(E) } 270001$

Solution: We can find the sum using the following method. We break it down into cases. The first case is the numbers $1$ to $9$ . The second case is the numbers $10$ to $99$ . The third case is the numbers $100$ to $999$ . The fourth case is the numbers $1,000$ to $9,999$ . And lastly, the sum of the digits in $10,000$ . The first case is just the sum of the numbers $1$ to $9$ which is, using $\frac{n(n+1)}{2}$ $45$ . In the second case, every number $1$ to $9$ is used $19$ times. $10$ times in the tens place, and $9$ times in the ones place. So the sum is just $19(45)$ . Similarly, in the third case, every number $1$ to $9$ is used $100$ times in the hundreds place, $90$ times in the tens place, and $90$ times in the ones place, for a total sum of $280(45)$ . By the same method, every number $1$ to $9$ is used $1,000$ times in the thousands place, $900$ times in the hundreds place, $900$ times in the tens place, and $900$ times in the ones place, for a total of $3700(45)$ . Thus, our final sum is $45+19(45)+280(45)+3700(45)+1=4000(45)+1=\boxed{180001}.$

Solution:
Consider the numbers from $0000-9999$ . We have $40000$ digits and each has equal an probability of being $0,1,2....9$ .
Thus, our desired sum is $4000\left( \frac{9 \cdot 10}{2} \right)+1=4000(45)+1=\boxed{180001}.$

Solution:
As in Solution 2, we consider the four digit numbers from $0000-9999.$ We see that we have $10000\times4=40000$ digits with each digit  appearing equally.
Thus, the digit sum will be the average of the digits multiplied by $40000.$ The digit average comes out to be $\frac{0+9}{2},$ since all digits are consecutive.
So, our answer will be $\frac{9}{2} \times 40000 = 180000.$ However, since we purposely did not include $10000,$ we add one to get our final answer as $\boxed{180001}.$


Answer: 3030.0

Find the sum of the arithmetic series \[20+20\frac{1}{5}+20\frac{2}{5}+\cdots+40\] $\textbf{(A)}\ 3000 \qquad\textbf{(B)}\ 3030 \qquad\textbf{(C)}\ 3150 \qquad\textbf{(D)}\ 4100 \qquad\textbf{(E)}\ 6000$

Solution: Our first term is $20$ and our last term is $40$ . To find the number of terms, $n$ , we note that the common difference between each term is $\frac{1}{5}$ . So we have \[20+\frac{1}{5}(n-1)=40\implies n-1=100\implies n=101.\] Using our formula, our sum is \[101\left(\frac{20+40}{2}\right)=101\times 30=\boxed{3030.}\]


Answer: 846.0

Find the sum of the digits in the answer to
$\underbrace{9999\cdots 99}_{94\text{ nines}} \times \underbrace{4444\cdots 44}_{94\text{ fours}}$
where a string of $94$ nines is multiplied by a string of $94$ fours.
$\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072$

Solution:
Notice that:
$9 \cdot 4 = 36$ and $3+6 = 9 = 9 \cdot 1$
$99 \cdot 44 = 4356$ and $4+5+3+6 = 18 = 9 \cdot 2$
So the sum of the digits of $x$ 9s times $x$ 4s is simply $x \cdot 9$ (Try to find the proof that it works for all values of $x$ ~MATHWIZARD10).
Therefore the answer is $94 \cdot 9 = \boxed{846.}$

Solution:
\[\underbrace{9999\cdots 99}_{94\text{ nines}} \cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = (10^{94}-1)\cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{94\text{ fours}} \underbrace{000\cdots 0}_{94\text{ zeros}} - \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{93\text{ fours}} 3 \underbrace{555\cdots 5}_{93\text{ zeros}}6\]
\[4 \cdot 93 + 3 + 5 \cdot 93 + 6 = 9 \cdot 94 = \boxed{846}\]


Answer: 25.0

Find the sum of the digits of the largest even three digit number (in base ten representation)
which is not changed when its units and hundreds digits are interchanged.
$\textbf{(A) }22\qquad \textbf{(B) }23\qquad \textbf{(C) }24\qquad \textbf{(D) }25\qquad \textbf{(E) }26$

Solution:
Solution by e_power_pi_times_i
Since the number doesn't change when the units and hundreds digits are switched, the number must be of the form $aba$ . We want to create the largest even $3$ -digit number, so $a = 8$ and $b = 9$ . The sum of the digits is $8+9+8 = \boxed{25}$


Answer: 500.0

Find the sum of the roots , real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots.

Solution:
From Vieta's formulas , in a polynomial of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0$ , then the sum of the roots is $\frac{-a_{n-1}}{a_n}$
From the Binomial Theorem , the first term of $\left(\frac 12-x\right)^{2001}$ is $-x^{2001}$ , but $x^{2001}+-x^{2001}=0$ , so the term with the largest degree is $x^{2000}$ . So we need the coefficient of that term, as well as the coefficient of $x^{1999}$
\begin{align*}\binom{2001}{1} \cdot (-x)^{2000} \cdot \left(\frac{1}{2}\right)^1&=\frac{2001x^{2000}}{2}\\ \binom{2001}{2} \cdot (-x)^{1999} \cdot \left(\frac{1}{2}\right)^2 &=\frac{-x^{1999}*2001*2000}{8}=-2001 \cdot 250x^{1999} \end{align*}
Applying Vieta's formulas, we find that the sum of the roots is $-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}$

Solution: We find that the given equation has a $2000^{\text{th}}$ degree polynomial. Note that there are no multiple roots. Thus, if $\frac{1}{2} - x$ is a root, $x$ is also a root. Thus, we pair up $1000$ pairs of roots that sum to $\frac{1}{2}$ to get a sum of $\boxed{500}$

Solution: Note that if $r$ is a root, then $\frac{1}{2}-r$ is a root and they sum up to $\frac{1}{2}.$ We make the substitution $y=x-\frac{1}{4}$ so \[(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.\] Expanding gives \[2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots\] so by Vieta, the sum of the roots of $y$ is 0.  Since $x$ has a degree of 2000, then $x$ has 2000 roots so the sum of the roots is \[2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.\]


Answer: 8.0

Find the sum of the squares of all real numbers satisfying the equation $x^{256}-256^{32}=0$
$\textbf{(A) }8\qquad \textbf{(B) }128\qquad \textbf{(C) }512\qquad \textbf{(D) }65,536\qquad \textbf{(E) }2(256^{32})$

Solution:
Solution by e_power_pi_times_i
Notice that the solutions to the equation $x^{256}-1=0$ are the $256$ roots of unity. Then the solutions to the equation $x^{256}-256^{32}=0$ are the $256$ roots of unity dilated by $\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2$ . However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are $\pm1$ , and dilating by $2$ gives $\pm2$ . The sum of the squares is $(2)^2+(-2)^2 = \boxed{8}$


Answer: 906.0

Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$ , where $x$ is measured in degrees and $100< x< 200.$

Solution:
Observe that $2\cos 4x\cos x = \cos 5x + \cos 3x$ by the sum-to-product formulas. Defining $a = \cos 3x$ and $b = \cos 5x$ , we have $a^3 + b^3 = (a+b)^3 \rightarrow ab(a+b) = 0$ . But $a+b = 2\cos 4x\cos x$ , so we require $\cos x = 0$ $\cos 3x = 0$ $\cos 4x = 0$ , or $\cos 5x = 0$
Hence we see by careful analysis of the cases that the solution set is $A = \{150, 126, 162, 198, 112.5, 157.5\}$ and thus $\sum_{x \in A} x = \boxed{906}$


Answer: 227.0

Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.

Solution:
We are given that \[100a + 10b + c = 81b + 9c + a,\] which rearranges to \[99a = 71b + 8c.\] Taking both sides modulo $71,$ we have \begin{align*} 28a &\equiv 8c \pmod{71} \\ 7a &\equiv 2c \pmod{71}. \end{align*} The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$
Therefore, the requested three-digit positive integer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$

Solution:
As shown in Solution 1, we get $99a = 71b+8c$
Note that $99$ and $71$ are large numbers comparatively to $8$ , so we hypothesize that $a$ and $b$ are equal and $8c$ fills the gap between them. The difference between $99$ and $71$ is $28$ , which is a multiple of $4$ . So, if we multiply this by $2$ , it will be a multiple of $8$ and thus the gap can be filled. Therefore, the only solution is $(a,b,c)=(2,2,7)$ , and the answer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$

Solution:
As shown in Solution 1, we get $99a = 71b+8c.$
We list a few multiples of $99$ out: \[99,198,297,396.\] Of course, $99$ can't be made of just $8$ 's. If we use one $71$ , we get a remainder of $28$ , which can't be made of $8$ 's either. So $99$ doesn't work. $198$ can't be made up of just $8$ 's. If we use one $71$ , we get a remainder of $127$ , which can't be made of $8$ 's. If we use two $71$ 's, we get a remainder of $56$ , which can be made of $8$ 's.
Therefore we get $99\cdot2=71\cdot2+8\cdot7$ so $a=2,b=2,$ and $c=7$ . Plugging this back into the original problem shows that this answer is indeed correct. Therefore, $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$

Solution:
As shown in Solution 1, we get $99a = 71b+8c$
We can see that $99$ is $28$ larger than $71$ , and we have an $8c$ . We can clearly see that $56$ is a multiple of $8$ , and any larger than $56$ would result in $c$ being larger than $9$ . Therefore, our only solution is $a = 2, b = 2, c = 7$ . Our answer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$

Solution:
As shown in Solution 1, we have that $99a = 71b + 8c$
Note that by the divisibility rule for $9$ , we have $a+b+c \equiv a \pmod{9}$ .  Since $b$ and $c$ are base- $9$ digits, we can say that $b+c = 0$ or $b+c=9$ .  The former possibility can be easily eliminated, and thus $b+c=9$ .  Next, we write the equation from Solution 1 as $99a = 63b + 8(b+c)$ , and dividing this by $9$ gives $11a = 7b+8$ .  Taking both sides modulo $7$ , we have $4a \equiv 1 \pmod{7}$ .  Multiplying both sides by $2$ gives $a\equiv 2 \pmod{7}$ , which implies $a=2$ .  From here, we can find that $b=2$ and $c=7$ , giving an answer of $\boxed{227}$


Answer: 9.0

Find the units digit of the decimal expansion of \[\left(15 + \sqrt{220}\right)^{19} + \left(15 + \sqrt{220}\right)^{82}.\]
$\textbf{(A)}\ 0\qquad  \textbf{(B)}\ 2\qquad  \textbf{(C)}\ 5\qquad  \textbf{(D)}\ 9\qquad  \textbf{(E)}\ \text{none of these}$

Solution:
Let $A=15+\sqrt{220}$ and $B=15-\sqrt{220}.$ Note that $A^{19}+B^{19}$ and $A^{82}+B^{82}$ are both integers: When we expand (Binomial Theorem) and combine like terms for each expression, the rational terms are added and the irrational terms are canceled.
We have \begin{align*} A^{19}+B^{19} &= \left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{1}15^{18}\sqrt{220}^1+\cdots+\binom{19}{19}15^0\sqrt{220}^{19}\right] + \left[\binom{19}{0}15^{19}\sqrt{220}^0-\binom{19}{1}15^{18}\sqrt{220}^1+\cdots-\binom{19}{19}15^0\sqrt{220}^{19}\right] \\ &= 2\left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{2}15^{17}\sqrt{220}^2+\cdots+\binom{19}{18}15^1\sqrt{220}^{18}\right] \\ &= 2\left[\binom{19}{0}15^{19}+\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9\right]. \end{align*} Similarly, we have \[A^{82}+B^{82}=2\left[\binom{82}{0}15^{82}+\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}\right].\] We add the two equations and take the sum modulo $10:$ \begin{align*} \left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right) &= 2\Biggl[\binom{19}{0}15^{19}+\phantom{ }\underbrace{\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9}_{0\pmod{10}}\phantom{ }\Biggr]+2\Biggl[\binom{82}{0}15^{82}+\phantom{ }\underbrace{\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}}_{0\pmod{10}}\phantom{ }\Biggr] \\ &\equiv 2\left[\binom{19}{0}15^{19}\right]+2\left[\binom{82}{0}15^{82}\right] \\ &\equiv 2\left[5\right]+2\left[5\right] \\ &\equiv 0\pmod{10}. \end{align*} It is clear that $0<B^{82}<B^{19}<B<0.5,$ from which $0<B^{19}+B^{82}<1.$ We conclude that the units digit of the decimal expansion of $B^{19}+B^{82}$ is $0.$ Since the units digit of the decimal expansion of $\left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right)$ is $0,$ the units digit of the decimal expansion of $A^{19}+A^{82}$ is $\boxed{9}.$


Answer: 828.0

Solution:
Suppose that $52+6\sqrt{43}$ is in the form of $(a + b\sqrt{43})^2$ FOILing yields that $52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}$ . This implies that $a$ and $b$ equal one of $\pm3, \pm1$ . The possible sets are $(3,1)$ and $(-3,-1)$ ; the latter can be discarded since the square root must be positive. This means that $52 + 6\sqrt{43} = (\sqrt{43} + 3)^2$ . Repeating this for $52-6\sqrt{43}$ , the only feasible possibility is $(\sqrt{43} - 3)^2$
Rewriting, we get $(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3$ . Using the difference of cubes , we get that $[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]$ $= (6)(3 \cdot 43 + 9) = \boxed{828}$ .
Note: You can also just use the formula $(a + b)^2 = a^2 + 2ab + b^2$ instead of foiling

Solution:
The $3/2$ power is quite irritating to work with so we look for a way to eliminate that. Notice that squaring the expression will accomplish that.
Let $S$ be the sum of the given expression. \[S^2= ((52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2})^2\] \[S^2 = (52+6\sqrt{43})^{3} + (52-6\sqrt{43})^{3} - 2((52+6\sqrt{43})(52-6\sqrt{43}))^{3/2}\] After doing the arithmetic (note that the first two terms will have some cancellation and that the last term will simplify quickly using difference of squares), we arrive at $S^2 = 685584$ which gives $S=\boxed{828}$

Solution: Factor as a difference of cubes. \[\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[\left(\left(\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\right)^2+\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}}+\left(\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right)^2\right)\right] =\] \[\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+\left(52^2-\left(36\right)\left(43\right)\right)^{\frac{1}{2}}\right] =\] \[\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+34\right].\] We can simplify the left factor as follows. \[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x\] \[104-2\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x^2\] \[104-68 = x^2\] \[36 = x^2.\] Since $\left(52+6\sqrt{43}\right)^{\frac{1}{2}} > \left(52-6\sqrt{43}\right)^{\frac{1}{2}}$ , we know that $x=6$ , so our final answer is $(6)(138) = \boxed{828}$

Solution: Let $x=52+6\sqrt{43}$ $y=52-6\sqrt{43}$ . Similarly to solution 2, we let \[S=x^{\frac{3}{2}}+y^{\frac{3}{2}}\] \begin{align*} S^2&=(x^{\frac{3}{2}}+y^{\frac{3}{2}})^2\\ &=x^3+y^3+2x^{\frac{3}{2}}y^{\frac{3}{2}} \end{align*} The expression can be simplified as follow \begin{align*} S^2&=x^3+y^3+2x^{\frac{3}{2}}y^{\frac{3}{2}}\\ &=(x+y)(x^2-xy+y^2)+2(xy)^{\frac{3}{2}}\\ &=(x+y)((x+y)^2-xy)+2\sqrt{xy}^3\\ &=(x+y)((x+y)^2-\sqrt{xy}^2)+\sqrt{xy}^3\\ &=(x+y)(x+y+\sqrt{xy})(x+y-\sqrt{xy})+2\sqrt{xy}^3\\ &=104((104+34)(104-34)+2\cdot34^3\\ &=685584 \end{align*} Thus $S=\sqrt{685584}=\boxed{828}$

Solution:
(Similar to Solution 3, but with substitution)
Let $a=\sqrt{52+6\sqrt{43}}$ and $b=\sqrt{52-6\sqrt{43}}.$ We want to find $a^3-b^3=(a-b)(a^2+ab+b^2).$
We have \[a^2+b^2=102,\text{ and}\] \[ab=\sqrt{(52+6\sqrt{43})(52-6\sqrt{43})}=\sqrt{1156}=34.\] Then, $(a-b)^2=a^2+b^2-2ab=104-2\cdot 34= 36\implies a-b=6.$
Our answer is \[a^3-b^3=(a-b)(a^2+b^2+ab)=6\cdot 138=\boxed{828.}\]

Solution:
(Similar to Solution 1, but expanding the cubes instead)
Like in Solution 1, we have $\sqrt{52 + 6\sqrt{43}} = \sqrt{43} + 3$ and $\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3.$
Therefore we have that $(52 + 6\sqrt{43})^{3/2} - (52 + 6\sqrt{43})^{3/2}$ $= \sqrt{52 + 6\sqrt{43}}^3 - \sqrt{52 - 6\sqrt{43}}^3$ $= (\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3.$
From here, we use the formula $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ . Applying them to our problem we get that $(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3 = (27 + 27\sqrt{43} + 9 \cdot 43 + 43\sqrt{43}) - (-27 + 27\sqrt{43} - 9*43 + 43\sqrt{43}).$ We see that all the terms with square roots cancel, leaving us with $2 (27 + 9 \cdot 43) = 2 \cdot 414 = \boxed{828}.$


Answer: 15.0

Solution:
We know that $\tan(\arctan(x)) = x$ so we can repeatedly apply the addition formula, $\tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ . Let $a = \cot^{-1}(3)$ $b=\cot^{-1}(7)$ $c=\cot^{-1}(13)$ , and $d=\cot^{-1}(21)$ . We have
so
and
so
Thus our answer is $10\cdot\frac{3}{2}=\boxed{015}$

Solution: On the coordinate plane, let $O=(0,0)$ $A_1=(3,0)$ $A_2=(3,1)$ $B_1=(21,7)$ $B_2=(20,10)$ $C_1=(260,130)$ $C_2=(250,150)$ $D_1=(5250,3150)$ $D_2=(5100,3400)$ , and $H=(5100,0)$ . We see that $\cot^{-1}(\angle A_2OA_1)=3$ $\cot^{-1}(\angle B_2OB_1)=7$ $\cot^{-1}(\angle C_2OC_1)=13$ , and $\cot^{-1}(\angle D_2OD_1)=21$ . The sum of these four angles forms the angle of triangle $OD_2H$ , which has a cotangent of $\frac{5100}{3400}=\frac{3}{2}$ , which must mean that $\cot( \cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21)=\frac{3}{2}$ . So the answer is $10\cdot\left(\frac{3}{2}\right)=\boxed{015}.$

Solution:
Recall that $\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta$ and that $\arg(a + bi) = \tan^{-1}\frac{b}{a}$ . Then letting $w = 1 + 3i, x = 1 + 7i, y = 1 + 13i,$ and $z = 1 + 21i$ , we are left with
\[10\cot(\frac{\pi}{2} - \arg w + \frac{\pi}{2} - \arg x + \frac{\pi}{2} - \arg y + \frac{\pi}{2} - \arg z) = 10\cot(2\pi - \arg wxyz)\] \[= -10\cot(\arg wxyz).\]
Expanding $wxyz$ , we are left with \[(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)\] \[= (2+i)(13+i)(21+i)\] \[= (25+15i)(21+i)\] \[= (5+3i)(21+i)\] \[= (102+68i)\] \[= (3+2i)\] \[= 10\cot \tan^{-1}\frac{2}{3}\] \[= 10 \cdot \frac{3}{2} = \boxed{015}\]


Answer: 93.0

Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ $a_2$ $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$

Solution:
One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of $a_1$ , then use that to calculate $a_2$ and sum another arithmetic series to get our answer.
A somewhat quicker method is to do the following: for each $n \geq 1$ , we have $a_{2n - 1} = a_{2n} - 1$ .  We can substitute this into our given equation to get $(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137$ .  The left-hand side of this equation is simply $2(a_2 + a_4 + \ldots + a_{98}) - 49$ , so our desired value is $\frac{137 + 49}{2} = \boxed{093}$

Solution:
We want to find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ , which can be rewritten as $a_1+1+a_2+2+a_3+\ldots+a_{49}+49 \implies a_1+a_2+a_3+\ldots+a_{49}+\frac{49 \cdot 50}{2}$ .
We can split $a_1+a_2+a_3+\ldots+a_{98}$ into two parts: \[a_1+a_2+a_3+\ldots+a_{49}\] and \[a_{50}+a_{51}+a_{52}+\ldots+a_{98}\] Note that each term in the second expression is $49$ greater than the corresponding term, so, letting the first equation be equal to $x$ , we get $a_1+a_2+a_3+\ldots+a_{98}=137=2x+49^2 \implies x=\frac{137-49^2}{2}$ . Calculating $49^2$ by sheer multiplication is not difficult, but you can also do $(50-1)(50-1)=2500-100+1=2401$ . We want to find the value of $x+\frac{49 \cdot 50}{2}=x+49 \cdot 25=x+1225$ . Since $x=\frac{137-2401}{2}$ , we find $x=-1132$ $-1132+1225=\boxed{93}$

Solution: Since we are dealing with an arithmetic sequence, \[a_2+a_4+a_6+a_8+\ldots+a_{98} = 49a_{50}\] We can also figure out that \[a_1+a_2+a_3+\ldots+a_{98} = a_1 + 97a_{50} = 137\] \[a_1 = a_{50}-49 \Rightarrow 98a_{50}-49 = 137\] Thus, $49a_{50} = \frac{137 + 49}{2} = \boxed{093}$


Answer: 3.0

Find the value of $x$ that satisfies the equation $25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}.$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 9$

Solution:
Manipulate the powers of $5$ in order to get a clean expression.
\[\frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 25^{\frac{17}{x}}} = \frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 5^{\frac{34}{x}}} = 5^{\frac{48}{x}-(\frac{26}{x}+\frac{34}{x})} = 5^{-\frac{12}{x}}\] \[25^{-2} = (5^2)^{-2} = 5^{-4}\] \[5^{-4} = 5^{-\frac{12}{x}}\]
If two numbers are equal, and their bases are equal, then their exponents are equal as well.  Set the two exponents equal to each other.
\begin{align*}-4&=\frac{-12}{x}\\ -4x&=-12\\ x&=\boxed{3}


Answer: 50.0

Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\] $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$

Solution: We can group each subtracting pair together: \[(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).\] After subtracting, we have: \[2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).\] There are $50$ even numbers, therefore there are $\dfrac{50}{2}=25$ even pairs. Therefore the sum is $2 \cdot 25=\boxed{50}$

Solution: Since our list does not end with one, we divide every number by 2 and we end up with \[50-49+48-47+ \ldots +4-3+2-1\] We can group each subtracting pair together: \[(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).\] There are now $25$ pairs of numbers, and the value of each pair is $1$ .  This sum is $25$ . However, we divided by $2$ originally so we will multiply $2*25$ to get the final answer of $\boxed{50}$


Answer: 32.0

Find the value(s) of $x$ such that $8xy - 12y + 2x - 3 = 0$ is true for all values of $y$
$\textbf{(A) } \frac23 \qquad \textbf{(B) } \frac32 \text{ or } -\frac14 \qquad \textbf{(C) } -\frac23 \text{ or } -\frac14 \qquad \textbf{(D) } \frac32 \qquad \textbf{(E) } -\frac32 \text{ or } -\frac14$

Solution:
We have $8xy - 12y + 2x - 3 = 4y(2x - 3) + (2x - 3) = (4y + 1)(2x - 3)$
As $(4y + 1)(2x - 3) = 0$ must be true for all $y$ , we must have $2x - 3 = 0$ , hence $\boxed{32}$


Answer: 1.0

Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square.

Solution:
We will first take the expression modulo $3$ . We get $2^n+12^n+2011^n \equiv -1^n+1^n \pmod 3$
Lemma 1: All perfect squares are equal to $0$ or $1$ modulo $3$ .
We can prove this by testing the residues modulo $3$ . We have $0^2 \equiv 0 \pmod 3$ $1^2 \equiv 1 \pmod 3$ , and $2^2 \equiv 1 \pmod 3$ , so the lemma is true.
We know that if $n$ is odd, $-1^n+1^n \equiv 0 \pmod 3$ , which satisfies the lemma's conditions. However, if $n$ is even, we get $2 \pmod 3$ , which does not satisfy the lemma's conditions. So, we can conclude that $n$ is odd.
Now, we take the original expression modulo $4$ . For right now, we will assume that $n>1$ , and test $n=1$ later. For $n>1$ $2^n \equiv 0 \pmod 4$ , so $2^n+12^n+2011^n=-1^n \pmod 4$
Lemma 2: All perfect squares are equal to $0$ or $1$ modulo $4$ .
We can prove this by testing the residues modulo $4$ . We have $0^2 \equiv 0 \pmod 4$ $1^2 \equiv 1 \pmod 4$ $2^2 \equiv 0 \pmod 4$ , and $3^2 \equiv 1 \pmod 4$ , so the lemma is true.
We know that if $n$ is even, $-1^n \equiv 0 \pmod 4$ , which satisfies the lemma's conditions. However, if $n$ is odd, $-1^n \equiv -1 \equiv 3 \pmod 4$ , which does not satisfy the lemma's conditions. Therefore, $n$ must be even.
However, a number cannot be even and odd at the same time, so this is impossible. Now, we only have to test $n=1$ . We know that $2^1+12^1+2011^1=45^2$ , so the only integer is $\boxed{1}$


Answer: 6097392.0

Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,...,N\}$ , one can still find $2016$ distinct numbers among the remaining elements with sum $N$

Solution:
Since any $2016$ elements are removed, suppose we remove the integers from $1$ to $2016$ . Then the smallest possible sum of $2016$ of the remaining elements is \[2017+2018+\cdots + 4032 = 1008 \cdot 6049 = 6097392\] so clearly $N\ge 6097392$ . We will show that $N=6097392$ works.
$\vspace{0.2 in}$
$\{1,2\cdots 6097392\}$ contain the integers from $1$ to $6048$ , so pair these numbers as follows:
\[1, 6048\]
\[2, 6047\]
\[3, 6046\]
\[\cdots\]
\[3024, 3025\]
When we remove any $2016$ integers from the set $\{1,2,\cdots N\}$ , clearly we can remove numbers from at most $2016$ of the $3024$ pairs above, leaving at least $1008$ complete pairs. To get a sum of $N$ , simply take these $1008$ pairs, all of which sum to $6049$ . The sum of these $2016$ elements is $1008 \cdot 6049 = 6097392$ , as desired.
$\vspace{0.2 in}$
We have shown that $N$ must be at least $6097392$ , and that this value is attainable. Therefore our answer is $\boxed{6097392}$


Answer: 2.2

Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpositions?
$(\textbf{A})\: 1.6\qquad(\textbf{B}) \: 1.8\qquad(\textbf{C}) \: 2.0\qquad(\textbf{D}) \: 2.2\qquad(\textbf{E}) \: 2.4$

Solution:
After the first swap, we do casework on the next swap.
Case 1: Silva swaps the two balls that were just swapped
There is only one way for Silva to do this, and it leaves 5 balls occupying their original position.
Case 2: Silva swaps one ball that has just been swapped with one that hasn't been swapped
There are two ways for Silva to do this, and it leaves 2 balls occupying their original positions.
Case 3: Silva swaps two balls that have not been swapped
There are two ways for Silva to do this, and it leaves 1 ball occupying their original position.
Our answer is the average of all 5 possible swaps, so we get \[\frac{5+2\cdot2+2\cdot1}{5}=\frac{11}5=\boxed{2.2}.\]

Solution: The "expected value" in the question tips us off to this technique. Consider any ball. The probability it returns to the same position is the probability of being swapped twice plus the probability of never being swapped: \[\frac{2}{5} \cdot \frac{1}{5} + \left(\frac{3}{5}\right)^2 = \frac{11}{25}.\] Multiply by 5 for 5 balls to get $\boxed{2.2}.$


Answer: 3.0

Five cards are lying on a table as shown.
\[\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\  \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}\]

Solution: Using the answer choices, we see that P and Q are logically equivalent (both are non-vowel letters) and so are $4$ and $6$ (both are even numbers). Thus, if one of these is correct, then the other option in its pair must also be correct. There can't be 2 answers. So, the only remaining answer choice is $\boxed{3}$


Answer: 150.0

Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$

Solution:
Firstly, observe that it is not possible for a single student to receive $4$ or $5$ awards because this would mean that one of the other students receives no awards. Thus, each student must receive either $1$ $2$ , or $3$ awards. If a student receives $3$ awards, then the other two students must each receive $1$ award; if a student receives $2$ awards, then another student must also receive $2$ awards and the remaining student must receive $1$ award. We consider each of these two cases in turn.
If a student receives three awards, there are $3$ ways to choose which student this is, and $\binom{5}{3}$ ways to give that student $3$ out of the $5$ awards. Next, there are $2$ students left and $2$ awards to give out, with each student getting one award. There are clearly just $2$ ways to distribute these two awards out, giving $3\cdot\binom{5}{3}\cdot 2=60$ ways to distribute the awards in this case.
In the other case, two students receive $2$ awards and one student recieves $1$ award . We know there are $3$ choices for which student gets $1$ award. There are $\binom{3}{1}$ ways to do this. Then, there are $\binom{5}{2}$ ways to give the first student his two awards, leaving $3$ awards yet to distribute. There are then $\binom{3}{2}$ ways to give the second student his $2$ awards. Finally, there is only $1$ student and $1$ award left, so there is only $1$ way to distribute this award. This results in $\binom{5}{2}\cdot\binom{3}{2}\cdot 1\cdot 3 =90$ ways to distribute the awards in this case. Adding the results of these two cases, we get $60+90=\boxed{150}$

Solution:
Upon inspection (specified in the above solution), there are two cases of the distribution of awards to the students: one student gets 3 awards and the other each get 1 award or one student gets 1 award and the other two get 2 awards.
In the first case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 3 awards. From here, there are $\binom{5}{3} = 10$ ways to choose the 3 awards from the 5 total awards. Now, one person has $2$ choices for awards and the other has $1$ choice for the award. Thus, the total number of ways to choose awards in this case is $3 \cdot 10 \cdot 2 \cdot 1 = 60$
In the other case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 1 award, and $5$ choices for his/her award. Then, one person has $\binom{4}{2} = 6$ ways to have his/her awards and the other person has $\dbinom{2}{2} = 1$ ways to have his/her awards. This gives $3 \cdot 5 \cdot 6 \cdot 1  = 90$ ways for this case.
Adding these cases together, we get $60 + 90 = 150$ ways to distribute the awards, or choice $\boxed{150}$

Solution:
Without the restriction that each student receives at least one award, we could take each of the awards and choose one of the $3$ students to give it to. This would be $3^5$ ways to distribute the awards in total. Now we need to subtract the cases where at least one student doesn't receive an award. If a student doesn't receive an award, there are $3$ choices for which student that is, so $2^5$ ways of choosing a student to receive each of the awards; in total, $3\cdot32=96$
However, if $2$ students both don't receive an award, then this case would be counted twice among the $96$ , so we need to add back in these cases. Said in other words, $2$ students not receiving an award is equivalent to $1$ student receiving $5$ awards, and there are $3$ choices for whom that student would be. To finish, the total number of ways to distribute the awards is $243 - 96+3$ , or $\boxed{150}$


Answer: 2.0

Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

Solution:
Assume that Edie and Dee were originally in seats 3 and 4. If this were so, there is no possible position for which Bea can move 2 seats to the right. The same applies for seats 2 and 3. This means that either Edie or Dee was originally in an edge seat. If Edie and Dee were in seats 1 and 2, then Bea must have been in seat 3, which would mean that seat 5 would now be occupied and the positioning would not work. So, Edie and Dee are in seats 4 and 5. This means that Bea was originally in seat 1. Ceci must have been in seat 3 to keep seat 1 open, which leaves seat 2.
Thus, Ada was in seat $\boxed{2}$


Answer: 2.0

Solution: Note that the person (out of A,B,C) that moves the most, moves the amount equal to the sum of what the other 2 move. They essentially make a cycle. D & E are seat fillers and can be ignored. A,B,C take up either seats 1,2,3 or 2,4,5. In each case you find A was originally in seat $\boxed{2}$

Solution: Note that the net displacements in the right direction sum up to $0$ . The sum of the net displacements of Bea, Ceci, Dee, Edie is $2-1 = 1$ , so Ada moved exactly $1$ place to the left. Since Ada ended on an end seat, she must have started on seat $\boxed{2}$


Answer: 2.0

Solution:
Assume that Edie and Dee were originally in seats 3 and 4. If this were so, there is no possible position for which Bea can move 2 seats to the right. The same applies for seats 2 and 3. This means that either Edie or Dee was originally in an edge seat. If Edie and Dee were in seats 1 and 2, then Bea must have been in seat 3, which would mean that seat 5 would now be occupied and the positioning would not work. So, Edie and Dee are in seats 4 and 5. This means that Bea was originally in seat 1. Ceci must have been in seat 3 to keep seat 1 open, which leaves seat 2.
Thus, Ada was in seat $\boxed{2}$


Answer: 2.0

Solution: Note that the person (out of A,B,C) that moves the most, moves the amount equal to the sum of what the other 2 move. They essentially make a cycle. D & E are seat fillers and can be ignored. A,B,C take up either seats 1,2,3 or 2,4,5. In each case you find A was originally in seat $\boxed{2}$

Solution: Note that the net displacements in the right direction sum up to $0$ . The sum of the net displacements of Bea, Ceci, Dee, Edie is $2-1 = 1$ , so Ada moved exactly $1$ place to the left. Since Ada ended on an end seat, she must have started on seat $\boxed{2}$


Answer: 191.0

Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution:
For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are $14!$ arrangements without restrictions.
First, there are $\binom{7}{5}$ ways to choose the man-woman diameters. Then, there are $10\cdot8\cdot6\cdot4\cdot2$ ways to place the five men each in a man-woman diameter. Finally, there are $9!$ ways to place the nine women without restrictions.
Together, the requested probability is \[\frac{\tbinom{7}{5}\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{21\cdot(10\cdot8\cdot6\cdot4\cdot2)}{14\cdot13\cdot12\cdot11\cdot10} = \frac{48}{143},\] from which the answer is $48+143 = \boxed{191}.$

Solution:
We can simply just loop through each of the men and find the probability that the person opposite from him is a woman.
Start by sitting down the $1$ st man. The probability that the person opposite to him is a woman is $\frac{9}{13}$ since out of the $13$ people who can sit opposite to him, $9$ can be a woman. With the $2$ nd man, we can use the same logic: there are $11$ people who can sit opposite to him, but only $8$ of them are a woman, so the probability is $\frac{8}{11}.$ We use the same logic for the $3$ rd, $4$ th and $5$ th men to get probabilities of $\frac{7}{9}$ $\frac{6}{7}$ and $\frac{5}{5},$ respectively.
Multiplying these probabilities, we get a final answer of \[\frac{9}{13}\cdot\frac{8}{11}\cdot\frac{7}{9}\cdot\frac{6}{7}\cdot\frac{5}{5}=\frac{48}{143}\longrightarrow\boxed{191}.\]

Solution:
This problem is equivalent to solving for the probability that no man is standing diametrically opposite to another man. We can simply just construct this.
We first place the $1$ st man anywhere on the circle, now we have to place the $2$ nd man somewhere around the circle such that he is not diametrically opposite to the first man. This can happen with a probability of $\frac{12}{13}$ because there are $13$ available spots, and $12$ of them are not opposite to the first man.
We do the same thing for the $3$ rd man, finding a spot for him such that he is not opposite to the other $2$ men, which would happen with a probability of $\frac{10}{12}$ using similar logic. Doing this for the $4$ th and $5$ th men, we get probabilities of $\frac{8}{11}$ and $\frac{6}{10}$ respectively.
Multiplying these probabilities, we get, \[\frac{12}{13}\cdot\frac{10}{12}\cdot\frac{8}{11}\cdot\frac{6}{10}=\frac{48}{143}\longrightarrow\boxed{191}.\]

Solution:
Assume that rotations and reflections are distinct arrangements, and replace men and women with identical M's and W's, respectively. (We can do that because the number of ways to arrange $5$ men in a circle and the number of ways to arrange $9$ women in a circle, are constants.) The total number of ways to arrange $5$ M's and $9$ W's is $\binom{14}{5} = 2002.$
To count the number of valid arrangements (i.e. arrangements where every M is diametrically opposite a W), we notice that exactly $2$ of the pairs of diametrically opposite positions must be occupied by $2$ W's. There are $\binom{7}{2} = 21$ ways to choose these $2$ pairs. For the remaining $5$ pairs, we have to choose which position is occupied by an M and which is occupied by a W. This can be done in $2^{5} = 32$ ways. Therefore, there are $21*32 = 672$ valid arrangements.
Therefore, the probability that an arrangement is valid is $\frac{672}{2002} = \frac{48}{143}$ for an answer of $\boxed{191}.$

Solution:
To start off, we calculate the total amount of ways to organize all $14$ people irrespective of any constraints. This is simply ${14\choose5} = 2002$ , because we just count how many ways we can place all $5$ men in any of the $14$ slots.
Since men cannot be diametrically opposite with each other, because of the constraints, placing down one man in any given spot will make another spot on the opposite side of the circle unable to hold any men. This means that placing down one man will effectively take away $2$ spots.
There are $14$ possible slots the first man can be placed. Once that man was placed, the next man only have $12$ possible slots because the slot that the first man is in is taken and the diametrically opposite spot to the first man can't have any men. Similar logic applies for the third man, who has $10$ possible slots. The fourth man has $8$ possible slots, and the fifth man has $6$ possible slots.
This means the number of ways you can place all $5$ men down is $14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$ . However, since the men are all indistinct from each other, you also have to divide that value by $5! = 120$ , since there are $120$ ways to arrange the $5$ men in each possible positioning of the men on the circle. This means the total number of ways to arrange the men around the circle so that none of them are diametrically opposite of each other is: $\frac{14 \cdot 12 \cdot 10 \cdot 8 \cdot 6}{5!} = 672$ . The women simply fill in the rest of the available slots in each arrangement of men.
Thus, the final probability is $\frac{672}{2002} = \frac{48}{143}$ , meaning the answer is $48 + 143 = \boxed{191}$

Solution:
We will first assign seats to the men. The first man can be placed in any of the $14$ slots. The second man can be placed in any of the remaining $13$ seats, except for the one diametrically opposite to the first man. So, there are $13 - 1 = 12$ ways to seat him. With a similar argument, the third man can be seated in $10$ ways, the fourth man in $8$ ways and the last man in $6$ ways.
So, the total number of ways to arrange the men is $14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$
The women go to the remaining $9$ spots. Note that since none of the seats diametrically opposite to the men is occupied, each man is opposite a woman. The number of ways to arrange the women is therefore, simply $9!$ , meaning that the total number of ways to arrange the people with restrictions is $14 \cdot 12 \cdot 10 \cdot 8 \cdot 6 \cdot 9!$ In general, there are $14!$ ways to arrange the people without restrictions. So, the probability is \[\frac{14 \cdot 12 \cdot 10 \cdot 8 \cdot 6 \cdot 9!}{14!} = \frac{8 \cdot 6}{13 \cdot 11} = \frac{48}{143}.\] The answer is $48 + 143 = \boxed{191}$

Solution:
First pin one man on one seat (to ensure no rotate situations). Then there are $13!$ arrangements.
Because $5$ men must have women at their opposite side, we consider the $2$ nd man and the woman opposite as one group and name it $P_2.$ There are $4$ groups, $P_1, P_2, P_3, P_4$ except the first man pinned on the same point. And for the rest $4$ women, name them $P_5$ and $P_6.$ First to order $P_1, P_2, P_3, P_4, P_5, P_6,$ there are $6!$ ways. For the $1$ st man, there are $9$ women to choose, $8$ for the $2$ nd, $\ldots,$ $5$ for the $5$ th, and then for the $2$ women pairs $3$ and $1.$ Because every $2$ person in the group have chance to change their position, there are $2^6$ possibilities.
So the possibility is \[P=\frac{6!\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 3\cdot 1 \cdot 2^6}{13!}=\frac{48}{143}.\]
The answer is $48+143=\boxed{191}.$

Solution: We get around the condition that each man can't be opposite to another man by simply considering all $7$ diagonals, and choosing $5$ where there will be a single man. For each diagonal, the man can go on either side, and there are $\binom{14}{5}$ ways to arrange the men and the women in total. Thus our answer is $\frac{\binom{7}{5}\cdot 2^5}{\binom{14}{5}} = \frac{48}{143}.$ We get $48 + 143 = \boxed{191}$

Solution:
We can find the probability of one arrangement occurring, and multiply it by the total number of arrangements.
The probability of a man being in any specific position is $\frac{5}{14}.$ The probability of a woman being across from him is $\frac{9}{13}.$ The probability of a man being in any valid position is now $\frac{4}{12},$ and the probability of a woman being across from him is $\frac{8}{11},$ and so forth. We stop when there are no more men left. Multiplying these probabilities together,
\[P(\mathrm{One\ successful\ outcome})=\frac{5}{14}\cdot \frac{9}{13}\cdot \frac{4}{12}\cdot \frac{8}{11}\cdot \frac{3}{10}\cdot \frac{7}{9}\cdot \frac{2}{8}\cdot \frac{1}{6}\cdot \frac{6}{7}\cdot \frac{5}{5} = \frac{1}{2002}.\] To find the total number of successful outcomes, we consider the diagonals; the total number of diagonals to be made is $\binom75$ , since there are $7$ total diagonals, and we want to choose $5$ of them to connect a man to a woman. For each of these diagonals, the man can be on either side of the diagonal.
It follows that there are $2$ possibilities for each diagonal (man on one side, woman on the other, and vice versa). There are $5$ diagonals with a man and a woman, so there are $2^5$ different ways for these diagonals to appear.
There are $\binom75$ successful diagonals, and for each of these diagonals, there are $2^5$ ways to seat the men and the women, there are $\binom75$ $\cdot 2^5$ successful outcomes.
Recall that \[P(\mathrm{Any\ successful\ outcome})=P(\mathrm{One\ successful\ outcome})\cdot P(\mathrm{Total\ number\ of\ successful\ outcomes}).\] Therefore, \[P(\mathrm{Any\ successful\ outcome}) = \frac{1}{2002}\cdot 2^5\cdot \binom75 = \frac{1}{2002}\cdot 2^5\cdot 21 = \frac{2^5\cdot 21}{2002} = \frac{48}{143}.\] The requested sum is $48+143=\boxed{191}.$


Answer: 8.0

Five people are sitting at a round table. Let $f\geq 0$ be the number of people sitting next to at least 1 female and $m\geq0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is
$\mathrm{(A) \ 7 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 9 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 11 }$

Solution:
Suppose there are more men than women; then there are between zero and two women.
If there are no women, the pair is $(0,5)$ . If there is one woman, the pair is $(2,5)$
If there are two women, there are two arrangements: one in which they are together, and one in which they are apart, giving the pairs $(4,5)$ and $(3,4)$
All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so $\boxed{8}$

Solution: Denote $T_n$ as the number of such pairs for $n$ people. Then for $T_{n-1}$ , when we add an extra spot, we can either have a male or female giving two options. Note that these two options however double the value of $T_{n-1}$ . Now if we note that $T_2=1$ , we have that $T_5=8$ , so that the answer is $\boxed{8}$


Answer: 171.0

Five test scores have a mean (average score) of $90$ , a median (middle score) of $91$ and a mode (most frequent score) of $94$ .  The sum of the two lowest test scores is
$\text{(A)}\ 170 \qquad \text{(B)}\ 171 \qquad \text{(C)}\ 176 \qquad \text{(D)}\ 177 \qquad \text{(E)}\ \text{not determined by the information given}$

Solution: Because there was an odd number of scores, $91$ must be the middle score. Since there are two scores above $91$ and $94$ appears the most frequent (so at least twice) and $94>91$ $94$ appears twice. Also, the sum of the five numbers is $90 \times 5 =450$ . Thus, the sum of the lowest two scores is $450-91-94-94= \boxed{171}$


Answer: 544.0

Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way).

Solution:
It is obvious that any configuration of one-way roads which contains a town whose roads all lead into it or lead out of it cannot satisfy the given. We claim that any configuration which does not have a town whose roads all lead into it or lead out of it does satisfy the given conditions. Now we show that a loop of $3$ or more towns exist. Pick a town, then choose a neighboring town to travel to $5$ times. Of these $6$ towns visited, at least two of them must be the same; therefore there must exist a loop of $3$ or more towns because a loop of $2$ towns cannot exist. We want to show that the loop can be reached from any town, and any town can be reached from the loop.
$\textbf{Case 1}$ . The loop has $5$ towns.
Clearly every town can be reached by going around the loop.
$\textbf{Case 2}$ . The loop has $4$ towns.
The town not on the loop must have a road leading to it. This road comes from a town on the loop. Therefore this town can be reached from the loop. This town not on the loop must also have a road leading out of it. This road leads to a town on the loop. Therefore the loop can be reached from the town.
$\textbf{Case 3}$ . The loop has $3$ towns.
There are two towns not on the loop; call them Town $A$ and Town $B$ . Without loss of generality assume $A$ leads to $B$ . Because a road must lead to $A$ , the town where this road comes from must be on the loop. Therefore $A$ and therefore $B$ can be reached from the loop. Because a road must lead out of $B$ , the town it leads to must be on the loop. Therefore the loop can be reached from $B$ and also $A$
The number of good configurations is the total number of configurations minus the number of bad configurations. There are $2^{{5\choose2}}$ total configurations. To find the number of bad configurations in which a town exists such that all roads lead to it, there are $5$ ways to choose this town and $2^6$ ways to assign the six other roads that do not connect to this town. The same logic is used to find the number of bad configurations in which a town exists such that all roads lead out of it. It might be tempting to conclude that there are $5 \cdot 2^6+5 \cdot 2^6$ bad configurations, but the configurations in which there exists a town such that all roads lead to it and a town such that all roads lead out of it are overcounted. There are $5$ ways to choose the town for which all roads lead to it, $4$ ways to choose the town for which all roads lead out of it, and $2^3$ ways to assign the remaining $3$ roads not connected to either of these towns. Therefore, the answer is $2^{{5\choose2}}-(5 \cdot 2^6+5 \cdot 2^6-5\cdot 4 \cdot 2^3)=\boxed{544}$

Solution:
The only way a town does not meet the conditions in the question is if the town has either all roads leading towards it or all roads leading away from it. For example, if all roads lead away from Town $A$ , there is no way to reach the town starting from Towns $B$ $C$ $D$ , or $E$ . If all roads lead towards Town $A$ , there is no way to reach any other town starting from Town $A$ . Thus, we will count the ways this occurs. (To show this, first note that there must always be a directed path that visits every town once, say $A \rightarrow C \rightarrow B \rightarrow E \rightarrow D.$ (Redei's Theorem). Taking cases, it can be seen that if $A$ does not have all its roads leading, then $E$ must have its roads leading into it.)
$\textbf{Case 1}$ . WLOG, Town $A$ has all roads leading towards it.
In this case, the four roads leading to Town $A$ have already been determined. There are $6$ roads that have not been given directions. Each of these roads have $2$ options: it can lead towards one town or the other town. Thus, there are $2^6$ arrangements of roads.
However, we must consider the $5$ towns in which this scenario can occur: there are $5 \cdot 2^6=320$ arrangements.
$\textbf{Case 2}$ . WLOG, Town $A$ has all roads leading away.
Notice that this case is symmetrical to the first case. Thus, there are $320$ arrangements.
$\textbf{Case 3}$ . WLOG, Town $A$ has all roads leading towards it and Town $B$ has all roads leading away.
We must check for double-counted cases. Drawing a flow diagram, we see that this case determines $7$ roads. For the undetermined $3$ roads, there are $2^3=8$ arrangements. However, we must again consider the different ways that this case can occur. There are $5$ choices for towns with roads leading towards it $4$ choices for roads leading away. Thus, the total double-counted cases are $5 \cdot 4 \cdot 8=160$ arrangements.
We must subtract the cases we counted from the total. There are a total of $10$ roads with $2$ possible arrangements each.
Therefore the total number of cases is $2^{10} =1024$ , and the number of cases that meet the conditions is $1024 - (320+320-160) = \boxed{544}$ arrangements.

Solution: Assume the five towns are named A, B, C, D, and E. Draw roads that connect each of the five towns. First, figure out how many total roads there are with no restrictions. This would be $2^{10}$ , or $1024$ . Next, add together the number of restrictions (or incorrect cases) there are. One incorrect case happens if we can't get to one of the five towns. When that happens, any roads that connect that town with any of the other four towns must be one way. Four roads are invalidated (because they are one way) as, assuming we can't get to town A, roads AB, AC, AD, and AE would all be one way. Since there are ${{5\choose1}}$ ways of choosing which town to restrict, there are $5\cdot2^{10-4}$ , or $320$ restrictions in this case. The next case is when we can't get to two of the towns. Let's assume we don't want to get to towns A and B. Then, roads AC, AD, AE, BC, BD, and BE are all invalidated.  Since there are ${{5\choose2}}$ of choosing which two towns to restrict, there are $10\cdot2^{10-6}$ , or $160$ restrictions in this case. Notice that when we restrict three towns, it is the same thing as restricting two towns. For example, if we restricted entry to towns C, D, and E from towns A and B, it's the same thing as if we restricted entry to towns A and B from towns C, D, and E, as the exact same roads are invalidated in both cases. Thus, our final answer is $1024-(320+160) = \boxed{544}$ arrangements.

Solution: As noted before, you can see that there are 2 ways that the condition cannot be met; it either has all roads leading into one city, or all roads leading out of one city. We then use complementary counting to count these cases, with PIE. Obviously there are $2^5*2^5$ ways to make roads (just draw a pentagon with all of the diagonals and assume that they are roads, and count the sides and then the diagonals). Then, make all roads leading in to another city. There are 4 roads leading into that city, so there are $2^6$ ways to designate that. Given that there are 5 cities, and it can also be the ways to go out of the city, you get $10*2^6$ ways NOT to get into the city. However, this disregards the case that 2 cities are like this, which is the PIE case. Quick inspection shows that there are no cases where 3 are like that, and inspection with le pentagon can also show you that there are $\binom{5}{2}*2$ (because they can all lead in and all lead out) $*2^3$ (because there are 3 roads left to choose). In the end, this comes out to $2^5*2^5-(10*2^6-2*\binom{5}{2}*2^3)=1024-(640-160)=\boxed{544}$

Solution:
We claim the following conditions:
Call the vertices $\omega_1, \omega_2, \omega_3, \omega_4, \omega_5$ . WLOG, let $\omega_1 \to \omega_2$ . Then, WLOG let $\omega_2 \to \omega_3$ . We have two ways to go from here:
If $\omega_3 \to \omega_1$ , we call it a "cycle".
We will prove (3):
Notice that once $\omega_4, \omega_5$ can reach one of the vertices on the "cycle", it can reach any of the other vertices on the same cycle simply by travelling along the arcs of the cycle. Additionally, one of $\omega_4, \omega_5$ must be able to travel to the cycle. Consider that it cannot, then $\omega_3 \to \omega_4$ and $\omega_5 \to \omega_4$ are forced. Now, $\omega_4 \to \omega_1, \omega_2, \omega_3$ which is on the cycle. Now, if $\omega_5 \to \omega_4$ , it will also be able to travel to the cycle.  Otherwise, the directed arc will directly travel to the cycle. Also, each of the vertices $\omega_1, \omega_2, \omega_3$ will travel to $\omega_4, \omega_5$
If our graph does not satisfy (1) and (2), $\omega_4, \omega_5$ will connect to the cycle on both directions. This can be shown below:
Clearly $\omega_4, \omega_5$ are connected. Assuming the graph does not satisfy (1) and (2), $\omega_4, \omega_5$ are connected to the cycle consisting of $\omega_1, \omega_2, \omega_3$ . WLOG, let $\omega_4 \to \omega_5$
Again, to avoid (2), $\omega_5 \to \omega_a$ where $a\in{1,2,3}$ . Now to avoid (1), one of $\omega_a \to \omega_4, a\in{1,2,3}$
Now it is clear that each vertex can go to any other vertex. This concludes our proof.
Say we try to avoid (3) so that WLOG $\omega_5 \to \omega_1$ . If we keep trying to avoid a "cycle", $\omega_3 \to \omega_4$ and $\omega_4 \to \omega_5$ which is a cycle.
Therefore, if there is any path $\omega_a \to \omega_b \to \omega_c$ , it will lead to a cycle and ultimately a valid path. The only way to avoid this is to have (1) or (2).
Consider (1):
WLOG, let $\omega_1$ have no arcs leading to it. Then all other vertices will have an arc directed toward them away from $\omega_1$ . Therefore, one path can have only 1 unique vertex that has no arcs leading to it. There are 5 ways to choose the vertex and $2^{\binom{5}{2}-4}=2^6$ ways for the remaining 6 arcs since they can be directed in any way. There are $2^6\cdot5=320$ paths for this condition.
Consider (2):
WLOG, let $\omega_1$ have all arcs leading to it. Similarly, all other vertices will have an arc that is directed away from them and points to $\omega_1$ . Therefore, only one unique vertex can have all arcs leading to it. There are 5 ways to choose the vertex and $2^{\binom{5}{2}-4}=2^6$ ways for the remaining arcs. There are also $2^6\cdot5=320$ paths for this condition.
However, we have overcounted paths that satisfy both (1) and (2). These paths will have one vertex that has all arcs leading to it and another that has all arcs directed away from it. There are $5\cdot4=20$ ways to choose these 2 vertices, and the remaining $\binom{3}{2}=3$ arcs each have 2 ways to be directed toward each other, so there are a total of $2^3\cdot 20=160$ overlapping paths.
By PIE, the number of invalid cases is $320+320-160=480$
The total number of cases is $2^{\binom{5}{2}}=2^{10}=1024$ since there are 10 edges and each can go 2 ways. By complementary counting, the number of valid paths is just $1024-480=\boxed{544}$


Answer: 23.0

Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt));  fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray);  xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4));  draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3));  label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$

Solution:
For $c\geq 1.5$ the shaded area is at most $1.5$ , which is too little. Hence $c<1.5$ , and therefore the point $(2,1)$ is indeed inside the shaded part, as shown in the picture.
Then the area of the shaded part is one less than the area of the triangle with vertices $(c,0)$ $(3,0)$ , and $(3,3)$ . The area of the entire triangle is $\frac{3(3-c)}2$ , therefore the area of the shaded part is $\frac{7-3c}{2}$
The entire figure has area $5$ , hence we want the shaded part to have area $\frac 52$ . Solving for $c$ , we get $c=\boxed{23}$

Solution:
The unit square is of area 1, so the five unit squares have area 5.
Therefore the shaded space must occupy 2.5. The missing unit square is of area 1, and if reconstituted the original triangle would be of area 3.5.
It can then be inferred: $(3-c) * 3 = 7$
$3-c=\frac{7}{3}$ , so $3-\frac{7}{3}=c$
$3-\frac{7}{3} = \frac{9-7}{3} = \frac{2}{3}$ $\boxed{23}$

Solution:
The shaded space of the object can become a triangle by adding a unit square to its bottom. This triangle has a base length of $3-c$ and a height of $3$ . The area of this triangle region is now (using the formula $A=bh/2$ for a triangle) $(9-3c)/2$ . But, remember that we have to subtract the area of the extra unit square from this area to get the area of the shaded region.
If you add 3 unit squares to the top of the unshaded portion, the figure is now a right trapezoid. Using the formula for a trapezoid, you can derive that its area is $(9+3c)/2$ . After you subtract the area of the 3 extra unit squares, we have derived the area of the unshaded portion.
Since these areas are both equal, we set them equal to each other and solve for $c$ $(9-3c)/2 -1 = (9+3c)/2 -3$ $c$ is now solved to be $\frac{2}{3}$ $\boxed{23}$

Solution:
We are looking for the area of the shaded region to be $\frac 52$ . We start by testing $(A) \frac 12$ . The area of the shaded region would be $\frac{(3-\frac 12)(3) }{2}-1=\frac {11}{4}$ when $c=\frac 12$ . This does not match our wanted answer.
We try $(B) \frac 35$ next. The area of the shaded region would be $\frac{(3-\frac 35)(3) }{2}-1=\frac {13}{5}$ when $c=\frac 35$ . This also does not match our desired answer.
We then try $(C) \frac 23$ . The area of the shaded region would be $\frac{(3-\frac 23)(3) }{2}-1=\frac {5}{2}$ when $c=\frac 23$ . Therefore, our answer is $\boxed{23}$


Answer: 23.0

Solution:
For $c\geq 1.5$ the shaded area is at most $1.5$ , which is too little. Hence $c<1.5$ , and therefore the point $(2,1)$ is indeed inside the shaded part, as shown in the picture.
Then the area of the shaded part is one less than the area of the triangle with vertices $(c,0)$ $(3,0)$ , and $(3,3)$ . The area of the entire triangle is $\frac{3(3-c)}2$ , therefore the area of the shaded part is $\frac{7-3c}{2}$
The entire figure has area $5$ , hence we want the shaded part to have area $\frac 52$ . Solving for $c$ , we get $c=\boxed{23}$

Solution:
The unit square is of area 1, so the five unit squares have area 5.
Therefore the shaded space must occupy 2.5. The missing unit square is of area 1, and if reconstituted the original triangle would be of area 3.5.
It can then be inferred: $(3-c) * 3 = 7$
$3-c=\frac{7}{3}$ , so $3-\frac{7}{3}=c$
$3-\frac{7}{3} = \frac{9-7}{3} = \frac{2}{3}$ $\boxed{23}$

Solution:
The shaded space of the object can become a triangle by adding a unit square to its bottom. This triangle has a base length of $3-c$ and a height of $3$ . The area of this triangle region is now (using the formula $A=bh/2$ for a triangle) $(9-3c)/2$ . But, remember that we have to subtract the area of the extra unit square from this area to get the area of the shaded region.
If you add 3 unit squares to the top of the unshaded portion, the figure is now a right trapezoid. Using the formula for a trapezoid, you can derive that its area is $(9+3c)/2$ . After you subtract the area of the 3 extra unit squares, we have derived the area of the unshaded portion.
Since these areas are both equal, we set them equal to each other and solve for $c$ $(9-3c)/2 -1 = (9+3c)/2 -3$ $c$ is now solved to be $\frac{2}{3}$ $\boxed{23}$

Solution:
We are looking for the area of the shaded region to be $\frac 52$ . We start by testing $(A) \frac 12$ . The area of the shaded region would be $\frac{(3-\frac 12)(3) }{2}-1=\frac {11}{4}$ when $c=\frac 12$ . This does not match our wanted answer.
We try $(B) \frac 35$ next. The area of the shaded region would be $\frac{(3-\frac 35)(3) }{2}-1=\frac {13}{5}$ when $c=\frac 35$ . This also does not match our desired answer.
We then try $(C) \frac 23$ . The area of the shaded region would be $\frac{(3-\frac 23)(3) }{2}-1=\frac {5}{2}$ when $c=\frac 23$ . Therefore, our answer is $\boxed{23}$


Answer: 12.0

Flora the frog starts at 0 on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance $m$ with probability $\frac{1}{2^m}$
What is the probability that Flora will eventually land at 10?
$\textbf{(A)}~\frac{5}{512}\qquad\textbf{(B)}~\frac{45}{1024}\qquad\textbf{(C)}~\frac{127}{1024}\qquad\textbf{(D)}~\frac{511}{1024}\qquad\textbf{(E)}~\frac{1}{2}$

Solution: Initially, the probability of landing at $10$ and landing past $10$ (summing the infinte series) are exactly the same. Landing before 10 repeats this initial condition, with a different irrelevant scaling factor. Therefore, the probability must be $\boxed{12}$


Answer: 336.0

For $-1<r<1$ , let $S(r)$ denote the sum of the geometric series \[12+12r+12r^2+12r^3+\cdots .\] Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a)=2016$ . Find $S(a)+S(-a)$

Solution: The sum of an infinite geometric series is $\frac{a}{1-r}\rightarrow \frac{12}{1\mp a}$ . The product $S(a)S(-a)=\frac{144}{1-a^2}=2016$ $\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}$ , so the answer is $\frac{2016}{6}=\boxed{336}$


Answer: 863.0

For $1 \leq i \leq 215$ let $a_i = \dfrac{1}{2^{i}}$ and $a_{216} = \dfrac{1}{2^{215}}$ . Let $x_1, x_2, ..., x_{216}$ be positive real numbers such that $\sum_{i=1}^{216} x_i=1$ and $\sum_{1 \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}$ . The maximum possible value of $x_2=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Note that \begin{align*}\sum_{1 \leq i < j \leq 216} x_ix_j &= \frac12\left(\left(\sum_{i=1}^{216} x_i\right)^2-\sum_{i=1}^{216} x_i^2\right)\\&=\frac12\left(1-\sum x_i^2\right).\end{align*} Substituting this into the second equation and collecting $x_i^2$ terms, we find \[\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}=\frac{1}{215}.\] Conveniently, $\sum_{i=1}^{216} 1-a_i=215$ so we find \[\left(\sum_{i=1}^{216} 1-a_i\right)\left(\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}\right)=1=\left(\sum_{i=1}^{216} x_i\right)^2.\] This is the equality case of the Cauchy-Schwarz Inequality, so $x_i=c(1-a_i)$ for some constant $c$ . Summing these equations and using the facts that $\sum_{i=1}^{216} x_i=1$ and $\sum_{i=1}^{216} 1-a_i=215$ , we find $c=\frac{1}{215}$ and thus $x_2=c(1-a_2)=\frac{1}{215}\cdot \left(1-\frac{1}{4}\right)=\frac{3}{860}$ . Hence the desired answer is $860+3=\boxed{863}$


Answer: 36.0

For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
Noticing the $\sin$ and $\cos$ in both $P$ and $Q,$ we think of the angle addition identities:
\[\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b - \sin a \sin b\]
With this in mind, we multiply $P$ by $\sin \theta$ and $Q$ by $\cos \theta$ to try and use some angle addition identities. Indeed, we get \begin{align*} P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\ &= \cos \theta - \dfrac{1}{4} \cos \theta + \dfrac{1}{8} \sin{2 \theta} + \dfrac{1}{16} \cos{3 \theta} + \cdots \\ &= \cos \theta - \dfrac{1}{2}P \end{align*} after adding term-by-term. Similar term-by-term adding yields \[P \cos \theta + Q \sin \theta = -2(Q - 1).\] This is a system of equations; rearrange and rewrite to get \[P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta\] and \[P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.\] Subtract the two and rearrange to get \[\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.\] Then, square both sides and use Pythagorean Identity to get a quadratic in $\sin \theta.$ Factor that quadratic and solve for $\sin \theta = -17/19, 1/3.$ Since we're given $\pi\leq\theta<2\pi,$ $\sin\theta$ is nonpositive. We therefore use the negative solution, and our desired answer is $17 + 19 = \boxed{036}.$

Solution:
Use sum to product formulas to rewrite $P$ and $Q$
\[P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ...\]
Therefore, \[P \sin \theta - Q \cos \theta = -2P\]
Using \[\frac{P}{Q} =  \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P\]
Plug in to the previous equation and cancel out the "P" terms to get: \[\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2\]
Then use the pythagorean identity to solve for $\sin\theta$ \[\sin\theta = -\frac{17}{19} \implies \boxed{036}\]

Solution:
Note that \[e^{i\theta}=\cos(\theta)+i\sin(\theta)\]
Thus, the following identities follow immediately: \[ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)\] \[i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)\] \[i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)\]
Consider, now, the sum $Q+iP$ . It follows fairly immediately that:
\[Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}\] \[Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\]
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:
\[Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)\] \[Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}\]
Comparing real and imaginary parts, we find: \[\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}\]
Squaring this equation and letting $\sin^2(\theta)=x$
$\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}$
Clearing denominators and solving for $x$ gives sine as $x=-\frac{17}{19}$
$017+019=\boxed{036}$

Solution:
A bit similar to Solution 3. We use $\phi = \theta+90^\circ$ because the progression cycles in $P\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)$ . So we could rewrite that as $P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)$
Similarly, $Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, \cos 3\phi\dots)$
Setting complex $z=q_1+p_1i$ , we get $z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)$
$(Q,P)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}=\frac{1}{1-\frac{1}{2}\cos\phi-\frac{i}{2}\sin\phi}=\frac{1-0.5\cos\phi+0.5i\sin\phi}{\dots}$
The important part is the ratio of the imaginary part $i$ to the real part. To cancel out the imaginary part from the denominator, we must add $0.5i\sin\phi$ to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find $\frac{P}{Q}=\tan\text{arg}(\Sigma)$ a PROPORTION of values. So denominators would cancel out.
$\frac{P}{Q}=\frac{\text{Re}\Sigma}{\text{Im}\Sigma}=\frac{0.5\sin\phi}{1-0.5\cos\phi}=\frac{\sin\phi}{2-\cos\phi}=\frac{2\sqrt{2}}{7}$
Setting $\sin\theta=y$ , we obtain \[\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}\] \[7\sqrt{1-y^2}=2\sqrt{2}(2+y)\] \[49-49y^2=8y^2+32y+32\] \[57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}\] \[y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}\]
Since $y<0$ because $\pi<\theta<2\pi$ $y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}$ . Adding up, $17+19=\boxed{036}$

Solution:
We notice $\sin\theta=-\frac{i}{2}(e^{i\theta}-e^{-i\theta})$ and $\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$
We observe that both $P$ and $Q$ can be split into $2$ parts, namely the terms which contain the $\cos$ and the terms which contain the $\sin .$
The $\cos$ part of $P$ can be expressed as: \begin{align*}\frac12\cos\theta-\frac18\cos3\theta+\cdots&=\frac14\left(e^{i\theta}\left(1-\frac{e^{i2\theta}}{4}+\cdots\right)+e^{-i\theta}\left(1-\frac{e^{-i2\theta}}{4}+\cdots\right)\right) \\ &= \frac{1}{4}\left(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}}\right)\\  &= \frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}.\end{align*}
Repeating the above process, we find that the $\sin$ part of $P$ is \[\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}},\] the $\cos$ part of $Q$ is \[\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}},\] and finally, the $\sin$ part of $Q$ is \[\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}.\]
Converting back to trigonometric form, we have \begin{align*}\frac{2\sqrt{2}}{7}&=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}\\ &=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}.\end{align*} Using the $\sin$ double identity and simplifying, we have \[\frac{2\sqrt2}{7}=\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}.\] Factoring the denominator, we have \[10-4\sin^2{\theta}-3\sin{\theta}=(5-4\sin\theta)(2+\sin\theta).\] Simplifying \begin{align*}\frac{2\sqrt2}{7}&= \frac{\cos{\theta}(5-4\sin{\theta})}{(5-4\sin\theta)(2+\sin\theta)}\\ &=\frac{\cos\theta}{2+\sin\theta}.\end{align*} We set $\sin\theta$ as $x$ , and by the Pythagorean Identity, we have $57x^2+32x-17=0$ . This factors into $(19x+17)(3x-1)=0$ , which yields the 2 solutions $x=-\frac{17}{19}, x=\frac{1}{3}$ . As $\pi\leq\theta<2\pi$ , the latter root is erroneous, and we are left with $\sin\theta=-\frac{17}{19}$ . Thus, our final answer is $17+19=\boxed{036}$

Solution:
Follow solution 3, up to the point of using the geometric series formula \[Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}\]
Moving everything to the other side, and considering only the imaginary part, we get \[Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0\]
We can then write $P = 2 \sqrt{2} k$ , and $Q = 7k$ , ( $k \neq 0$ ). Thus, we can substitute and divide out by k. \[2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0\] \[2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\sqrt{1-\sin^{2}\theta}=\ 0\] \[2\sqrt{2}+\sqrt{2}\sin\theta\ =\frac{7}{2}\left(\sqrt{1-\sin^{2}\theta}\right)\] \[8+8\sin\theta+2\sin^{2}\theta=\frac{49}{4}-\frac{49}{7}\sin^{2}\theta\] \[\frac{57}{4}\sin^{2}\theta+8\sin\theta-\frac{17}{4} = 0\] \[57\sin^{2}\theta+32\sin\theta-17 = 0\] \[\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0\]
Since $\pi \le \theta < 2\pi$ , we get $\sin \theta < 0$ , and thus, $\sin\theta = \frac{-17}{19} \implies \boxed{036}$