Answer: 8.0

For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$

Solution:
Let $p$ and $q$ be the roots of $x^{2}+kx+36.$ By Vieta's Formulas , we have $p+q=-k$ and $pq=36.$
It follows that $p$ and $q$ must be distinct factors of $36.$ The possibilities of $\{p,q\}$ are \[\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.\] Each unordered pair gives a unique value of $k.$ Therefore, there are $\boxed{8}$ values of $k,$ namely $\pm37,\pm20,\pm15,\pm13.$

Solution:
Note that $k$ must be an integer. Using the quadratic formula $x=\frac{-k \pm \sqrt{k^2-144}}{2}.$ Since $4$ divides $144$ evenly, $k$ and $k^2-144$ have the same parity, so $x$ is an integer if and only if $k^2-144$ is a perfect square.
Let $k^2-144=n^2.$ Then, $(k+n)(k-n)=144.$ Since $k$ is an integer and $144$ is even, $k+n$ and $k-n$ must both be even. Assuming that $k$ is positive, we get $5$ possible values of $k+n$ , namely $2, 4, 8, 6, 12$ , which will give distinct positive values of $k$ , but $k+n=12$ gives $k+n=k-n$ and $n=0$ , giving $2$ identical integer roots. Therefore, there are $4$ distinct positive values of $k.$ Multiplying that by $2$ to take the negative values into account, we get $4\cdot2=\boxed{8}$ values of $k$

Solution:
Proceed similar to Solution 2 and deduce that the discriminant of $x^{2}+kx+36$ must be a perfect square greater than $0$ to satisfy all given conditions. Seeing something like $k^2-144$ might remind us of a right triangle, where $k$ is the hypotenuse, and $12$ is a leg. There are four ways we could have this: a $9$ $12$ $15$ triangle, a $12$ $16$ $20$ triangle, a $5$ $12$ $13$ triangle, and a $12$ $35$ $37$ triangle.
Multiply by $2$ to account for negative $k$ values (since $k$ is being squared), and our answer is $\boxed{8}$

Solution: Since $36 = 2^2\cdot3^2$ , that means there are $(2+1)(2+1) = 9$ possible factors of $36$ . Since $6 \cdot 6$ violates the distinct root condition, subtract $1$ from $9$ to get $8$ . Each sum is counted twice, and we count of those twice for negatives. This cancels out, so we get $\boxed{8}$


Answer: 8.0

Solution:
Let $p$ and $q$ be the roots of $x^{2}+kx+36.$ By Vieta's Formulas , we have $p+q=-k$ and $pq=36.$
It follows that $p$ and $q$ must be distinct factors of $36.$ The possibilities of $\{p,q\}$ are \[\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.\] Each unordered pair gives a unique value of $k.$ Therefore, there are $\boxed{8}$ values of $k,$ namely $\pm37,\pm20,\pm15,\pm13.$

Solution:
Note that $k$ must be an integer. Using the quadratic formula $x=\frac{-k \pm \sqrt{k^2-144}}{2}.$ Since $4$ divides $144$ evenly, $k$ and $k^2-144$ have the same parity, so $x$ is an integer if and only if $k^2-144$ is a perfect square.
Let $k^2-144=n^2.$ Then, $(k+n)(k-n)=144.$ Since $k$ is an integer and $144$ is even, $k+n$ and $k-n$ must both be even. Assuming that $k$ is positive, we get $5$ possible values of $k+n$ , namely $2, 4, 8, 6, 12$ , which will give distinct positive values of $k$ , but $k+n=12$ gives $k+n=k-n$ and $n=0$ , giving $2$ identical integer roots. Therefore, there are $4$ distinct positive values of $k.$ Multiplying that by $2$ to take the negative values into account, we get $4\cdot2=\boxed{8}$ values of $k$

Solution:
Proceed similar to Solution 2 and deduce that the discriminant of $x^{2}+kx+36$ must be a perfect square greater than $0$ to satisfy all given conditions. Seeing something like $k^2-144$ might remind us of a right triangle, where $k$ is the hypotenuse, and $12$ is a leg. There are four ways we could have this: a $9$ $12$ $15$ triangle, a $12$ $16$ $20$ triangle, a $5$ $12$ $13$ triangle, and a $12$ $35$ $37$ triangle.
Multiply by $2$ to account for negative $k$ values (since $k$ is being squared), and our answer is $\boxed{8}$

Solution: Since $36 = 2^2\cdot3^2$ , that means there are $(2+1)(2+1) = 9$ possible factors of $36$ . Since $6 \cdot 6$ violates the distinct root condition, subtract $1$ from $9$ to get $8$ . Each sum is counted twice, and we count of those twice for negatives. This cancels out, so we get $\boxed{8}$


Answer: 103.0

For integers $a$ and $b$ consider the complex number \[\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i\]
Find the number of ordered pairs of integers $(a,b)$ such that this complex number is a real number.

Solution:
We consider two cases:
Case 1: $ab \ge -2016$
In this case, if \[0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = -\frac{\sqrt{|a+b|}}{ab+100}\] then $ab \ne -100$ and $|a + b| = 0 = a + b$ .  Thus $ab = -a^2$ so $a^2 < 2016$ .  Thus $a = -44,-43, ... , -1, 0, 1, ..., 43, 44$ , yielding $89$ values. However since $ab = -a^2 \ne -100$ ,  we have $a \ne \pm 10$ .  Thus there are $87$ allowed tuples $(a,b)$ in this case.
Case 2: $ab < -2016$
In this case, we want \[0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = \frac{\sqrt{-ab-2016} - \sqrt{|a+b|}}{ab+100}\] Squaring, we have the equations $ab \ne -100$ (which always holds in this case) and \[-(ab + 2016)= |a + b|.\] Then if $a > 0$ and $b < 0$ , let $c = -b$ .  If $c > a$ \[ac - 2016 = c - a \Rightarrow (a - 1)(c + 1) = 2015 \Rightarrow (a,b) = (2, -2014), (6, -402), (14, -154), (32, -64).\] Note that $ab < -2016$ for every one of these solutions.  If $c < a$ , then \[ac - 2016 = a - c \Rightarrow (a + 1)(c - 1) = 2015 \Rightarrow (a,b) = (2014, -2), (402, -6), (154, -14), (64, -32).\] Again, $ab < -2016$ for every one of the above solutions. This yields $8$ solutions.  Similarly, if $a < 0$ and $b > 0$ , there are $8$ solutions.  Thus, there are a total of $16$ solutions in this case.
Thus, the answer is $87 + 16 = \boxed{103}$


Answer: 510.0

For integers $a,b,c$ and $d,$ let $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d.$ Find the number of ordered triples $(a,b,c)$ of integers with absolute values not exceeding $10$ for which there is an integer $d$ such that $g(f(2))=g(f(4))=0.$

Solution:
There can be two different cases for this problem, either $f(2)=f(4)$ or not. If it is, note that by Vieta's formulas $a = -6$ . Then, $b$ can be anything. However, $c$ can also be anything, as we can set the root of $g$ (not equal to $f(2) = f(4)$ ) to any integer, producing a possible integer value of $d$ . Therefore there are $21^2 = 441$ in this case*. If it isn't, then $f(2),f(4)$ are the roots of $g$ . This means by Vieta's, that:
\[f(2)+f(4) = -c \in [-10,10]\] \[20 + 6a + 2b \in [-10,10]\] \[3a + b \in [-15,-5].\]
Solving these inequalities while considering that $a \neq -6$ to prevent $f(2) = f(4)$ , we obtain $69$ possible tuples and adding gives $441+69=\boxed{510}$ .
~awang11

Solution:
Define $h(x)=x^2+cx$ . Since $g(f(2))=g(f(4))=0$ , we know $h(f(2))=h(f(4))=-d$ . Plugging in $f(x)$ into $h(x)$ , we get $h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)$ . Setting $h(f(2))=h(f(4))$ \[16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc\] . Simplifying and cancelling terms, \[240+112a+24b+12a^2+4ab+12c+2ac=0\] \[120+56a+12b+6a^2+2ab+6c+ac=0\] \[6a^2+2ab+ac+56a+12b+6c+120=0\] \[6a^2+2ab+ac+20a+36a+12b+6c+120=0\] \[a(6a+2b+c+20)+6(6a+2b+c+20)=0\] \[(a+6)(6a+2b+c+20)=0\]
Therefore, either $a+6=0$ or $6a+2b+c=-20$ . The first case is easy: $a=-6$ and there are $441$ tuples in that case. In the second case, we simply perform casework on even values of $c$ , to get $77$ tuples, subtracting the $8$ tuples in both cases we get $441+77-8=\boxed{510}$


Answer: 564.0

For nonnegative integers $a$ and $b$ with $a + b \leq 6$ , let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$ . Let $S$ denote the sum of all $T(a, b)$ , where $a$ and $b$ are nonnegative integers with $a + b \leq 6$ . Find the remainder when $S$ is divided by $1000$

Solution: Let $c=6-(a+b)$ , and note that $\binom{6}{a + b}=\binom{6}{c}$ . The problem thus asks for the sum $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$ . Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately, which is equal to $\binom{18}{6}=18564$ . Therefore, the answer is $\boxed{564}$

Solution: Alternatively, one can note that we can consider groups where $a+b$ is constant, say $c$ . Fix any value of $c$ . Then the sum of all of the values of $T(a,b)$ such that $a+b=c$ is $\binom{6}{a+b} \sum_{a+b=c} \binom{6}{a}\binom{6}{b}$ which by Vandermonde's is $\binom{6}{a+b}\binom{12}{a+b}$ . Remember, that expression is the resulting sum for a fixed $a+b$ . So, for $a+b\le 6$ , we want $\sum_{c=0}^{6} \binom{6}{c}\binom{12}{c}$ . This is (by Vandermonde's or committee forming) $\binom{18}{6} = 18564 \implies \boxed{564}$ ~ firebolt360

Solution: Treating $a+b$ as $n$ , this problem asks for \[\sum_{n=0}^{6} \left[\binom{6}{n}\sum_{m=0}^{n}\left[\binom{6}{m}\binom{6}{n-m}\right]\right].\] But \[\sum_{m=0}^{n} \left[\binom{6}{m} \binom{6}{n-m}\right]\] can be computed through the following combinatorial argument.  Choosing $n$ elements from a set of size $12$ is the same as splitting the set into two sets of size $6$ and choosing $m$ elements from one, $n-m$ from the other where $0\leq m\leq n$ .  The number of ways to perform such a procedure is simply $\binom{12}{n}$ . Therefore, the requested sum is \[\sum_{n=0}^{6} \left[\binom{6}{n} \binom{12}{n}\right] = 18564.\] As such, our answer is $\boxed{564}$

Solution:
Case 1: $a<b$
Subcase 1: $a=0$ \[\binom{6}{0}\binom{6}{1}\binom{6}{1}=36\] \[\binom{6}{0}\binom{6}{2}\binom{6}{2}=225\] \[\binom{6}{0}\binom{6}{3}\binom{6}{3}=400\] \[\binom{6}{0}\binom{6}{4}\binom{6}{4}=225\] \[\binom{6}{0}\binom{6}{5}\binom{6}{5}=36\] \[\binom{6}{0}\binom{6}{6}\binom{6}{6}=1\] \[36+225+400+225+36+1=923\] Subcase 2: $a=1$ \[\binom{6}{1}\binom{6}{2}\binom{6}{3}=1800 \equiv 800 \pmod {1000}\] \[\binom{6}{1}\binom{6}{3}\binom{6}{4}=1800 \equiv 800 \pmod {1000}\] \[\binom{6}{1}\binom{6}{4}\binom{6}{5}=540\] \[\binom{6}{1}\binom{6}{5}\binom{6}{6}=36\] \[800+800+540+36=2176 \equiv 176 \pmod {1000}\] Subcase 3: $a=2$ \[\binom{6}{2}\binom{6}{3}\binom{6}{5}=1800\equiv800\pmod{1000}\] \[\binom{6}{2}\binom{6}{4}\binom{6}{6}=225\] \[800+225=1025\equiv25\pmod{1000}\]
\[923+176+25=1124\equiv124\pmod{1000}\]
Case 2: $b<a$
By just switching $a$ and $b$ in all of the above cases, we will get all of the cases such that $b>a$ is true. Therefore, this case is also $124\pmod{1000}$
Case 3: $a=b$ \[\binom{6}{0}\binom{6}{0}\binom{6}{0}=1\] \[\binom{6}{1}\binom{6}{1}\binom{6}{2}=540\] \[\binom{6}{2}\binom{6}{2}\binom{6}{4}=3375\equiv375\pmod{1000}\] \[\binom{6}{3}\binom{6}{3}\binom{6}{6}=400\] \[1+540+375+400=1316\equiv316\pmod{1000}\]
\[316+124+124=\boxed{564}\]

Solution:
We begin as in solution 1 to rewrite the sum as $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$ .
Consider the polynomial $P(x)=\left(\binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \cdot \cdot \cdot + \binom{6}{6}x^6\right)^3$ .
We can see the sum we wish to compute is just the coefficient of the $x^6$ term. However $P(x)=((1+x)^6)^3=(1+x)^{18}$ . Therefore, the coefficient of the $x^6$ term is just $\binom{18}{6} = 18564$ so the answer is $\boxed{564}$

Solution: Let $c=6-(a+b)$ . Then $\binom{6}{a+b}=\binom{6}{c}$ , and $a+b+c=6$ . The problem thus asks for \[\sum_{a+b+c=6}\binom{6}{a}\binom{6}{b}\binom{6}{c} \pmod {1000}.\] Suppose we have $6$ red balls, $6$ green balls, and $6$ blue balls lined up in a row, and we want to choose $6$ balls from this set of $18$ balls by considering each color separately. Over all possible selections of $6$ balls from this set, there are always a nonnegative number of balls in each color group. The answer is $\binom{18}{6} \pmod {1000}=18\boxed{564}$

Solution:
Since $\binom{6}{n}=\binom{6}{6-n}$ , we can rewrite $T(a,b)$ as $\binom{6}{a}\binom{6}{b}\binom{6}{6-(a+b)}$ . Consider the number of ways to choose a committee of 6 people from a group of 6 democrats, 6 republicans, and 6 independents. We can first pick $a$ democrats, then pick $b$ republicans, provided that $a+b \leq 6$ . Then we can pick the remaining $6-(a+b)$ people from the independents. But this is just $T(a,b)$ , so the sum of all $T(a,b)$ is equal to the number of ways to choose this committee.
On the other hand, we can simply pick any 6 people from the $6+6+6=18$ total politicians in the group. Clearly, there are $\binom{18}{6}$ ways to do this. So the desired quantity is equal to $\binom{18}{6}$ . We can then compute (routinely) the last 3 digits of $\binom{18}{6}$ as $\boxed{564}$


Answer: 275.0

For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Note that all the coefficients of odd-powered terms is an odd number of odd degree terms multiplied together, and all coefficients of even-powered terms have an even number of odd degree terms multiplied together. Since every odd degree term is negative, and every even degree term is positive, the sum is just equal to $Q(-1)=P(-1)^{5}=\left( \dfrac{3}{2}\right)^{5}=\dfrac{243}{32}$ , so the desired answer is $243+32=\boxed{275}$

Solution:
We are looking for the sum of the absolute values of the coefficients of $Q(x)$ . By defining $P'(x) = 1 + \frac{1}{3}x+\frac{1}{6}x^2$ , and defining $Q'(x) = P'(x)P'(x^3)P'(x^5)P'(x^7)P'(x^9)$ , we have made it so that all coefficients in $Q'(x)$ are just the positive/absolute values of the coefficients of $Q(x)$ . .
To find the sum of the absolute values of the coefficients of $Q(x)$ , we can just take the sum of the coefficients of $Q'(x)$ . This sum is equal to \[Q'(1) = P'(1)P'(1)P'(1)P'(1)P'(1) = \left(1+\frac{1}{3}+\frac{1}{6}\right)^5 = \frac{243}{32},\]
so our answer is $243+32 = \boxed{275}$

Solution: Multiply $P(x)P(x^3)$ and notice that the odd degree terms have a negative coefficient. Observing that this is probably true for all polynomials like this (including $P(x)P(x^3)P(x^5)P(x^7)P(x^9)$ ), we plug in $-1$ to get $\frac{243}{32} \implies \boxed{275}$


Answer: 695.0

For positive integer $n$ , let $s(n)$ denote the sum of the digits of $n$ .  Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$

Solution:
You know whatever $n$ is, it has to have 3 digits, because if it had only two, the maximum of $s(n)$ is 18.
Now let $n=100a_2+10a_1+a_0$
So first we know, $a_2+a_1+a_0=20$ . Okay now we have to split into cases based on which digit gets carried. This meaning, when you add a 3 digit number to 864, we have to know when to carry the digits. Note that if you don't understand any of the steps I take, just try adding any 3-digit number to 864 regularly (using the old-fashioned "put one number over the other" method, not mental calculation), and observe what you do at each step.
(1) $\textcolor{red}{*}$ None of the digits get carried over to the next space:
So this means $a_2<2, a_1<4$ and $a_0<6$ . So
$s(864+n)=(8+a_2)+(6+a_1)+(4+a_0)=38$ So it doesn't work. Now:
(2) $a_2+8$ is the only one that carries over
So this means $a_2>1, a_1<4$ and $a_0<6$ . So
$s(864+n)=1+(8+a_2-10)+(6+a_1)+(a_0+4)=29$
(3) $\textcolor{red}{*}$ $a_0+4$ is the only one that carries over.
So
$s(864+n)=(8+a_2)+(6+a_1+1)+(4+a_0-10)=29$
(4)The first and second digit carry over (but not the third)
$s(864+n)=1+(8+a_2-10+1)+(6+a_1-10)+(4+a_0)=20$
Aha! This case works but we still have to make sure it's possible for $a_2+a_1+a_0=20$ (We assumed this is true, so we have to find a number that works.) Since only the second and first digit carry over, $a_2>0, a_1>3$ and $a_0<6$ . The smallest value we can get with this is 695. Let's see if we can find a smaller one:
(5)The first and third digit carry over (but not the second)
$s(864+n)=1+(8+a_2-10)+(7+a_1)+(4+a_0-10)=20$
The largest value for the middle digit is 2, so the other digits have to be both 9's. So the smallest possible value is 929
(6) All the digits carry over
$s(864+n)=1+(9+a_2-10)+(7+a_1-10)+(4+a_0-10)=\text{Way less than 20}$
So the answer is $\boxed{695}$ which after a quick test, does indeed work.

Solution:
First, it is easy to verify that $695$ works and that no other numbers beginning with the digit 6 work (i.e. $686, 677, 668, 659$ do not work).
Suppose by contradiction that there is a smaller valid $n$ , where the leading digit of the three-digit number $n$ is 5 or less. (Two-digit $n$ obviously do not work because 9 + 9 < 20.) Clearly $n > 200$ because the smallest three-digit number whose digits sum to 20 is $299$ . Also, because the second digit is at most 9, the units digit is at least 6, which means that the addition $N = n + 864$ regroups in the ones place. Then the units digit of $N$ is clearly less than 4. But as $1000 < 200 + 864 < N < 600 + 864 = 1464$ , the sum of the thousands digit and the hundredth digit is at most 5. Because the second digit is at most 9, the sum of the digits of $N$ is at most $5 + 9 + 4 < 20$ , contradiction. Hence $\boxed{695}$ is the answer.

Solution: First of all, notice that the smallest $n$ with $s(n) = 20$ is $299$ . Also, if $s(n + 864) = 20$ $s(n - 136) = 19$ (because subtracting $1000$ from the number removes the $1$ in the thousands place). After checking $s(n - 136)$ for various $n$ with $s(n) = 20$ , we see that we need to have a carry when subtracting $136$ . To have this, we must either have a $2$ in the tens place or a $5$ in the units place. The minimum $n$ for the former is $929$ , and for the latter it is $695$ . We check and see that $s(695-136) = s(559) = 19$ , so our answer is $\boxed{695}$

Solution:
Observation (Lemma) : If r is the number of regroups in the addition of n+k, $S(n+k) = S(n)+S(k)-9r$
Proof :  When you add two numbers, and you do a carry, you are taking away 10 from 1 column, and adding 1 to another column, giving a net loss of 9 to the total.
Thus, we can see that we need to regroup exactly twice when we add 864. And, the lowest possible n is 299, so let's start from there.
299 gets three regroups, so we are going to need to take away from digits, and dump the excess in the hundred's place, since the hundreds are going to regroup anyways.
So, if we take away from the tens digit, we need to take away until we get 2 in the tens digit(since the ones will regroup). So, we get the number 929, which works (929+864 = 1793), but is not the smallest.
If we take away from the ones digit, we only have to take away 4, turning the unit's place to 5. 5+4 is 9, so it won't regroup. Dump the ones into the hundred's place, and we get the number $\boxed{695}$

Solution:
Although this solution doesn't directly solve the problem, it greatly hastens the bashing process.
Call the three digits a, b, and c.
When you add each of 8, 6, and 4 to a, b, and c the resultant will either get smaller or larger, depending on the original number.
e.g. If c is 7, then adding 4 will reduce the 7 to a 1, whilst leaving a one for b.
If c is 3, then adding 4 will simply add four to the total, and make the 3 a 7.
Each of 8, 6, and 4 all can reduce the original number by a certain amount and can increase the original number by a certain amount.
8 can reduce by 2 for all numbers greater than 1, 6 can reduce numbers by 4, and 4 can reduce numbers by 6.
Possibilities:
Also, realize that if the number is reduced, then a one will be carried to the following decimal place on the left, consequently reducing that amount they reduce. It's like a puzzle! Within no time you should find that if you add 4 to c, subtract 4 from b, subtract 1 from a, and leave a 1 in the thousands place, the total is equated to zero. This is optimal because most of the addition is kept to the left, where the effect to real value is less. (e.g. 299 is smaller than 992)
Now you have +1, -1, -4, +4 in the decimals, and a VERY fast trial and error gives $\boxed{695}.$

Solution:
First, note that to compute $s(m+n)$ (for any positive integers $m$ and $n$ ), one can simply find the sum of $s(m)$ and $s(n)$ minus 9 times the number of times one regroups when adding $m$ to $n$ .  One can see why this is by noticing that if one were to "forget" to regroup, and leave, say, a 10 in  the ones' place, the sum of the digits would be 9 higher than if one did regroup.  Anyway, one can see that the smallest 3-digit number (on AIME, all the answers are integers from 0 to 999) whose digits sum to 20 is 299.  If we add this number to 864, we have to regroup 3 times, so $s(299+864)$ will be $9=9\cdot3-(8+6+4)$ smaller than $s(299)$ .  We want this difference to be 0, so we need to find a way to only regroup two times.
We now notice that regrouping the hundreds is inevitable, so we must either prevent regrouping the ones or the tens.  Preventing regrouping the tens would require moving many of the tens to the hundreds' place (the ones' place is already full), which is bad when we are trying to minimize the number, but preventing regrouping the ones requires moving fewer ones to the hundreds' place.
We find that to preventing regrouping the ones, the ones' place of our number must be at most 5 (a larger number would sum to ten when added to 4).  Because we want to minimize the number of ones we move to the hundreds' place, we leave exactly 5 by moving four ones to the hundreds' place: $299\to\boxed{695}$

Solution: Bashing out modulo $9$ and getting lucky we get that if the $8$ and $6$ carry over when adding $n$ and $864$ , that $100a+10b+c \equiv 1+100(a-1)+10(b-4)+c+4 \pmod{9}$ such that $n=100a+10b+c$ and after maximizing $b$ and $c$ such that $c<6$ to not make the $4$ carry over to minimize $a$ we get that $\boxed{695}$ is our answer after confirming there are no lesser solutions.


Answer: 12.0

For positive integer $n_{}^{}$ , define $S_n^{}$ to be the minimum value of the sum $\sum_{k=1}^n \sqrt{(2k-1)^2+a_k^2},$ where $a_1,a_2,\ldots,a_n^{}$ are positive real numbers whose sum is 17. There is a unique positive integer $n^{}_{}$ for which $S_n^{}$ is also an integer. Find this $n^{}_{}$

Solution: Consider $n$ right triangles joined at their vertices, with bases $a_1,a_2,\ldots,a_n$ and heights $1,3,\ldots, 2n - 1$ . The sum of their hypotenuses is the value of $S_n$ . The minimum value of $S_n$ , then, is the length of the straight line connecting the bottom vertex of the first right triangle and the top vertex of the last right triangle, so \[S_n \ge \sqrt {\left(\sum_{k = 1}^n (2k - 1)\right)^2 + \left(\sum_{k = 1}^n a_k\right)^2}.\] Since the sum of the first $n$ odd integers is $n^2$ and the sum of $a_1,a_2,\ldots,a_n$ is 17, we get \[S_n \ge \sqrt {17^2 + n^4}.\] If this is an integer, we can write $17^2 + n^4 = m^2$ , for an integer $m$ . Thus, $(m - n^2)(m + n^2) = 289\cdot 1 = 17\cdot 17 = 1\cdot 289.$ The only possible value, then, for $m$ is $145$ , in which case $n^2 = 144$ , and $n = \boxed{012}$


Answer: 749.0

For positive integers $N$ and $k$ , define $N$ to be $k$ -nice if there exists a positive integer $a$ such that $a^{k}$ has exactly $N$ positive divisors. Find the number of positive integers less than $1000$ that are neither $7$ -nice nor $8$ -nice.

Solution: We claim that an integer $N$ is only $k$ -nice if and only if $N \equiv 1 \pmod k$ . By the number of divisors formula, the number of divisors of $\prod_{i=1}^n p_i^{a_i}$ is $\prod_{i=1}^n (a_i+1)$ . Since all the $a_i$ s are divisible by $k$ in a perfect $k$ power, the only if part of the claim follows. To show that all numbers $N \equiv 1 \pmod k$ are $k$ -nice, write $N=bk+1$ . Note that $2^{kb}$ has the desired number of factors and is a perfect kth power. By PIE, the number of positive integers less than $1000$ that are either $1 \pmod 7$ or $1\pmod 8$ is $142+125-17=250$ , so the desired answer is $999-250=\boxed{749}$

Solution: All integers $a$ will have factorization $2^a3^b5^c7^d...$ . Therefore, the number of factors in $a^7$ is $(7a+1)(7b+1)...$ , and for $a^8$ is $(8a+1)(8b+1)...$ . The most salient step afterwards is to realize that all numbers $N$ not $1 \pmod{7}$ and also not $1 \pmod{8}$ satisfy the criterion. The cycle repeats every $56$ integers, and by PIE, $7+8-1=14$ of them are either $7$ -nice or $8$ -nice or both. Therefore, we can take $\frac{42}{56} * 1008 = 756$ numbers minus the $7$ that work between $1000-1008$ inclusive, to get $\boxed{749}$ positive integers less than $1000$ that are not nice for $k=7, 8$


Answer: 188.0

For positive integers $a$ $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ $b$ , and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$

Solution:
This should be $\lfloor \frac{1000}{c} \rfloor$ . The current function breaks when $c \mid 1000$ and $b \mid c$ . Take $c = 200$ and $b = 20$ . Then, we have $\lfloor \frac{999}{200} \rfloor = 4$ stamps of value 200, $\lfloor \frac{199}{20} \rfloor = 9$ stamps of value b, and 19 stamps of value 1. The maximum such a collection can give is $200 \cdot 4 + 20 \cdot 9 +19 \cdot 1 = 999$ , just shy of the needed 1000. As for the rest of solution, proceed similarly, except use $1000$ instead of $999$
Also, some explanation: $b-1$ one cent stamps cover all residues module $b$ . Having $\lfloor \frac{c-1}{b} \rfloor$ stamps of value b covers all residue classes modulo $c$ . Finally, we just need $\lfloor \frac{1000}{c} \rfloor$ to cover everything up to 1000.
In addition, note that this function sometimes may not always minimize the number of stamps required. This is due to the fact that the stamps of value $b$ and of value $1$ have the capacity to cover values greater than or equal to $c$ (which occurs when $c-1$ has a remainder less than $b-1$ when divided by $b$ ). Thus, in certain cases, not all $\lfloor \frac{1000}{c} \rfloor$ stamps of value c may be necessary, because the stamps of value $b$ and 1 can replace one $c$
CrazyVideoGamez
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Therefore using $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$ $\lfloor \frac{c-1}{b} \rfloor$ stamps of value $b$ , and $b-1$ stamps of value $1$ , all values up to $1000$ can be represented in sub-collections, while minimizing the number of stamps.
So, $f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1$
$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$ . We can get the answer by solving this equation.
$c > \lfloor \frac{c-1}{b} \rfloor + b-1$
$\frac{999}{c} + c > \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$
$c^2 - 97c + 999 > 0$ $c > 85.3$ or $c < 11.7$
$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 > \frac{999}{c}$
$97 > \frac{999}{c}$ $c>10.3$
The $3$ least values of $c$ are $11$ $88$ $89$ $11 + 88+ 89 = \boxed{188}$


Answer: 21.0

For positive integers $m$ and $n$ such that $m+10<n+1$ , both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$ . What is $m+n$
$\textbf{(A)}20\qquad\textbf{(B)}21\qquad\textbf{(C)}22\qquad\textbf{(D)}23\qquad\textbf{(E)}24$

Solution: The mean and median are \[\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,\] so $3m+17=2n$ and $m+11=n$ . Solving this gives $\left(m,n\right)=\left(5,16\right)$ for $m+n=\boxed{21}$ . (trumpeter)

Solution: This is an alternate solution if you don't want to solve using algebra. First, notice that the median $n$ is the average of $m+10$ and $n+1$ . Therefore, $n=m+11$ , so the answer is $m+n=2m+11$ , which must be odd. This leaves two remaining options: ${(B) 21}$ and ${(D) 23}$ . Notice that if the answer is $(B)$ , then $m$ is odd, while $m$ is even if the answer is $(D)$ . Since the average of the set is an integer $n$ , the sum of the terms must be even. $4+10+1+2+2n$ is odd by definition, so we know that $3m+2n$ must also be odd, thus with a few simple calculations $m$ is odd. Because all other answers have been eliminated, $(B)$ is the only possibility left. Therefore, $m+n=\boxed{21}$ . ∎ --anna0kear

Solution: Since the median is $n$ , then $\frac{m+10+n+1}{2} = n \Rightarrow m+11 = n$ , or $m= n-11$ . Plug this in for $m$ values to get $\frac{7n-16}{6} = n \Rightarrow 7n-16 = 6n \Rightarrow n= 16$ . Plug it back in to get $m = 5$ , thus $16 + 5 = \boxed{21}$


Answer: 26.0

For positive integers $n$ , denote $D(n)$ by the number of pairs of different adjacent digits in the binary (base two) representation of $n$ . For example, $D(3) = D(11_{2}) = 0$ $D(21) = D(10101_{2}) = 4$ , and $D(97) = D(1100001_{2}) = 2$ . For how many positive integers less than or equal $97$ to does $D(n) = 2$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$

Solution:
If $D(n)$ is even, then the binary expansion of $n$ will both begin and end with a $1$ , because all positive binary numbers begin with a $1$ , and if you switch digits twice, you will have a $1$ at the end.  Thus, we are only concerned with the $49$ odd numbers between $1$ and $98$ inclusive.
All of these odd numbers will have an even $D(n)$ $D(n) = 0$ will be given by the numbers $1, 11, 111, 1111, 11111, 111111$ , which is a total of $6$ numbers.
We skip $D(n) = 2$ for now, and move to $D(n) = 4$ , which is easier to count.  The smallest $D(n) = 4$ happens when $n = 10101$ .  To get another number such that $D(n) = 4$ , we may extend any of the five blocks of zeros or ones by one digit.  This will form $110101, 100101, 101101, 101001, 101011$ , all of which are odd numbers that have $D(n) = 4$ .  To find seven digit numbers that have $D(n) = 4$ , we can again extend any block by one, so long as it remains less than $1100001$ or under.  There are five cases.
1)  Extending $110101$ is impossible without going over $1100001$
2)  Extending $100101$ by putting a $1$ at the beginning will go over $1100001$ , but the other four extensions work, giving $1000101, 1001101, 1001001, 1001011$
3)  Extending $101101$ by putting a $1$ at the beginning will go over $1100001$ , but the other four extensions give $1001101, 1011101, 1011001, 1011011$ .  However, $1001101$ already appeared in #2, giving only three new numbers.
4)  Extending $101001$ at the first group is impossible.  The other four extensions are $1001001, 1011001, 1010001, 1010011$ , but the first two are repeats.  Thus, there are only two new numbers.
5)  Extending $101011$ at the first group is impossible.  The other four extensions give $1001011, 1011011, 1010011, 1010111$ , but only the last number is new.
Thus, there is $1$ five digit number, $5$ six digit numbers, and $4 + 3 + 2 + 1 = 10$ seven digit numbers under $1100001$ for which $D(n) = 4$ .  That gives a total of $16$ numbers.
There smallest number for which $D(n) = 6$ is $1010101$ , which is under $98$ .  Further extensions, as well as cases where $D(n) > 6$ , are not possible.
Thus, we know that there are $6$ odd numbers that have $D(n) = 0$ , and $16$ odd numbers that have $D(n) = 4$ , and $1$ number that has $D(n) = 6$ .  The remaining odd numbers must have $D(n) = 2$ .  This means there are $49 - 6 - 16 - 1 = 26$ numbers that have $D(n) = 2$ , which is option $\boxed{26}$

Solution:
For $D(n)$ to be $2$ $n$ must be in the form $1...10...01...1$ . Note that $n$ must be at least $3$ digits for $D(n)$ to be $2$
Case $1$ $n$ has $3$ digits
$n$ $101_2$ , there is only $1$ possible value for $n$ when $n$ has $3$ digits.
Case $2$ $n$ has $4$ digits
$n = 1$ _ _ $1_2$ , there could be $1$ zero or $2$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $2$ possible places. If there are $2$ zeros, there is only $1$ possible placement. Therefore, there are $2+1=3$ possible values for $n$ when $n$ has $4$ digits.
Case $3$ $n$ has $5$ digits
$n = 1$ _ _ _ $1_2$ , there could be $1$ $2$ or $3$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $3$ possible places. If there are $2$ zeros, there are $2$ possible placements. If there are $3$ zeros, there is only $1$ possible placement. Therefore, there are $3+2+1=6$ possible values for $n$ when $n$ has $5$ digits.
Case $4$ $n$ has $6$ digits
$n = 1$ _ _ _ _ $1_2$ , there could be $1$ $2$ $3$ or $4$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $4$ possible places. If there are $2$ zeros, there are $3$ possible placements. If there are $3$ zeros, there are $2$ possible placements. If there are $4$ zeros, there is only $1$ possible placement. Therefore, there are $4+3+2+1=10$ possible values for $n$ when $n$ has $6$ digits.
Case $4$ $n$ has $7$ digits
Given by the problem, $n \le 1100001_2$ . We could first count the number of $n$ that is in the form of $10$ _ _ _ _ $1_2$ , where the digits between the second and last digit are in the form of $0...01...1$
There could be $0$ $1$ $2$ $3$ or $4$ zeros. Therefore, there are $5$ possible values for $n$ when $n=10$ _ _ _ _ $1_2$
There is also the case were $n = 1100001_2$ . Therefore, there are $5+1=6$ possible values for $n$ when $n$ has $7$ digits.
Therefore, the answer is $1+3+6+10+6=\boxed{26}$


Answer: 512.0

For positive integers $n$ and $k$ , let $f(n, k)$ be the remainder when $n$ is divided by $k$ , and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$ . Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$

Solution:
We can find that
$20\equiv 6 \pmod{7}$
$21\equiv 5 \pmod{8}$
$22\equiv 6 \pmod{8}$
$23\equiv 7 \pmod{8}$
$24\equiv 6 \pmod{9}$
$25\equiv 7 \pmod{9}$
$26\equiv 8 \pmod{9}$
Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get
$99\equiv 31 \pmod{34}$
$100\equiv 32 \pmod{34}$
So the sum is $6+3\times(6+...+31)+31+32=1512$ ,it is also 17+20+23+...+95, so the answer is $\boxed{512}$ .
By: Kris17

Solution:
$Lemma:$ Highest remainder when $n$ is divided by $1\leq k\leq n/2$ is obtained for $k_0 = (n + (3 - n \pmod{3}))/3$ and the remainder thus obtained is $(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]$
$Note:$ This is the second highest remainder when $n$ is divided by $1\leq k\leq n$ and the highest remainder occurs when $n$ is divided by $k_M$ $(n+1)/2$ for odd $n$ and $k_M$ $(n+2)/2$ for even $n$
Using the lemma above:
$\sum\limits_{n=20}^{100} F(n) = \sum\limits_{n=20}^{100} [(n - 6)/3 + (2/3)*n \pmod{3}]$ $= [(120*81/2)/3 - 2*81 + (2/3)*81] = 1512$
So the answer is $\boxed{512}$


Answer: 25.0

For positive integers $n,$ let $\tau (n)$ denote the number of positive integer divisors of $n,$ including 1 and $n.$ For example, $\tau (1)=1$ and $\tau(6) =4.$ Define $S(n)$ by $S(n)=\tau(1)+ \tau(2) + \cdots + \tau(n).$ Let $a$ denote the number of positive integers $n \leq 2005$ with $S(n)$ odd , and let $b$ denote the number of positive integers $n \leq 2005$ with $S(n)$ even . Find $|a-b|.$

Solution: Notice that the difference between consecutive squares are consecutively increasing odd numbers. Thus, there are $3$ numbers between $1$ (inclusive) and $4$ (exclusive), $5$ numbers between $4$ and $9$ , and so on. The number of numbers from $n^2$ to $(n + 1)^2$ is $(n + 1 - n)(n + 1 + n) = 2n + 1$ . Whenever the lowest square beneath a number is odd, the parity will be odd, and the same for even. Thus, $a = [2(1) + 1] + [2(3) + 1] \ldots [2(43) + 1] = 3 + 7 + 11 \ldots 87$ $b = [2(2) + 1] + [2(4) + 1] \ldots [2(42) + 1] + 70 = 5 + 9 \ldots 85 + 70$ , the $70$ accounting for the difference between $2005$ and $44^2 = 1936$ , inclusive. Notice that if we align the two and subtract, we get that each difference is equal to $2$ . Thus, the solution is $|a - b| = |b - a| = |2 \cdot 21 + 70 - 87| = \boxed{025}$

Solution:
Similarly, $b = (3^2 - 2^2) + (5^2 - 4^2) + \ldots + (45^2 - 44^2) - 19$ , where the $-19$ accounts for those numbers between $2005$ and $2024$
Thus $a = (2^2 - 1^2) + (4^2 - 3^2) + \ldots + (44^2 - 43^2)$
Then, $|a - b| = |2(2^2 + 4^2 + \ldots + 44^2) - 2(1^2 + 3^2 + 5^2 + \ldots 43^2) + 1^2 - 45^2 + 19|$ .
We can apply the formula $1^2 + 2^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$ .  From this formula, it follows that $2^2 + 4^2 + \ldots + (2n)^2 = \frac{2n(n + 1)(2n + 1)}{3}$ and so that
$|a - b| = \left| 2\cdot \frac{44\cdot23\cdot45}{3} - 2\cdot \frac{22 \cdot 43 \cdot 45}{3} - 45^2 + 20\right| = |-25| =\boxed{025}$

Solution:
Let $\Delta n$ denote the sum $1+2+3+ \dots +n-1+n$ . We can easily see from the fact "It is well-known that $\tau(n)$ is odd if and only if $n$ is a perfect square.", that
$a = (2^2-1^2) + (4^2-3^2) \dots (44^2 - 43^2) = (2+1)(2-1)+(4+3)(4-3) \dots (44+43)(44-43) = 1+2+3...44 = \Delta 44$
$b = 3^2-2^2+5^2-4^2...2006-44^2 = (3+2)(3-2) \dots (43+42)(43-42) + 1 + (2005 - 44^2) = \Delta 43 + 69$
$a-b = \Delta 44-\Delta 43-69 = 44-69 = -25$ . They ask for $|a-b|$ , so our answer is $|-25| = \boxed{025}$


Answer: 736.0

For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$

Solution: We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given $4$ and $10$ as the sides, so we know that the $3$ rd side is between $6$ and $14$ , exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the $3$ rd side. So the triangles' sides are between $6$ and $\sqrt{84}$ exclusive, and the larger bound is between $\sqrt{116}$ and $14$ , exclusive. The area of these triangles are from $0$ (straight line) to $2\sqrt{84}$ on the first "small bound" and the larger bound is between $0$ and $20$ $0 < s < 2\sqrt{84}$ is our first equation, and $0 < s < 20$ is our $2$ nd equation. Therefore, the area is between $\sqrt{336}$ and $\sqrt{400}$ , so our final answer is $\boxed{736}$

Solution:
If $a,b,$ and $c$ are the side-lengths of an obtuse triangle with $a\leq b\leq c,$ then both of the following must be satisfied:
For one such obtuse triangle, let $4,10,$ and $x$ be its side-lengths and $K$ be its area. We apply casework to its longest side:
Case (1): The longest side has length $\boldsymbol{10,}$ so $\boldsymbol{0<x<10.}$
By the Triangle Inequality Theorem, we have $4+x>10,$ from which $x>6.$
By the Pythagorean Inequality Theorem, we have $4^2+x^2<10^2,$ from which $x<\sqrt{84}.$
Taking the intersection produces $6<x<\sqrt{84}$ for this case.
At $x=6,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\sqrt{84},$ the obtuse triangle degenerates into a right triangle with area $K=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.$ Together, we obtain $0<K<2\sqrt{84},$ or $K\in\left(0,2\sqrt{84}\right).$
Case (2): The longest side has length $\boldsymbol{x,}$ so $\boldsymbol{x\geq10.}$
By the Triangle Inequality Theorem, we have $4+10>x,$ from which $x<14.$
By the Pythagorean Inequality Theorem, we have $4^2+10^2<x^2,$ from which $x>\sqrt{116}.$
Taking the intersection produces $\sqrt{116}<x<14$ for this case.
At $x=14,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\sqrt{116},$ the obtuse triangle degenerates into a right triangle with area $K=\frac12\cdot4\cdot10=20.$ Together, we obtain $0<K<20,$ or $K\in\left(0,20\right).$
Answer
It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers $s$ are in exactly one of $\left(0,2\sqrt{84}\right)$ or $\left(0,20\right).$ By the exclusive disjunction, the set of all such $s$ is \[[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),\] from which $a^2+b^2=\boxed{736}.$

Solution:
We have the diagram below.

We proceed by taking cases on the angles that can be obtuse, and finding the ranges for $s$ that they yield .
If angle $\theta$ is obtuse, then we have that $s \in (0,20)$ . This is because $s=20$ is attained at $\theta = 90^{\circ}$ , and the area of the triangle is strictly decreasing as $\theta$ increases beyond $90^{\circ}$ . This can be observed from \[s=\frac{1}{2}(4)(10)\sin\theta\] by noting that $\sin\theta$ is decreasing in $\theta \in (90^{\circ},180^{\circ})$
Then, we note that if $\alpha$ is obtuse, we have $s \in (0,4\sqrt{21})$ . This is because we get $x=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}$ when $\alpha=90^{\circ}$ , yileding $s=4\sqrt{21}$ . Then, $s$ is decreasing as $\alpha$ increases by the same argument as before.
$\angle{ACB}$ cannot be obtuse since $AC>AB$
Now we have the intervals $s \in (0,20)$ and $s \in (0,4\sqrt{21})$ for the cases where $\theta$ and $\alpha$ are obtuse, respectively. We are looking for the $s$ that are in exactly one of these intervals, and because $4\sqrt{21}<20$ , the desired range is \[s\in [4\sqrt{21},20)\] giving \[a^2+b^2=\boxed{736}\Box\]

Solution:
Note: Archimedes15 Solution which I added an answer
here are two cases. Either the $4$ and $10$ are around an obtuse angle or the $4$ and $10$ are around an acute triangle. If they are around the obtuse angle, the area of that triangle is $<20$ as we have $\frac{1}{2} \cdot 40 \cdot \sin{\alpha}$ and $\sin$ is at most $1$ . Note that for the other case, the side lengths around the obtuse angle must be $4$ and $x$ where we have $16+x^2 < 100 \rightarrow x < 2\sqrt{21}$ . Using the same logic as the other case, the area is at most $4\sqrt{21}$ . Square and add $4\sqrt{21}$ and $20$ to get the right answer \[a^2+b^2= \boxed{736}\Box\]

Solution:
For $\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\overline{AB}.$
As shown below, all locations for $C$ at which $\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints.  Note that:
For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$  Let the brackets denote areas:
Finally, the set of all such $s$ is $[a,b)=\left[2\sqrt{84},20\right),$ from which $a^2+b^2=\boxed{736}.$

Solution:
Let a triangle in $\tau(s)$ be $ABC$ , where $AB = 4$ and $BC = 10$ . We will proceed with two cases:
Case 1: $\angle ABC$ is obtuse. If $\angle ABC$ is obtuse, then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $(0,10)$ ; therefore, the area of the triangle will fall in the range of $(0, 20)$
Case 2: $\angle BAC$ is obtuse. Then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $\left(0, \sqrt{10^{2} - 4^{2}}\right)$ . Therefore, the area of the triangle will fall in the range of $\left(0, 2 \sqrt{84}\right)$
If $s < 2 \sqrt{84}$ , there will exist two types of triangles in $\tau(s)$ - one type with $\angle ABC$ obtuse; the other type with $\angle BAC$ obtuse. If $s \geq 2 \sqrt{84}$ , as we just found, $\angle BAC$ cannot be obtuse, so therefore, there is only one type of triangle - the one in which $\angle ABC$ is obtuse. Also, if $s > 20$ , no triangle exists with lengths $4$ and $10$ . Therefore, $s$ is in the range $\left[ 2 \sqrt{84}, 20\right)$ , so our answer is $\left(2 \sqrt{84}\right)^{2} + 20^{2} = \boxed{736}$

Solution:
Let's rephrase the condition. It is required to find such values of the area of an obtuse triangle with sides $4$ and $10,$ when there is exactly one such obtuse triangle. In the diagram, $AB = 4, AC = 10.$
The largest area of triangle with sides $4$ and $10$ is $20$ for a right triangle with legs $4$ and $10$ $AC\perp AB$ ).
The diagram shows triangles with equal heights. The yellow triangle $ABC'$ has the longest side $BC',$ the blue triangle $ABC$ has the longest side $AC.$ If $BC\perp AB,$ then $BC = \sqrt {AC^2 – AB^2} = 2 \sqrt{21}$ the area is equal to $4\sqrt{21}.$ In the interval, the blue triangle $ABC$ is acute-angled, the yellow triangle $ABC'$ is obtuse-angled. Their heights and areas are equal. The condition is met.
If the area is less than $4\sqrt{21},$ both triangles are obtuse, not equal, so the condition is not met.
Therefore, $s$ is in the range $\left[ 4 \sqrt{21}, 20\right)$ , so answer is $\left(4 \sqrt{21}\right)^{2} + 20^{2} = \boxed{736}$

Solution:
If $4$ and $10$ are the shortest sides and $\angle C$ is the included angle, then the area is \[\frac{4\cdot10\cdot\sin\angle C}{2} = 20\sin\angle C.\] Because $0\leq\sin\angle C\leq1$ , the maximum value of $20\sin\angle C$ is $20$ , so $s\leq20$
If $4$ is a shortest side and $10$ is the longest side, the length of the other short side is $4\cos\angle C+2\sqrt{4\cos^2 \angle C+21}$ by law of cosines, and the area is $2\left(4\cos\angle C+2\sqrt{4\cos^2\angle C+21}\right)\sqrt{1-\cos\angle C}$ . Because $-1\le \cos\angle C\le 0$ , this is minimized if $\cos\angle C=0$ , where $s=4\sqrt{21}$
So, the answer is $20^2+\left(4\sqrt{21}\right)^2=\boxed{736}$


Answer: 0.0

For real numbers $x$ , let \[P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)\] where $i = \sqrt{-1}$ . For how many values of $x$ with $0\leq x<2\pi$ does \[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\  1 \qquad\textbf{(C)}\  2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution: Let $a=\cos(x)+i\sin(x)$ . Now $P(a)=1+a-a^2+a^3$ $P(-1)=-2$ and $P(0)=1$ so there is a real root $a_1$ between $-1$ and $0$ . The other $a$ 's must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex $a$ 's squared is $\frac{1}{a_1}$ which is greater than $1$ . If $x$ is real number then $a$ must have magnitude of $1$ , but none of the solutions for $a$ have magnitude of $1$ , so the answer is $\boxed{0}$ ~lopkiloinm

Solution:
For $\textrm{Im}(P(x))=0$ , we get \[\sin(2x)=\sin(x)+\sin(3x)=2\sin(2x)\cos(x)\] So either $\sin(2x)=0$ , i.e. $x\in\{0,\pi\}$ or $\cos(x)=\tfrac 12$ , i.e. $x\in \{\pi/3, 5\pi/3\}$
For none of these values do we get $\textrm{Re}(P(x))=0$
Therefore, the answer is $\boxed{0}$

Solution:
We have \begin{align*} P \left( x \right) & = 1 + e^{ix} - e^{i 2x} + e^{i 3x} . \end{align*}
Denote $y = e^{i x}$ . Hence, this problem asks us to find the number of $y$ with $| y| = 1$ that satisfy \[ 1 + y - y^2 + y^3 = 0 . \hspace{1cm} (1) \]
Taking imaginary part of both sides, we have \begin{align*} 0 & = {\rm Im} \ \left( 1 + y - y^2 + y^3 \right) \\ & = \frac{1}{2i} \left( y - \bar y - y^2 + \bar y^2 + y^3 - \bar y^3 \right) \\ & = \frac{y - \bar y}{2i} \left( 1 - y - \bar y + y^2 + y \bar y + \bar y^2 \right) \\ & = {\rm Im} \ y \left( 1 - \left( y + \bar y \right) + \left( y + \bar y \right)^2 - y \bar y \right) \\ & = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - |y|^2 \right) \\ & = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - 1 \right) \\ & = 2 {\rm Im} \ y \cdot {\rm Re} \ y \left( 2 {\rm Re} \ y - 1 \right) \\ \end{align*} The sixth equality follows from the property that $|y| = 1$
Therefore, we have either ${\rm Re} \ y = 0$ or ${\rm Im} \ y = 0$ or $2 {\rm Re} \ y - 1 = 0$
Case 1: ${\rm Re} \ y = 0$
Because $|y| = 1$ $y = \pm i$
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
Case 2: ${\rm Im} \ y = 0$
Because $|y| = 1$ $y = \pm 1$
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
Case 3: $2 {\rm Re} \ y - 1 = 0$
Because $|y| = 1$ $y = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i = e^{i \pm \frac{\pi}{3}}$
However, these solutions fail to satisfy Equation (1).
Therefore, there is no solution in this case.
All cases above imply that there is no solution in this problem.
Therefore, the answer is $\boxed{0}$

Solution:
Let $a=\cos(x)+i\sin(x)$ , so by De Moivre $P(x)=a^3-a^2+a+1$ . The problem essentially asks for the number of real roots of $P$ which lie on the complex unit circle.
Let $|r|=1$ be a root of $P$ , and note that we can't have $r^3-r^2+r=0$ , else $P(r)=0$ . Thus, suppose henceforth that $r^3-r^2+r \neq 0$ . We then have $r^3-r^2+r=r^2(r+\tfrac{1}{r}-1)=a^2(2\mathrm{Re}(r)-1)$ , hence the argument of $r^3-r^2+r$ is either the argument of $a^2$ , or the argument of $-a^2$ . Since $r^3-r^2+r=-1$ is real, it follows that $a^2=\pm 1 \implies a \in \{1,i,-1,-i\}$ . Now, we can check all of these values and find that none of them work, yielding an answer of $\boxed{0}$

Solution: $P(x)$ can be written equivalently as $P(x) = 1 + cis(x) - cis(2x) + cis(3x).$ Thus, we aim to find $x$ such that the sum of the vectors $cis(x)$ $-cis(2x)$ , and $cis(3x)$ is -1. Notice that $cis(x)$ $-cis(2x)$ $cis(3x)$ all lie on the unit circle in the complex plane, and the vector $cis(x) + cis(3x)$ is collinear with $-cis(2x).$ Since $|-cis(2x)| = 1$ and we want the three vectors to sum to -1, we either have $-cis(2x) = 1$ and $cis(x) + cis(3x) = -2$ , or $-cis(2x) = -1$ and $cis(x) + cis(3x) = 0.$ If the first condition is true, $cis(x) = cis(3x)=-1.$ This will imply that $x= \pi.$ But then $-cis(2x) = -1$ , which violates the condition. Similarly, we can show that the second condition cannot be met either. Thus $P(x)$ does not have any solutions on the interval $[0, 2\pi].$ Therefore, the answer is $\boxed{0}$


Answer: 72.0

For real numbers $x$ and $y$ , define $x \spadesuit y = (x+y)(x-y)$ . What is $3 \spadesuit (4 \spadesuit 5)$
$\mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72$

Solution:
Since $x \spadesuit y = (x+y)(x-y)$
$3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) =  3 \spadesuit (-9) = (3+(-9))(3-(-9)) = \boxed{72}$


Answer: 98.0

For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ .  Find $|a| + |b| + |c|$

Solution:
From Vieta's formulas, we know that $a+b+c = 0$ , and $ab+bc+ac = -2011$ . Thus $a = -(b+c)$ . All three of $a$ $b$ , and $c$ are non-zero: say, if $a=0$ , then $b=-c=\pm\sqrt{2011}$ (which is not an integer). $\textsc{wlog}$ , let $|a| \ge |b| \ge |c|$ . If $a > 0$ , then $b,c < 0$ and if $a < 0$ , then $b,c > 0$ . We have \[-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc\] Thus $a^2 = 2011 + bc$ . We know that $b$ $c$ have the same sign. So $|a| \ge 45 = \lceil \sqrt{2011} \rceil$
Also, if we fix $a$ $b+c$ is fixed, so $bc$ is maximized when $b = c$ . Hence, \[2011 = a^2 - bc > \tfrac{3}{4}a^2 \qquad \Longrightarrow \qquad a ^2 < \tfrac{4}{3}\cdot 2011 = 2681+\tfrac{1}{3}\] So $|a| \le 51$ . Thus we have bounded $a$ as $45\le |a| \le 51$ , i.e. $45\le |b+c| \le 51$ since $a=-(b+c)$ . Let's analyze $bc=(b+c)^2-2011$ . Here is a table:
We can tell we don't need to bother with $45$
$105 = (3)(5)(7)$ , So $46$ won't work. $198/47 > 4$
$198$ is not divisible by $5$ $198/6 = 33$ , which is too small to get $47$
$293/48 > 6$ $293$ is not divisible by $7$ or $8$ or $9$ , we can clearly tell that $10$ is too much.
Hence, $|a| = 49$ $a^2 -2011 = 390$ $b = 39$ $c = 10$
Answer: $\boxed{098}$

Solution:
Starting off like the previous solution, we know that $a + b + c = 0$ , and $ab + bc + ac = -2011$
Therefore, $c = -b-a$
Substituting, $ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011$
Factoring the perfect square, we get: $ab-(b+a)^2=-2011$ or $(b+a)^2-ab=2011$
Therefore, a sum ( $a+b$ ) squared minus a product ( $ab$ ) gives $2011$ ..
We can guess and check different $a+b$ ’s starting with $45$ since $44^2 < 2011$
$45^2 = 2025$ therefore $ab = 2025-2011 = 14$
Since no factors of $14$ can sum to $45$ $1+14$ being the largest sum), a + b cannot equal $45$
$46^2 = 2116$ making $ab = 105 = 3 * 5 * 7$
$5 * 7 + 3 < 46$ and $3 * 5 * 7 > 46$ so $46$ cannot work either.
We can continue to do this until we reach $49$
$49^2 =  2401$ making $ab = 390 = 2 * 3 * 5* 13$
$3 * 13 + 2* 5 = 49$ , so one root is $10$ and another is $39$ . The roots sum to zero, so the last root must be $-49$
$|-49|+10+39 = \boxed{098}$

Solution: Let us first note the obvious that is derived from Vieta's formulas: $a+b+c=0, ab+bc+ac=-2011$ . Now, due to the first equation, let us say that $a+b=-c$ , meaning that $a,b>0$ and $c<0$ . Now, since both $a$ and $b$ are greater than 0, their absolute values are both equal to $a$ and $b$ , respectively. Since $c$ is less than 0, it equals $-a-b$ . Therefore, $|c|=|-a-b|=a+b$ , meaning $|a|+|b|+|c|=2(a+b)$ . We now apply Newton's sums to get that $a^2+b^2+ab=2011$ ,or $(a+b)^2-ab=2011$ . Solving, we find that $49^2-390$ satisfies this, meaning $a+b=49$ , so $2(a+b)=\boxed{098}$

Solution:
We have
As a result, we have
$a+b+c=0$
$ab+bc+ac=-2011$
$abc=-m$
So, $a=-b-c$
As a result, $ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011$
Solve $b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}$ and $\Delta =8044-3c^2=k^2$ , where $k$ is an integer
Cause $89<\sqrt{8044}<90$
So, after we tried for $2$ times, we get $k=88$ and $c=10$
then $b=39$ $a=-b-c=-49$
As a result, $|a|+|b|+|c|=10+39+49=\boxed{098}$

Solution: First, derive the equations $a=-b-c$ and $ab+bc+ca=-2011\implies b^2+bc+c^2=2011$ . Since the product is negative, $a$ is negative, and $b$ and $c$ positive. Now, a simple mod 3 testing of all cases shows that $b\equiv \{1,2\} \pmod{3}$ , and $c$ has the repective value. We can choose $b$ not congruent to 0, make sure you see why. Now, we bash on values of $b$ , testing the quadratic function to see if $c$ is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for $b=10$ $c=39, -49$ . Choosing $c$ positive we get $a=-49$ , so $|a|+|b|+|c|=10+29+39=\boxed{098}$ ~firebolt360


Answer: 14.0

For some particular value of $N$ , when $(a+b+c+d+1)^N$ is expanded and like terms are combined, the resulting expression contains exactly $1001$ terms that include all four variables $a, b,c,$ and $d$ , each to some positive power. What is $N$
$\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution:
All the desired terms are in the form $a^xb^yc^zd^w1^t$ , where $x + y + z + w + t = N$ (the $1^t$ part is necessary to make stars and bars work better.)
Since $x$ $y$ $z$ , and $w$ must be at least $1$ $t$ can be $0$ ), let $x' = x - 1$ $y' = y - 1$ $z' = z - 1$ , and $w' = w - 1$ , so $x' + y' + z' + w' + t = N - 4$ . Now, we use stars and bars (also known as ball and urn) to see that there are $\binom{(N-4)+4}{4}$ or $\binom{N}{4}$ solutions to this equation. We notice that $1001=7\cdot11\cdot13$ , which leads us to guess that $N$ is around these numbers. This suspicion proves to be correct, as we see that $\binom{14}{4} = 1001$ , giving us our answer of $\boxed{14.}$

Solution: By the Hockey Stick Identity , the number of terms that have all $a,b,c,d$ raised to a positive power is $\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}$ . We now want to find some $N$ such that $\binom{N}{4} = 1001$ . As mentioned above, after noticing that $1001 = 7\cdot11\cdot13$ , and some trial and error, we find that $\binom{14}{4} = 1001$ , giving us our answer of $\boxed{14.}$

Solution:
The terms are in the form $a^xb^yc^zd^w1^t$ , where $x + y + z + w + t = N$ . The problem becomes distributing $N$ identical balls to $5$ different boxes $(x, y, z, w, t)$ such that each of the boxes $(x, y, z, w)$ has at least $1$ ball. The $N$ balls in a row have $N-1$ gaps among them. We are going to put $4$ or $3$ divisors into those $N-1$ gaps. There are $2$ cases of how to put the divisors.
Case $1$ :
Put 4 divisors into $N-1$ gaps. It corresponds to each of $(a, b, c, d, 1)$ has at least one term. There are $\binom{N-1}{4}$ terms.
Case $2$ :
Put 3 divisors into $N-1$ gaps. It corresponds to each of $(a, b, c, d)$ has at least one term. There are $\binom{N-1}{3}$ terms.
So, there are $\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}$ terms. $\binom{N}{4} = 1001$ , and since we have $\binom{14}{4} = 1001, N=\boxed{14.}$


Answer: 16.0

For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$

Solution:
We can expand the fraction $0.\overline{23}_k$ as follows: $0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + \cdots$
Notice that this is equivalent to \[2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + \cdots )\]
By summing the geometric series and simplifying, we have $\frac{2k+3}{k^2-1} = \frac{7}{51}$ . Solving this quadratic equation (or simply testing the answer choices) yields the answer $k = \boxed{16}$

Solution:
Let $a = 0.2323\dots_k$ . Therefore, $k^2a=23.2323\dots_k$
From this, we see that $k^2a-a=23_k$ , so $a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}$
Now, similar to in Solution 1, we can either test if $2k+3$ is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is $\boxed{16}$

Solution:
Just as in Solution 1, we arrive at the equation $\frac{2k+3}{k^2-1}=\frac{7}{51}$
Therefore now, we can rewrite this as $\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}$ . Notice that $2k+3=2(k+1)+1=2(k-1)+5$ . As $17$ is a prime number, we therefore must have that one of $k-1$ and $k+1$ is divisible by $17$ . Now, checking each of the answer choices, this will lead us to the answer $\boxed{16}$

Solution: Assuming you are familiar with the rules for basic repeating decimals, $0.232323... = \frac{23}{99}$ . Now we want our base, $k$ , to conform to $23\equiv7\pmod k$ and $99\equiv51\pmod k$ , the reason being that we wish to convert the number from base $10$ to base $k$ . Given the first equation, we know that $k$ must equal 9, 16, 23, or generally, $7n+2$ . The only number in this set that is one of the multiple choices is $16$ . When we test this on the second equation, $99\equiv51\pmod k$ , it comes to be true. Therefore, our answer is $\boxed{16}$

Solution:
Note that the LHS equals \[\bigg(\frac{2}{k} + \frac{2}{k^3} + \cdots \bigg) + \bigg(\frac{3}{k^2} + \frac{3}{k^4} + \cdots \bigg) = \frac{\frac{2}{k}}{1 - \frac{1}{k^2}} + \frac{\frac{3}{k^2}}{1 - \frac{1}{k^2}} = \frac{2k+3}{k^2-1},\] from which we see our equation becomes \[\frac{2k+3}{k^2-1} = \frac{7}{51}, \ \ \implies \ \ 51(2k+3) = 7(k^2-1).\]
Note that $17$ therefore divides $k^2 - 1,$ but as $17$ is prime this therefore implies \[k \equiv \pm 1 \pmod{17}.\] (Warning: This would not be necessarily true if $17$ were composite.) Note that $\boxed{16}$ is the only answer choice congruent satisfying this modular congruence, thus completing the problem. $\square$


Answer: 16.0

Solution:
We can expand the fraction $0.\overline{23}_k$ as follows: $0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + \cdots$
Notice that this is equivalent to \[2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + \cdots )\]
By summing the geometric series and simplifying, we have $\frac{2k+3}{k^2-1} = \frac{7}{51}$ . Solving this quadratic equation (or simply testing the answer choices) yields the answer $k = \boxed{16}$

Solution:
Let $a = 0.2323\dots_k$ . Therefore, $k^2a=23.2323\dots_k$
From this, we see that $k^2a-a=23_k$ , so $a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}$
Now, similar to in Solution 1, we can either test if $2k+3$ is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is $\boxed{16}$

Solution:
Just as in Solution 1, we arrive at the equation $\frac{2k+3}{k^2-1}=\frac{7}{51}$
Therefore now, we can rewrite this as $\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}$ . Notice that $2k+3=2(k+1)+1=2(k-1)+5$ . As $17$ is a prime number, we therefore must have that one of $k-1$ and $k+1$ is divisible by $17$ . Now, checking each of the answer choices, this will lead us to the answer $\boxed{16}$

Solution: Assuming you are familiar with the rules for basic repeating decimals, $0.232323... = \frac{23}{99}$ . Now we want our base, $k$ , to conform to $23\equiv7\pmod k$ and $99\equiv51\pmod k$ , the reason being that we wish to convert the number from base $10$ to base $k$ . Given the first equation, we know that $k$ must equal 9, 16, 23, or generally, $7n+2$ . The only number in this set that is one of the multiple choices is $16$ . When we test this on the second equation, $99\equiv51\pmod k$ , it comes to be true. Therefore, our answer is $\boxed{16}$

Solution:
Note that the LHS equals \[\bigg(\frac{2}{k} + \frac{2}{k^3} + \cdots \bigg) + \bigg(\frac{3}{k^2} + \frac{3}{k^4} + \cdots \bigg) = \frac{\frac{2}{k}}{1 - \frac{1}{k^2}} + \frac{\frac{3}{k^2}}{1 - \frac{1}{k^2}} = \frac{2k+3}{k^2-1},\] from which we see our equation becomes \[\frac{2k+3}{k^2-1} = \frac{7}{51}, \ \ \implies \ \ 51(2k+3) = 7(k^2-1).\]
Note that $17$ therefore divides $k^2 - 1,$ but as $17$ is prime this therefore implies \[k \equiv \pm 1 \pmod{17}.\] (Warning: This would not be necessarily true if $17$ were composite.) Note that $\boxed{16}$ is the only answer choice congruent satisfying this modular congruence, thus completing the problem. $\square$


Answer: 325.0

For some positive integer $n$ , the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3$ . How many positive integer divisors does the number $81n^4$ have?
$\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425$

Solution: Since the prime factorization of $110$ is $2 \cdot 5 \cdot 11$ , we have that the number is equal to $2 \cdot 5 \cdot 11 \cdot n^3$ .  This has $2 \cdot 2 \cdot 2=8$ factors when $n=1$ .  This needs a multiple of 11 factors, which we can achieve by setting $n=2^3$ , so we have $2^{10} \cdot 5 \cdot 11$ has $44$ factors.  To achieve the desired $110$ factors, we need the number of factors to also be divisible by $5$ , so we can set $n=2^3 \cdot 5$ , so $2^{10} \cdot 5^4 \cdot 11$ has $110$ factors.  Therefore, $n=2^3 \cdot 5$ .  In order to find the number of factors of $81n^4$ , we raise this to the fourth power and multiply it by $81$ , and find the factors of that number.  We have $3^4 \cdot 2^{12} \cdot 5^4$ , and this has $5 \cdot 13 \cdot 5=\boxed{325}$ factors.

Solution:
$110n^3$ clearly has at least three distinct prime factors, namely 2, 5, and 11.
The number of factors of $p_1^{n_1}\cdots p_k^{n_k}$ is $(n_1+1)\cdots(n_k+1)$ when the $p$ 's are distinct primes. This tells us that none of these factors can be 1. The number of factors is given as 110. The only way to write 110 as a product of at least three factors without $1$ s is $2\cdot 5\cdot 11$
We conclude that $110n^3$ has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order.  I.e., there are six different possible values of $n$ all of the form $n=p_1\cdot p_2^3$
$81n^4$ thus has prime factorization $81n^4=3^4\cdot p_1^4\cdot p_2^{12}$ and a factor count of $5\cdot5\cdot13=\boxed{325}$


Answer: 325.0

Solution: Since the prime factorization of $110$ is $2 \cdot 5 \cdot 11$ , we have that the number is equal to $2 \cdot 5 \cdot 11 \cdot n^3$ .  This has $2 \cdot 2 \cdot 2=8$ factors when $n=1$ .  This needs a multiple of 11 factors, which we can achieve by setting $n=2^3$ , so we have $2^{10} \cdot 5 \cdot 11$ has $44$ factors.  To achieve the desired $110$ factors, we need the number of factors to also be divisible by $5$ , so we can set $n=2^3 \cdot 5$ , so $2^{10} \cdot 5^4 \cdot 11$ has $110$ factors.  Therefore, $n=2^3 \cdot 5$ .  In order to find the number of factors of $81n^4$ , we raise this to the fourth power and multiply it by $81$ , and find the factors of that number.  We have $3^4 \cdot 2^{12} \cdot 5^4$ , and this has $5 \cdot 13 \cdot 5=\boxed{325}$ factors.

Solution:
$110n^3$ clearly has at least three distinct prime factors, namely 2, 5, and 11.
The number of factors of $p_1^{n_1}\cdots p_k^{n_k}$ is $(n_1+1)\cdots(n_k+1)$ when the $p$ 's are distinct primes. This tells us that none of these factors can be 1. The number of factors is given as 110. The only way to write 110 as a product of at least three factors without $1$ s is $2\cdot 5\cdot 11$
We conclude that $110n^3$ has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order.  I.e., there are six different possible values of $n$ all of the form $n=p_1\cdot p_2^3$
$81n^4$ thus has prime factorization $81n^4=3^4\cdot p_1^4\cdot p_2^{12}$ and a factor count of $5\cdot5\cdot13=\boxed{325}$


Answer: 31.0

For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ $AB=2$ , and $CD=AD$ .  How many different values of $p<2015$ are possible?
$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$

Solution:
Let $BC = x$ and $CD = AD = y$ be positive integers. Drop a perpendicular from $A$ to $CD$ to show that, using the Pythagorean Theorem, that \[x^2 + (y - 2)^2 = y^2.\] Simplifying yields $x^2 - 4y + 4 = 0$ , so $x^2 = 4(y - 1)$ . Thus, $y$ is one more than a perfect square.
The perimeter $p = 2 + x + 2y = 2y + 2\sqrt{y - 1} + 2$ must be less than 2015. Simple calculations demonstrate that $y = 31^2 + 1 = 962$ is valid, but $y = 32^2 + 1 = 1025$ is not. On the lower side, $y = 1$ does not work (because $x > 0$ ), but $y = 1^2 + 1$ does work. Hence, there are 31 valid $y$ (all $y$ such that $y = n^2 + 1$ for $1 \le n \le 31$ ), and so our answer is $\boxed{31}$

Solution:
Let $BC = x$ and $CD = AD = z$ be positive integers. Drop a perpendicular from $A$ to $CD$ . Denote the intersection point of the perpendicular and $CD$ as $E$
$AE$ 's length is $x$ , as well.
Call $ED$ $y$ .
By the Pythagorean Theorem, $x^2 + y^2 = (y + 2)^2$ .
And so: $x^2 = 4y + 4$ , or $y = (x^2-4)/4$
Writing this down and testing, it appears that this holds for all $x$ . However, since there is a dividend of 4, the numerator must be divisible by 4 to conform to the criteria that the side lengths are positive integers.
In effect, $x$ must be a multiple of 2 to let the side lengths be integers. We test, and soon reach 62. It gives us $p = 1988$ , which is less than 2015. However, 64 gives us $2116 > 2015$ , so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get $\boxed{31}$


Answer: 31.0

Solution:
Let $BC = x$ and $CD = AD = y$ be positive integers. Drop a perpendicular from $A$ to $CD$ to show that, using the Pythagorean Theorem, that \[x^2 + (y - 2)^2 = y^2.\] Simplifying yields $x^2 - 4y + 4 = 0$ , so $x^2 = 4(y - 1)$ . Thus, $y$ is one more than a perfect square.
The perimeter $p = 2 + x + 2y = 2y + 2\sqrt{y - 1} + 2$ must be less than 2015. Simple calculations demonstrate that $y = 31^2 + 1 = 962$ is valid, but $y = 32^2 + 1 = 1025$ is not. On the lower side, $y = 1$ does not work (because $x > 0$ ), but $y = 1^2 + 1$ does work. Hence, there are 31 valid $y$ (all $y$ such that $y = n^2 + 1$ for $1 \le n \le 31$ ), and so our answer is $\boxed{31}$

Solution:
Let $BC = x$ and $CD = AD = z$ be positive integers. Drop a perpendicular from $A$ to $CD$ . Denote the intersection point of the perpendicular and $CD$ as $E$
$AE$ 's length is $x$ , as well.
Call $ED$ $y$ .
By the Pythagorean Theorem, $x^2 + y^2 = (y + 2)^2$ .
And so: $x^2 = 4y + 4$ , or $y = (x^2-4)/4$
Writing this down and testing, it appears that this holds for all $x$ . However, since there is a dividend of 4, the numerator must be divisible by 4 to conform to the criteria that the side lengths are positive integers.
In effect, $x$ must be a multiple of 2 to let the side lengths be integers. We test, and soon reach 62. It gives us $p = 1988$ , which is less than 2015. However, 64 gives us $2116 > 2015$ , so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get $\boxed{31}$


Answer: 1.52

For some real number $r$ , the polynomial $8x^3-4x^2-42x+45$ is divisible by $(x-r)^2$ . Which of the following numbers is closest to $r$
$\text{(A)} \ 1.22 \qquad  \text{(B)} \ 1.32 \qquad  \text{(C)} \ 1.42 \qquad  \text{(D)} \ 1.52 \qquad  \text{(E)} \ 1.62$

Solution:
Solution by e_power_pi_times_i
Denote $s$ as the third solution. Then, by Vieta's, $2r+s = \dfrac{1}{2}$ $r^2+2rs = -\dfrac{21}{4}$ , and $r^2s = -\dfrac{45}{8}$ . Multiplying the top equation by $2r$ and eliminating, we have $3r^2 = r+\dfrac{21}{4}$ . Combined with the fact that $s = \dfrac{1}{2}-2r$ , the third equation can be written as $(\dfrac{r+\dfrac{21}{4}}{3})(\dfrac{1}{2}-2r) = -\dfrac{45}{8}$ , or $(4r+21)(4r-1) = 135$ . Solving, we get $r = \dfrac{3}{2}, -\dfrac{13}{2}$ . Plugging the solutions back in, we see that $-\dfrac{13}{2}$ is an extraneous solution, and thus the answer is $\boxed{1.52}$


Answer: 29.0

For the consumer, a single discount of $n\%$ is more advantageous than any of the following discounts:
(1) two successive $15\%$ discounts
(2) three successive $10\%$ discounts
(3) a $25\%$ discount followed by a $5\%$ discount
What is the smallest possible positive integer value of $n$
$\textbf{(A)}\ \ 27\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 29\qquad\textbf{(D)}\ 31\qquad\textbf{(E)}\ 33$

Solution: Let the original price be $x$ . Then, for option $1$ , the discounted price is $(1-.15)(1-.15)x = .7225x$ . For option $2$ , the discounted price is $(1-.1)(1-.1)(1-.1)x = .729x$ . Finally, for option $3$ , the discounted price is $(1-.25)(1-.05) = .7125x$ . Therefore, $n$ must be greater than $\max(x - .7225x, x-.729x, x-.7125x)$ . It follows $n/100$ must be greater than $.2875$ . We multiply this by $100$ to get the percent value, and then round up because $n$ is the smallest integer that provides a greater discount than $28.75$ , leaving us with the answer of $\boxed{29}$

Solution: Assume the original price was $100$ dollars. Thus, after a discount of $n\%$ , the price will be $100-n$ dollars. Using basic calculations, find out the value of the other discounts. This leaves us with the prices: $100-n$ $\frac{289}{4}$ $\frac{9^3}{10}$ , and $\frac{15\cdot19}{4}$ . Simplify these to get $100-n$ $72$ (rounded down), $72.9$ , and $71$ (rounded down). To have the greatest discount, we need the least price, which is $71$ dollars. Now we get the original fractional value of this, and the discount of this is $100-\frac{285}{4}$ , and we round this down to $28$ . Now, it's pretty easy. The integer value greater than this is $\boxed{29}$

Solution: Assume WLOG that the original price was $$100$ . Then, option 1 would cost $100 \cdot \frac{17}{20} \cdot \frac{17}{20} = $ 72.25$ . Option 2 would cost $100 \cdot \frac{9}{10} \cdot \frac{9}{10} \cdot \frac{9}{10} = $72.90$ . Option 3 would cost $100 \cdot \frac{3}{4} \cdot \frac{19}{20} = $71.25$ . Therefore, the largest integer smaller than all of these three is $71$ , so $100-71= \boxed{29}$


Answer: 97.0

For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$
Note: here $i = \sqrt { - 1}$
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98$

Solution:
We know that $i^x$ cycles every $4$ powers so we group the sum in $4$ s. \[i+2i^2+3i^3+4i^4=2-2i\] \[5i^5+6i^6+7i^7+8i^8=2-2i\]
We can postulate that every group of $4$ is equal to $2-2i$ .
For 24 groups we thus, get $48-48i$ as our sum.
We know the solution must lie near
The next term is the $24*4+1=97$ th term. This term is equal to $97i$ (first in a group of $4$ so $i^{97}=i$ ) and our sum is now $48+49i$ so $n=97\Rightarrow\boxed{97}$ is our answer

Solution:
Some may know the equation:
\[\sum_{k=1}^{n}kr^{k-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}\]
(For those curious, this comes from differentiating the equation for finite geometric sums)
Using this equation, we have
\[48+49i=i\frac{1-(n+1)i^n+ni^{n+1}}{(1-i)^2}\] \[=\frac{1-(n+1)i^n+ni^{n+1}}{-2}\] \[=-\frac{1}{2}+\frac{(n+1)i^n}{2}-\frac{ni^{n+1}}{2}\]
Since the imaginary and the real part must be positive, we know that $i^{n+1}=-1$ or $i^{n+1}=-i$ . By the same line of reason, $i^{n}=1,i$ . This only works for $n\equiv 1 \mod 4$ . Therefore, we have:
\[\frac{-1+n}{2}+\frac{(n+1)i}{2}=48+49i\]
Solving either the real or imaginary part gives $\boxed{97}$


Answer: 97.0

Solution:
Obviously, even powers of $i$ are real and odd powers of $i$ are imaginary.
Hence the real part of the sum is $2i^2 + 4i^4 + 6i^6 + \ldots$ , and
the imaginary part is $i + 3i^3 + 5i^5 + \cdots$
Let's take a look at the real part first. We have $i^2=-1$ , hence the real part simplifies to $-2+4-6+8-10+\cdots$ .
If there were an odd number of terms, we could pair them as follows: $-2 + (4-6) + (8-10) + \cdots$ , hence the result would be negative. As we need the real part to be $48$ , we must have an even number of terms. If we have an even number of terms, we can pair them as $(-2+4) + (-6+8) + \cdots$ . Each parenthesis is equal to $2$ , thus there are $24$ of them, and the last value used is $96$ . This happens for $n=96$ and $n=97$ . As $n=96$ is not present as an option, we may conclude that the answer is $\boxed{97}$


Answer: 256.0

Solution: Using the fact that $\log_{x^n}{y^n} = \log_{x}{y}$ , we see that the equation becomes $\log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} = 40$ . Thus, $5\log_{2}{x} = 40$ and $\log_{2}{x} = 8$ , so $x = 2^8 = 256$ , or $\boxed{256}$


Answer: 3.0

For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$ \[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\] Since the bases are equal, we can set the exponents equal, giving us $5x=15$ . Solving the equation gives us $x = \boxed{3}.$

Solution: We can rewrite this expression as $\log(10^x \cdot 100^{2x})=\log(1000^5)$ , which can be simplified to $\log(10^{x}\cdot10^{4x})=5\log(1000)$ , and that can be further simplified to $\log(10^{5x})=5\log(10^3)$ . This leads to $5x=15$ . Solving this linear equation yields $x = \boxed{3}.$


Answer: 3.0

Solution: We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$ \[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\] Since the bases are equal, we can set the exponents equal, giving us $5x=15$ . Solving the equation gives us $x = \boxed{3}.$

Solution: We can rewrite this expression as $\log(10^x \cdot 100^{2x})=\log(1000^5)$ , which can be simplified to $\log(10^{x}\cdot10^{4x})=5\log(1000)$ , and that can be further simplified to $\log(10^{5x})=5\log(10^3)$ . This leads to $5x=15$ . Solving this linear equation yields $x = \boxed{3}.$


Answer: 8.0

For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$

Solution: We have \begin{align*} 2021_b - 221_b &= (2021_b - 21_b) - (221_b - 21_b) \\ &= 2000_b - 200_b \\ &= 2b^3 - 2b^2 \\ &= 2b^2(b-1), \end{align*} which is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is $\boxed{8}.$

Solution:
Vertically subtracting $2021_b - 221_b,$ we see that the ones place becomes $0,$ and so does the $b^1$ place. Then, we perform a carry (make sure the carry is in base $b$ ). Let $b-2 = A.$ Then, we have our final number as \[1A00_b.\] Now, when expanding, we see that this number is simply $b^3 - (b - 2)^2.$
Now, notice that the final number will only be congruent to \[b^3-(b-2)^2\equiv0\pmod{3}.\] If either $b\equiv0\pmod{3},$ or if $b\equiv1\pmod{3}$ (because note that $(b - 2)^2$ would become $\equiv1\pmod{3},$ and $b^3$ would become $\equiv1\pmod{3}$ as well, and therefore the final expression would become $1-1\equiv0\pmod{3}.$ Therefore, $b$ must be $\equiv2\pmod{3}.$ Among the answers, only $8$ is $\equiv2\pmod{3},$ and therefore our answer is $\boxed{8}.$

Solution: By the definition of bases, we have \[2021_b - 221_b = \left(2b^3+2b+1\right) - \left(2b^2+2b+1\right).\] For values $b_1$ and $b_2$ such that $b_1\equiv b_2\pmod{3},$ we get \[\left(2b_1^3+2b_1+1\right) - \left(2b_1^2+2b_1+1\right) \equiv \left(2b_2^3+2b_2+1\right) - \left(2b_2^2+2b_2+1\right) \pmod{3}.\] Note that answer choices $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}$ are congruent to $0,1,0,1,2$ modulo $3,$ respectively. So, $\textbf{(A)}$ and $\textbf{(C)}$ are either both correct or both incorrect. Since there is only one correct answer, $\textbf{(A)}$ and $\textbf{(C)}$ are both incorrect. Similarly, $\textbf{(B)}$ and $\textbf{(D)}$ are both incorrect. This leaves us with $\boxed{8},$ the answer choice with a unique residue modulo $3.$


Answer: 5.0

For which of the following values of $k$ does the equation $\frac{x-1}{x-2} = \frac{x-k}{x-6}$ have no solution for $x$
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$

Solution:
The domain over which we solve the equation is $\mathbb{R} \setminus \{2,6\}$
We can now cross-multiply to get rid of the fractions, we get $(x-1)(x-6)=(x-k)(x-2)$
Simplifying that, we get $7x-6 = (k+2)x - 2k$ . Clearly for $k=5$ we get the equation $-6=-10$ which is never true. The answer is $\boxed{5}$


Answer: 1.0

For which real values of m are the simultaneous equations
\begin{align*}y &= mx + 3 \\  y& = (2m - 1)x + 4\end{align*}
satisfied by at least one pair of real numbers $(x,y)$
$\textbf{(A)}\ \text{all }m\qquad \textbf{(B)}\ \text{all }m\neq 0\qquad \textbf{(C)}\ \text{all }m\neq 1/2\qquad \textbf{(D)}\  \text{all }m\neq 1\qquad \textbf{(E)}\ \text{no values of }m$

Solution:
Solution by e_power_pi_times_i
Solving the systems of equations, we find that $mx+3 = (2m-1)x+4$ , which simplifies to $(m-1)x+1 = 0$ . Therefore $x = \dfrac{1}{1-m}$ $x$ is only a real number if $\boxed{1}$


Answer: 162.0

Forty slips are placed into a hat, each bearing a number $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , or $10$ , with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b \neq a$ . What is the value of $q/p$
$\textbf{(A) } 162 \qquad \textbf{(B) } 180 \qquad \textbf{(C) } 324 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 720$

Solution:
There are $10$ ways to determine which number to pick. There are $4!$ ways to then draw those four slips with that number, and $40 \cdot 39 \cdot 38 \cdot 37$ total ways to draw four slips. Thus $p = \frac{10\cdot 4!}{40 \cdot 39 \cdot 38 \cdot 37}$
There are ${10 \choose 2} = 45$ ways to determine which two numbers to pick for the second probability. There are ${4 \choose 2} = 6$ ways to arrange the order which we draw the non-equal slips, and in each order there are $4 \times 3 \times 4 \times 3$ ways to pick the slips, so $q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}$
Hence, the answer is $\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{162}$

Solution:
For probability $p$ , there are $\binom{10}{1}=10$ ways to choose the number you want to show up $4$ times.
Hence, the probability is $\frac{10}{\binom{40}{4}}$
For probability $q$ , there are $\binom{10}{2}=45$ ways to choose the $2$ numbers you want to show up twice. There are $\binom{4}{2}\cdot\binom{4}{2}$ ways to pick which slips you want out of the $4$ of each.
Hence, the probability is $\frac{45\cdot6\cdot6}{\binom{40}{4}}$
Hence, $\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=\boxed{162}$


Answer: 27.0

Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$

Solution:
Denote by $A$ and $B$ the numbers drawn by Alice and Bob, respectively.
Alice's sentence “I can't tell who has the larger number.” implies $A \in \left\{ 2 , \cdots , 39 \right\}$
Bob's sentence “I know who has the larger number.” implies $B \in \left\{ 1 , 2 , 39, 40 \right\}$
Their subsequent conversation that $B$ is prime implies $B = 2$
Then, Alice's next sentence  “In that case, if I multiply your number by 100 and add my number, the result is a perfect square.” implies $200 + A$ is a perfect square.
Hence, $A = 25$
Therefore, the answer is $\boxed{27}$

Solution: We see that $225$ is one such square that works. Bob gets $2$ and Alice gets $25$ which is valid. Thus, $2 + 25 = 27.$ So $\boxed{27}$ is our answer.


Answer: 27.0

Solution:
Let Alice have the number A, Bob B.
When Alice says that she can't tell who has the larger number, it means that $A$ cannot equal $1$ . Therefore, it makes sense that Bob has $2$ because he now knows that Alice has the larger number. $2$ is also prime. The last statement means that $200+A$ is a perfect square. The three squares in the range $200-300$ are $225$ $256$ , and $289$ . So, $A$ could equal $25$ $56$ , or $89$ , so $A+B$ is $27$ $58$ , or $91$ , of only $\boxed{27}$ is an answer choice.


Answer: 742.0

Forty teams play a tournament in which every team plays every other team exactly once.  No ties occur, and each team has a $50 \%$ chance of winning any game it plays.  The probability that no two teams win the same number of games is $\frac mn,$ where $m_{}$ and $n_{}$ are relatively prime positive integers.  Find $\log_2 n.$

Solution:
There are ${40 \choose 2} = 780$ total pairings of teams, and thus $2^{780}$ possible outcomes. In order for no two teams to win the same number of games, they must each win a different number of games. Since the minimum and maximum possible number of games won are 0 and 39 respectively, and there are 40 teams in total, each team corresponds uniquely with some $k$ , with $0 \leq k \leq 39$ , where $k$ represents the number of games the team won. With this in mind, we see that there are a total of $40!$ outcomes in which no two teams win the same number of games. Further, note that these are all the valid combinations, as the team with 1 win must beat the team with 0 wins, the team with 2 wins must beat the teams with 1 and 0 wins, and so on; thus, this uniquely defines a combination.
The desired probability is thus $\frac{40!}{2^{780}}$ . We wish to simplify this into the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime. The only necessary step is to factor out all the powers of 2 from $40!$ ; the remaining number is clearly relatively prime to all powers of 2.
The number of powers of 2 in $40!$ is $\left \lfloor \frac{40}{2} \right \rfloor + \left \lfloor \frac{40}{4} \right \rfloor + \left \lfloor \frac{40}{8} \right \rfloor + \left \lfloor \frac{40}{16} \right \rfloor + \left \lfloor \frac{40}{32} \right \rfloor = 20 + 10 + 5 + 2 + 1 = 38.$
$780-38 = \boxed{742}$


Answer: 520.0

Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$ . Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table under these conditions. Find the remainder when $N$ is divided by $1000$

Solution:
Let us first consider the $4$ ambassadors and the $6$ even-numbered chairs. If we consider only their relative positions, they can sit in one of $3$ distinct ways: Such that the $2$ empty even-numbered chairs are adjacent, such that they are separated by one occupied even-numbered chair, and such that they are opposite each other. For each way, there are $4!=24$ ways to assign the $4$ ambassadors to the $4$ selected seats.
In the first way, there are $6$ distinct orientations. The $4$ advisors can be placed in any of the $5$ odd-numbered chairs adjacent to the ambassadors, and for each placement, there is exactly one way to assign them to the ambassadors. This means that there are $24\cdot6\cdot\binom{5}{4}=720$ total seating arrangements for this way.
In the second way, there are $6$ distinct orientations. $3$ advisors can be placed in any of the $4$ chairs adjacent to the "chunk" of $3$ ambassadors, and $1$ advisor can be placed in either of the $2$ chairs adjacent to the "lonely" ambassador. Once again, for each placement, there is exactly one way to assign the advisors to the ambassadors. This means that there are $24\cdot6\cdot\binom{4}{3}\cdot\binom{2}{1}=1152$ total seating arrangements for this way.
In the third way, there are $3$ distinct orientations. For both "chunks" of $2$ ambassadors, $2$ advisors can be placed in any of the $3$ chairs adjacent to them, and once again, there is exactly one way to assign them for each placement. This means that there are $24\cdot3\cdot\binom{3}{2}\cdot\binom{3}{2}=648$ total seating arrangements for this way.
Totalling up the arrangements, there are $720+1152+648=2520$ total ways to seat the people, and the remainder when $2520$ is divided by $1000$ is $\boxed{520}$ . ~ emerald_block

Solution:
In the diagram, the seats are numbered 1...12.  Rather than picking seats for each person, however, each ambassador/assistant team picks a gap between the seats (A...L) and the ambassador sits in the even seat while the assistant sits in the odd seat.  For example, if team 1 picks gap C then Ambassador 1 will sit in seat 2 while assistant 1 will sit in seat 3.  No two teams can pick adjacent gaps.  For example, if team 1 chooses gap C then team 2 cannot pick gaps B or D.  In the diagram, the teams have picked gaps C, F, H and J.  Note that the gap-gaps - distances between the chosen gaps - (in the diagram, 2, 1, 1, 4) must sum to 8.  So, to get the number of seatings, we:
So, the total is $12\cdot35\cdot6=2520$
Therefore, the remainder is $\boxed{520}$

Solution: There are $6\cdot5\cdot4\cdot3\cdot2^4$ (placing the ambassadors and then choosing the locations of the advisors) total ways to let everyone sit. However, this may lead to advisors sitting in the same chair, leading to awkward situations. So we find how many ways this happens. There are $\dbinom{4}{2} = 6$ ways to choose which advisors end up sitting together, times $6\cdot2=12$ ways to find neighboring even seats (order matters), and for the other two ambassadors, there are $\dbinom{4}{2} \cdot 2 = 12$ ways to sit and 4 ways for their advisors to sit. This results in $6\cdot12\cdot12\cdot4=3456$ ways for this to happen. However, we overcounted the case when all four of the advisors run out of room to sit, where there are $\frac{6\cdot3}{2!}=9$ to choose the places the ambassadors sit and 4! ways to arrange the ambassadors. $9\cdot4!=216$ ways to happen. So the answer is $5760 - 3456 + 216 = 2520$ , which has a remainder of $\boxed{520}$ when divided by 1000.

Solution:
We see that for every 2 adjacent spots on the table, there is exactly one way for an ambassador and his or her partner to sit. There are 2 cases:
For the first case, there are 3 pairs and 4 empty spaces, which results in $\frac{7!}{3!4!} = 35$ ways to arrange the pairs.
For the second case, there are 4 pairs and 4 empty spaces, which results in $\frac{8!}{4!4!} = 70$ ways to arrange the pairs.
Finally, we have to assign ambassadors and their advisors to the pairs, which has $4!=24$ ways, so $N = (35+70)\cdot 24 = 2520$ , or $\boxed{520}$ ~PenguinMoosey


Answer: 13.0

Four boys bought a boat for $\textdollar 60$ . The first boy paid one half of the sum of the amounts paid by the other boys;
the second boy paid one third of the sum of the amounts paid by the other boys;
and the third boy paid one fourth of the sum of the amounts paid by the other boys. How much did the fourth boy pay?
$\textbf{(A) }\textdollar 10\qquad \textbf{(B) }\textdollar 12\qquad \textbf{(C) }\textdollar 13\qquad \textbf{(D) }\textdollar 14\qquad \textbf{(E) }\textdollar 15$

Solution:
If the first boy paid one half of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{3}$ of the total.
If the second boy paid one third of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{4}$ of the total.
If the third boy paid one fourth of the sum of the  amounts paid by the other boys, that means that he paid $\frac{1}{5}$ of the total.
Summing it up, we get $\textdollar 20 + \textdollar 15 + \textdollar 12 = \textdollar 47$ .
Therefore, our answer is $\textdollar 60 - \textdollar 47 = \boxed{13}$ ~awin


Answer: 389.0

Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution:
First, apply Heron's formula to find that $[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$ . The semiperimeter is $21$ , so the inradius is $\frac{A}{s} = \frac{84}{21} = 4$
Now consider the incenter $I$ of $\triangle ABC$ . Let the radius of one of the small circles be $r$ . Let the centers of the three little circles tangent to the sides of $\triangle ABC$ be $O_A$ $O_B$ , and $O_C$ . Let the center of the circle tangent to those three circles be $O$ . The homothety $\mathcal{H}\left(I, \frac{4-r}{4}\right)$ maps $\triangle ABC$ to $\triangle XYZ$ ; since $OO_A = OO_B = OO_C = 2r$ $O$ is the circumcenter of $\triangle XYZ$ and $\mathcal{H}$ therefore maps the circumcenter of $\triangle ABC$ to $O$ . Thus, $2r = R \cdot \frac{4 - r}{4}$ , where $R$ is the circumradius of $\triangle ABC$ . Substituting $R = \frac{abc}{4[ABC]} = \frac{65}{8}$ $r = \frac{260}{129}$ and the answer is $\boxed{389}$

Solution:
2007 AIME II-15b.gif
Consider a 13-14-15 triangle. $A=84.$ [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is $r=\frac{A}{s}=\frac{84}{21}=4$ , where $s$ is the semiperimeter.  Scale the triangle with the inradius by a linear scale factor, $u.$
The circumradius is $R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},$ where $a,$ $b,$ and $c$ are the side-lengths.  Scale the triangle with the circumradius by a linear scale factor, $v$
Cut and combine the triangles, as shown.  Then solve for $4u$
The solution is $260+129=\boxed{389}$

Solution:
Let $A'$ $B'$ $C'$ , and $O$ be the centers of circles $\omega_{A}$ $\omega_{B}$ $\omega_{C}$ $\omega$ , respectively, and let $x$ be their radius.
Now, triangles $ABC$ and $A'B'C'$ are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for $x$
Since $OA'=OB'=OC'=2x$ $O$ is the circumcenter of triangle $A'B'C'$ and its circumradius is $2x$ . Let $I$ denote the incenter of triangle $ABC$ and $r$ the inradius of $ABC$ . Then the inradius of $A'B'C'=r-x$ , so now we compute r. Computing the inradius by $A=rs$ , we find that the inradius of $ABC$ is $4$ . Additionally, using the circumradius formula $R=\frac{abc}{4K}$ where $K$ is the area of $ABC$ and $R$ is the circumradius, we find $R=\frac{65}{8}$ . Now we can equate the ratio of circumradius to inradius in triangles $ABC$ and $A'B'C'$
\[\frac{\frac{65}{8}}{4}=\frac{2x}{4-x}\]
Solving, we get $x=\frac{260}{129}$ , so our answer is $260+129=\boxed{389}$

Solution: According to the diagram, it is easily to see that there is a small triangle made by the center of three circles which aren't in the middle. The circumradius of them is $2r$ . Now denoting $AB=13;BC=14;AC=15$ , and centers of circles tangent to $AB,AC;AC,BC;AB,BC$ are relatively $M,N,O$ with $OJ,NK$ both perpendicular to $BC$ . It is easy to know that $tanB=\frac{12}{5}$ , so $tan\angle OBJ=\frac{2}{3}$ according to half angle formula. Similarly, we can find $tan\angle NCK=\frac{1}{2}$ . So we can see that $JK=ON=14-\frac{7x}{2}$ . Obviously, $\frac{2x}{14-\frac{7x}{2}}=\frac{65}{112}$ . After solving, we get $x=\frac{260}{129}$ , so our answer is $260+129=\boxed{389}$ . ~bluesoul

Solution:
Let $O_A, O_B, O_C, O$ be the centers of $w_A, w_B, w_C, w$ , respectively. Also, let $I$ be the incenter of $ABC$ and $r$ be the radius of circle $w$ . Since $AB||O_AO_B$ $BC||O_BO_C$ , and $CA||O_CO_A$ , we know that
\[\angle BAI = \angle O_BO_AI, \angle CBI = \angle O_CO_BI, \angle ACI = \angle O_AO_CI \text{ and }\angle CAI = \angle O_CO_AI, \angle ABI = \angle O_AO_BI, \angle BCI = \angle O_BO_CI.\]
That means $\angle ABC = \angle O_AO_BO_C$ $\angle BAC = \angle O_BO_AO_C$ , and $\angle ACB = \angle O_AO_CO_B$ . Thus, $\triangle ABC \sim \triangle O_AO_BO_C$ . We also know that we are scaling each side of $\triangle ABC$ (from $AB$ to $O_AO_B$ for instance), about $I$ (since A,O_A,I are collinear ; same apply with $B$ and $C$ ).
Now, let the homothety $\mathcal{H} (I, x)$ map $\triangle ABC$ to $\triangle O_AO_BO_C$ . To start off, we know the circumradius of $O_AO_BO_C$ is $O$ , since $OO_A = OO_B = OO_C = 2r$ . Since $O_AO_B = 13x$ $O_BO_C = 14x$ $O_CO_A = 15x$ , we can get an relationship involving $x$ and $r$ via another way to find the circumradius:
\[[\triangle O_AO_BO_C ] =\frac{abc}{4R} \Longrightarrow 84x^2 =\frac{13x\cdot 14x\cdot 15x}{4\cdot 2r} \Longrightarrow r=\frac{65x}{16}\]
Take notice of the inradius of $ABC$ . We get the inradius to be $[\triangle ABC ] = sr_0 \Longrightarrow r_0=4$ . Let the tangency point of the incircle and side $BC$ be $H$ . We know $IH = 4$ . We also know that we can cut off the part of $IH$ that is outside of $\triangle O_AO_BO_C$ to get the inradius of $\triangle O_AO_BO_C$ . To part that is outside $\triangle O_AO_BO_C$ turns out just to be the radius of circle $w_B$ (as seen in the picture). That means the inradius of $\triangle O_AO_BO_C$ is just $4-r$ . We can calculate that incradius in another way, though. We know that the inradius of $\triangle ABC$ is $4$ , which means the inradius of $\triangle O_AO_BO_C$ is just $4x$ (by our homethety ratio).
Thus, we have $4x = 4-r = 4-\dfrac{65x}{16} \Longrightarrow x = \dfrac{64}{129} \Longrightarrow r = \dfrac{260}{129}$ . That gives $\boxed{389}$ as our final answer.


Answer: 7.7

Solution: The side of the square of made up of two radii, and the area of the square is $(3+3)^2=36$ . The unshaded region of the square is composed of four quarter circles, and has the area of one circle, which is $9\pi$ . The area of the shaded region is $36-9\pi$ which less than $36-9(3)=9$ , closest to $\boxed{7.7}$