Answer: 996.0

What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?
$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$

Solution: The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$ , and we want this not to be a divisor of $n!$ (the product of the first $n$ positive integers). Notice that if and only if $n+1$ were composite, all of its factors would be less than or equal to $n$ , which means they would be able to cancel with the factors in $n!$ . Thus, the sum of $n$ positive integers would be a divisor of $n!$ when $n+1$ is composite. (Note: This is true for all positive integers except for 1 because 2 is not a divisor/factor of 1.) Hence in this case, $n+1$ must instead be prime. The greatest three-digit integer that is prime is $997$ , so we subtract $1$ to get $n=\boxed{996}$

Solution: As in Solution 1, we deduce that $n+1$ must be prime. If we can't immediately recall what the greatest three-digit prime is, we can instead use this result to eliminate answer choices as possible values of $n$ . Choices $A$ $C$ , and $E$ don't work because $n+1$ is even, and all even numbers are divisible by two, which makes choices $A$ $C$ , and $E$ composite and not prime. Choice $D$ also does not work since $999$ is divisible by $9$ , which means it's a composite number and not prime. Thus, the correct answer must be $\boxed{996}$

Solution:
The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$ and the product of the positive integers upto $n$ is $n!$ . The quotient of the two is -
$\frac{(2)(n!)}{(n)(n+1)}$
which simplifies to $\frac{(2)((n-1)!)}{n+1}$ . Thus, $n+1$ must be odd for the remainder to not be 0 (as $2$ will multiply with some number in $n!$ , cancelling out $n+1$ if it is even, which leaves us with the answer choices $n = 996$ and $n = 998$ . Notice that $n + 1$ must also be prime as otherwise there will be a factor of $n + 1$ in $2$ $n!$ somewhere. So either $997$ or $999$ must be prime - $999$ is obviously not prime as it is divisible by 9, so our answer should be $n$ where $n + 1 = 997$ , and so our answer is $n = 996$ or $\boxed{996}$


Answer: 6.0

What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$

Solution:
Since $2011 \equiv 11  \pmod{1000},$ we know that $2011^{2011} \equiv 11^{2011}  \pmod{1000}.$
To compute this, we use a clever application of the binomial theorem
$\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}$
In all of the other terms, the power of $10$ is greater than $2$ and so is equivalent to $0$ modulo $1000,$ which means we can ignore it. We have:
$\begin{aligned}11^{2011} &\equiv 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\&=611 \pmod{1000} \end{aligned}$
Therefore, the hundreds digit is $\boxed{6}.$

Solution:
We need to compute $2011^{2011} \pmod{1000}.$ By the Chinese Remainder Theorem , it suffices to compute $2011^{2011} \pmod{8}$ and $2011^{2011} \pmod{125}.$
In modulo $8,$ we have $2011^4 \equiv 1 \pmod{8}$ by Euler's Theorem, and also $2011 \equiv 3 \pmod{8},$ so we have \[2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.\]
In modulo $125,$ we have $2011^{100} \equiv 1 \pmod{125}$ by Euler's Theorem, and also $2011 \equiv 11 \pmod{125}.$ Therefore, we have $\begin{aligned} 2011^{2011} &= (2011^{100})^{20} \cdot 2011^{11} \\ &\equiv 1^{20} \cdot 11^{11} \\ &= 121^5 \cdot 11 \\ &= (-4)^5 \cdot 11 = -1024 \cdot 11 \\ &\equiv -24 \cdot 11 = -264 \\ &\equiv 111 \pmod{125}. \end{aligned}$
After finding the solution $2011^{2011} \equiv 611 \pmod{1000},$ we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is $\boxed{6}.$

Solution: Notice that the hundreds digit of $2011^{2011}$ won't be affected by $2000$ . Essentially we could solve the problem by finding the hundreds digit of $11^{2011}$ . Powers of $11$ are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of $11$ can be found by reading rows of the triangle and adding extra numbers up. [add source] For example, the sixth row of the triangle is $1, 5, 10, 10, 5,$ and $1$ . Adding all numbers from right to left, we get $161051$ , which is also $11^5$ . In other words, each number is $10^n$ steps from the right side of the row. The hundreds digit is $0$ . We can do the same for $11^{2011}$ , but we only need to find the $3$ digits from the right. Observing, every $3$ number from the right is $1 + 2 + 3... + n$ . So to find the third number from the right on the row of $11^{2011}$ \[f(11^n) = 1 + 2 + 3... + (n-1),\] or $\frac{(2010 \cdot 2011)}{2}$ , or $2021055$ . The last digit is five, but we must remember to add the number on the right of it, which, by observing other rows is obviously $2011$ . We must carry the $1$ in $2011$ 's tens digit to the $5$ in $2021055$ 's unit digit to get $\boxed{6}$ . The one at the very end of the row doesn't affect anything, so we can leave it alone.

Solution:
Since we are only looking at the last 3 digits and $2011^2$ has the same last 3 digits as $11^2$ , we can find $11^{2011}$ instead.
After this, we can repeatedly multiply the last 3 digits by 11 and take the last 3 digits of that product. We discover that $11^{51}$ 's last 2 digits are -11, the same as $11^1$
From this information, we can figure out $11^{11}$ and $11^{61}$ end in 611. Adding various multiples of 50 to the exponent gives us the fact that $11^{2011}$ 's last digits are 611. We get $\boxed{6}$ .
-ThisUsernameIsTaken

Solution:
We know that the hundreds digit of $2011^{2011}$ is just the hundreds digit of $11^{2011} \equiv 011 \pmod{1000}$ . If we actually take a look at the powers of $11 \pmod{1000}$ , we notice a pattern:
\[11^{1} \equiv 011 \pmod{1000}\] \[11^{2} \equiv 121 \pmod{1000}\] \[11^{3} \equiv 331 \pmod{1000}\] \[11^{4} \equiv 641 \pmod{1000}\] \[11^{5} \equiv 051 \pmod{1000}\] \[11^{6} \equiv 561 \pmod{1000}\] Notice how the units digit is always one, the tens digit is always the previous tens digit plus the ones digit (or one) and the hundreds digit is always the previous hundreds digit plus the previous tens digit. Knowing this, we confidently repeat this pattern without actually multiply the previous term by $11$ out (if you generally multiply a few numbers by $11$ , you can see why this pattern holds).
\[11^{7} \equiv 171 \pmod{1000}\] \[11^{8} \equiv 881 \pmod{1000}\] \[11^{9} \equiv 691 \pmod{1000}\] \[11^{10} \equiv 601 \pmod{1000}\] \[11^{11} \equiv 611 \pmod{1000}\] Since $11^{11}$ ends in _ $11$ , the pattern "cycles" every $10$ times. $\frac{2011}{10}=201$ remainder $1$ , meaning that $2011^{2011} \equiv 11^{2011} \equiv 611 \pmod{1000}$ , giving us a hundreds digit of $\boxed{6}$

Solution:
We know that the hundreds digit of $2011^{11}$ is just the hundreds digit of $11^{11}$ (mod 1000). If we actually take a look at the powers of 11, we notice a pattern:
\[11^{1} = 11\]
\[11^{2} = 121\]
\[11^{3} = 1331\]
\[11^{4} = 14641\]
We notice that the hundreds digit just increases by 1 starting from 1(the hundreds digit of $11^{1}$ ). All we have to do now is add all the numbers from 1 to 11 (since the hundreds digit increases by 1 with every power of 11 that comes after) to get the hundreds digit of $11^{11}$
$1 + 2 + 3 + 4.... + 11 = 66$ . Thus the answer is $\boxed{6}$


Answer: 61.0

What is the largest $2$ -digit prime factor of the integer $n = {200\choose 100}$

Solution: Expanding the binomial coefficient , we get ${200 \choose 100}=\frac{200!}{100!100!}$ . Let the required prime be $p$ ; then $10 \le p < 100$ . If $p > 50$ , then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$ . The largest such prime is $\boxed{061}$ , which is our answer.

Solution:
We know that \[{200\choose100}=\frac{200!}{100!100!}\] Since $p<100$ , there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\frac{200!}{100!100!}$ . (Note that here we assume the minimum because as $p$ goes larger in value, the number of factors of $p$ in a number decreases,)
So basically, $p$ is the largest prime number such that \[\left \lfloor\frac{200}{p}\right \rfloor>3\] Since $p<\frac{200}{3}=66.66...$ , the largest prime value for $p$ is $p=\boxed{61}$


Answer: 28.0

What is the largest difference that can be formed by subtracting two numbers chosen from the set $\{ -16,-4,0,2,4,12 \}$
$\text{(A)}\ 10 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 28 \qquad \text{(E)}\ 48$

Solution: To maximize anything of the form $a-b$ , we maximize $a$ and minimize $b$ .  The maximal element of the set is $12$ and the minimal element is $-16$ , so the maximal difference is \[12-(-16)=28\rightarrow \boxed{28}.\]


Answer: 38.0

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

Solution:
Take an even positive integer $x$ $x$ is either $0 \bmod{6}$ $2 \bmod{6}$ , or $4 \bmod{6}$ . Notice that the numbers $9$ $15$ $21$ , ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases:
If $x \ge 18$ and is $0 \bmod{6}$ $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$ . Note that $9$ and $9+6n$ are both odd composites.
If $x\ge 44$ and is $2 \bmod{6}$ $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$ . Note that $35$ and $9+6n$ are both odd composites.
If $x\ge 34$ and is $4 \bmod{6}$ $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$ . Note that $25$ and $9+6n$ are both odd composites.
Clearly, if $x \ge 44$ , it can be expressed as a sum of 2 odd composites. However, if $x = 42$ , it can also be expressed using case 1, and if $x = 40$ , using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$ . Therefore, $\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.

Solution: Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$ , then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$ , which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$ , yielding a maximal answer of 38. Since $38-25=13$ , which is prime, the answer is $\boxed{038}$

Solution:
The easiest method is to notice that any odd number that ends in a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...
For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9(ex. 34, 44 . . .). Hence the largest number that ends with a 4 that satisfies the conditions is 14.
If you list out all the numbers(15, 27, 9, 21, 33), you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and any number greater that ends with a 3 is bad(ex. 58, 68. . .), so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be $\boxed{38}$

Solution: Claim: The answer is $\boxed{038}$

Solution:
All numbers that could possibly work must be $2 \cdot p$ where $p$ is prime. As previous solutions stated, the maximum number that could possibly work by Chicken McNugget is $9 \cdot 25 - 9 - 25 = 225-34 = 191$ . We then bash from top to bottom:
1. $178 = 89 \cdot 2 => 87 + 91$ - refuted
2. $166 = 83 \cdot 2 => 81 + 85$ - refuted
3. $158 = 79 \cdot 2 => 77 + 81$ - refuted
4. $146 = 73 \cdot 2 => 69 + 77$ - refuted
5. $142 = 71 \cdot 2 => 65 + 77$ - refuted
6. $134 = 67 \cdot 2 => 65 + 69$ - refuted
7. $122 = 61 \cdot 2 => 57 + 65$ - refuted
8. $118 = 59 \cdot 2 => 55 + 63$ - refuted
9. $106 = 53 \cdot 2 => 51 + 55$ - refuted
10. $94 = 47 \cdot 2 => 45 + 49$ - refuted
11. $86 = 43 \cdot 2 => 35 + 51$ - refuted
12. $82 = 41 \cdot 2 => 33 + 49$ - refuted
13. $74 = 37 \cdot 2 => 35 + 39$ - refuted
14. $62 = 31 \cdot 2 => 27 + 35$ - refuted
15. $58 = 29 \cdot 2 => 25 + 33$ - refuted
16. $46 = 23 \cdot 2 => 21 + 25$ - refuted
17. $38 = 19 \cdot 2 => = 19 + 19$ - it works!
Because we did a very systematic bash as shown, we are confident the answer is $\boxed{038}$

Solution:
As stated above, all numbers that could possibly work must be $2 \cdot p$ where $p$ is prime. If $p$ > 30, we consider $p$ by modulo 30. $p$ could be 1,7,11,13,17,19,23,29 modulo 30. $2 \cdot p$ can be expressed as ( $p$ $q$ )+( $p$ $q$ ) for some positive, even $q$ less then $p$
If $p$ $1 \bmod{30}$ , p±4 would both be composite
If $p$ $7 \bmod{30}$ , p±2 would both be composite
If $p$ $11 \bmod{30}$ , p±14 would both be composite
If $p$ $13 \bmod{30}$ , p±8 would both be composite
If $p$ $17 \bmod{30}$ , p±8 would both be composite
If $p$ $19 \bmod{30}$ , p±14 would both be composite
If $p$ $23 \bmod{30}$ , p±2 would both be composite
If $p$ $29 \bmod{30}$ , p±4 would both be composite
So $p$ < 30
From here, just try all possible p and find the answer is $\boxed{038}$


Answer: 15.0

What is the largest integer that is a divisor of
\[(n+1)(n+3)(n+5)(n+7)(n+9)\]
for all positive even integers $n$
$\text {(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165$

Solution: For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is $3 \cdot 5 = 15$ , so ${\boxed{15}$ is the correct answer.

Solution:
We'll just test all the answer choices.
Note that for any 3 consecutive odd integers, there will be exactly one multiple of $3.$
Let's list all possibilities of 3 consecutive odd integers. (multiple of 3, not multiple of 3, not multiple of 3), (not multiple of 3, multiple of 3, not multiple of 3) and (not multiple of 3, not multiple of 3, multiple of 3)
To support this further, list the first few  consecutive lists of 3 consecutive odd integers.
We have $(1, 3, 5), (3, 5, 7), (5, 7, 9), (7, 9, 11), (11, 13, 15), \ldots$
So if the first two consecutive odd numbers aren't multiples of three, the last one must be a multiple of three.
Therefore for five consecutive odd integers, there must be at least one multiple of three.
In the same fashion as we did above, note that for any 5 consecutive integers, there will also be exactly one multiple of $5.$
Therefore, for any 5 consecutive odd integers, there must be exactly one multiple of five.
We can skip 7 since none of the answer choices are a multiple of 7.
Now we try $11.$ 11 doesn't work since as we see the first set of 5 consecutive odd integers doesn't fit, namely $(1, 3, 5, 7, 9).$
Since any 5 consecutive integers is divisible both by $3$ and $5$ , it also must be divisible by ${\boxed{15}$ and no higher since we saw that $11$ does not work and that there is no answer choice that is multiple of $7$


Answer: 15.0

Solution: For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is $3 \cdot 5 = 15$ , so ${\boxed{15}$ is the correct answer.

Solution:
We'll just test all the answer choices.
Note that for any 3 consecutive odd integers, there will be exactly one multiple of $3.$
Let's list all possibilities of 3 consecutive odd integers. (multiple of 3, not multiple of 3, not multiple of 3), (not multiple of 3, multiple of 3, not multiple of 3) and (not multiple of 3, not multiple of 3, multiple of 3)
To support this further, list the first few  consecutive lists of 3 consecutive odd integers.
We have $(1, 3, 5), (3, 5, 7), (5, 7, 9), (7, 9, 11), (11, 13, 15), \ldots$
So if the first two consecutive odd numbers aren't multiples of three, the last one must be a multiple of three.
Therefore for five consecutive odd integers, there must be at least one multiple of three.
In the same fashion as we did above, note that for any 5 consecutive integers, there will also be exactly one multiple of $5.$
Therefore, for any 5 consecutive odd integers, there must be exactly one multiple of five.
We can skip 7 since none of the answer choices are a multiple of 7.
Now we try $11.$ 11 doesn't work since as we see the first set of 5 consecutive odd integers doesn't fit, namely $(1, 3, 5, 7, 9).$
Since any 5 consecutive integers is divisible both by $3$ and $5$ , it also must be divisible by ${\boxed{15}$ and no higher since we saw that $11$ does not work and that there is no answer choice that is multiple of $7$


Answer: 3.0

What is the largest number of acute angles that a convex hexagon can have?
$\textbf{(A)}\  2 \qquad \textbf{(B)}\  3 \qquad \textbf{(C)}\  4\qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\  6$

Solution:
The sum of the interior angles of a hexagon is $720$ degrees.  In a convex polygon, each angle must be strictly less than $180$ degrees.
Six acute angles can only sum to less than $90\cdot 6 = 540$ degrees, so six acute angles could not form a hexagon.
Five acute angles and one obtuse angle can only sum to less than $90\cdot 5 + 180 = 630$ degrees, so these angles could not form a hexagon.
Four acute angles and two obtuse angles can only sum to less than $90\cdot 4 + 180\cdot 2 = 720$ degrees.  This is a strict inequality, so these angles could not form a hexagon.  (The limiting figure would be four right angles and two straight angles, which would really be a square with two "extra" points on two sides to form the straight angles.)
Three acute angles and three obtuse angles work.  For example, if you pick three acute angles of $80$ degrees, the three obtuse angles would be $160$ degrees and give a sum of $80\cdot 3 + 160\cdot 3 = 720$ degrees, which is a genuine hexagon.  Thus, the answer is $\boxed{3}$


Answer: 4.0

What is the largest number of solid $2\text{-in} \times 2\text{-in} \times 1\text{-in}$ blocks that can fit in a $3\text{-in} \times 2\text{-in}\times3\text{-in}$ box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution: We find that the volume of the larger block is $18$ , and the volume of the smaller block is $4$ . Dividing the two, we see that only a maximum of four $2$ by $2$ by $1$ blocks can fit inside the $3$ by $3$ by $2$ block. Drawing it out, we see that such a configuration is indeed possible. Therefore, the answer is $\boxed{4}$


Answer: 112.0

What is the largest positive integer $n$ for which there is a unique integer $k$ such that $\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}$

Solution: Flip the fractions and subtract one from all sides to yield \[\frac{7}{8}>\frac{k}{n}>\frac{6}{7}.\] Multiply both sides by $56n$ to get \[49n>56k>48n.\] This is equivalent to find the largest value of $n$ such that there is only one multiple of 56 within the open interval between $48n$ and $49n$ .  If $n=112,$ then $98>k>96$ and $k=97$ is the unique value. For $n\geq 113,$ there is at least $(49\cdot 113-48\cdot 113)-1=112$ possible numbers for $k$ and there is one $k$ every 56 numbers. Hence, there must be at least two values of $k$ that work.  So, the largest value of $n$ is $\boxed{112}$

Solution:
Notice that in order for $k$ to be unique, $\frac{n}{n + k+ 1} \le \frac{8}{15}$ and $\frac{n}{n+ k-1} \ge \frac{7}{13}$ must be true. Solving these inequalities for $k$ yields $\frac{7}{6}(k-1) \le n \le \frac{8}{7}(k+1)$
Thus, we want to find $k$ such that $\frac{7}{6}(k-1)\le \frac{8}{7}(k+1)$ . Solving this inequality yields $k \le 97$ , and plugging this into $\frac{n}{n+k} < \frac{7}{13}$ in the original equation yields $n \le 112$ so the answer is $\boxed{112}$


Answer: 215.0

What is the largest positive integer that is not the sum of a positive integral multiple of $42$ and a positive composite integer?

Solution:
Let our answer be $n$ . Write $n = 42a + b$ , where $a, b$ are positive integers and $0 \leq b < 42$ . Then note that $b, b + 42, ... , b + 42(a-1)$ are all primes.
If $b$ is $0\mod{5}$ , then $b = 5$ because $5$ is the only prime divisible by $5$ . We get $n = 215$ as our largest possibility in this case.
If $b$ is $1\mod{5}$ , then $b + 2 \times 42$ is divisible by $5$ and thus $a \leq 2$ . Thus, $n \leq 3 \times 42 = 126 < 215$
If $b$ is $2\mod{5}$ , then $b + 4 \times 42$ is divisible by $5$ and thus $a \leq 4$ . Thus, $n \leq 5 \times 42 = 210 < 215$
If $b$ is $3\mod{5}$ , then $b + 1 \times 42$ is divisible by $5$ and thus $a = 1$ . Thus, $n \leq 2 \times 42 = 84 < 215$
If $b$ is $4\mod{5}$ , then $b + 3 \times 42$ is divisible by $5$ and thus $a \leq 3$ . Thus, $n \leq 4 \times 42 = 168 < 215$
Our answer is $\boxed{215}$


Answer: 137.0

What is the largest possible distance between two points , one on the sphere of radius 19 with center $(-2,-10,5)$ and the other on the sphere of radius 87 with center $(12,8,-16)$

Solution: The distance between the two centers of the spheres can be determined via the distance formula in three dimensions: $\sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} = \sqrt{14^2 + 18^2 + 21^2} = 31$ . The largest possible distance would be the sum of the two radii and the distance between the two centers, making it $19 + 87 + 31 = \boxed{137}$


Answer: 32.0

What is the largest power of $2$ that is a divisor of $13^4 - 11^4$
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$

Solution: First, we use difference of squares on $13^4 - 11^4 = (13^2)^2 - (11^2)^2$ to get $13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2)$ . Using difference of squares again and simplifying, we get $(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)$ . Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of $2$ that is a divisor $13^4 - 11^4$ is $\boxed{32}$

Solution: Just like in the above solution, we use the difference-of-squares factorization, but only once to get $13^4-11^4=(13^2-11^2)(13^2+11^2).$ We can then compute that this is equal to $48\cdot290.$ Note that $290=2\cdot145$ (we don't need to factorize any further as $145$ is already odd) thus the largest power of $2$ that divides $290$ is only $2^1=2,$ while $48=2^4\cdot3,$ so the largest power of $2$ that divides $48$ is $2^4=16.$ Hence, the largest power of $2$ that is a divisor of $13^4-11^4$ is $2\cdot16=\boxed{32}.$

Solution:
Let $n=13^4-11^4.$ We wish to find the largest power of $2$ that divides $n$
Denote $v_p(k)$ as the largest exponent of $p$ in the prime factorization of $n$ . In this problem, we have $p=2$
By the Lifting the Exponent Lemma on $n$
\[v_2(13^4-11^4)=v_2(13-11)+v_2(4)+v_2(13+11)-1\] \[=v_2(2)+v_2(4)+v_2(24)-1\] \[=1+2+3-1=5.\]
Therefore, exponent of the largest power of $2$ that divids $13^4-11^4$ is $5,$ so the largest power of $2$ that divides this number is $2^5=\boxed{32}$

Solution:
We can simply take 13 to the 4th power, which is 28561. We subtract that by 11 to the 4th power, which is 14641 (You can use Pascal's Triangle to find this). Finally, subtract the numbers to get 13920.
To test the options, since we need the largest one, we can go from top down. Testing, we see that both D and E are decimals,
and 32 works. So, our answer is $\boxed{32}.$


Answer: 12.0

What is the largest quotient that can be formed using two numbers chosen from the set $\{ -24, -3, -2, 1, 2, 8 \}$
$\text{(A)}\ -24 \qquad \text{(B)}\ -3 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$

Solution:
Let the two chosen numbers be $a$ and $b$ .  To maximize the quotient, we first have either $a,b>0$ or $a,b<0$ , and from there we maximize $|a|$ and minimize $|b|$
For the case $a,b<0$ , we have $a=-24$ and $b=-2$ , which gives us $(-24)/(-2)=12$ .  For the case $a,b>0$ , we have $a=8$ and $b=1$ , which gives us $8/1=8$
Since $12>8$ , our answer is $\boxed{12}$


Answer: 1.0

What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$
$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$

Solution: Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$ . By the Trivial Inequality (all squares are nonnegative) the minimum value for this is $\boxed{1}$ , which can be achieved at $x=y=0$

Solution: We expand the original expression, then factor the result by grouping: \begin{align*} (xy-1)^2+(x+y)^2&=\left(x^2y^2-2xy+1\right)+\left(x^2+2xy+y^2\right) \\ &=x^2y^2+x^2+y^2+1 \\ &=x^2\left(y^2+1\right)+\left(y^2+1\right) \\ &=\left(x^2+1\right)\left(y^2+1\right). \end{align*} Clearly, both factors are positive. By the Trivial Inequality, we have \[\left(x^2+1\right)\left(y^2+1\right)\geq\left(0+1\right)\left(0+1\right)=\boxed{1}.\] Note that the least possible value of $(xy-1)^2+(x+y)^2$ occurs at $x=y=0.$

Solution: Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\nabla f(x, y) = \boldsymbol{0}$ . First, find the first partial derivative with respect to x and y and find where they are $0$ \[\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0\] \[\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0\] Thus, there is a local extremum at $(0, 0)$ . Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\implies \boxed{1}$


Answer: 1.0

Solution: Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$ . By the Trivial Inequality (all squares are nonnegative) the minimum value for this is $\boxed{1}$ , which can be achieved at $x=y=0$

Solution: We expand the original expression, then factor the result by grouping: \begin{align*} (xy-1)^2+(x+y)^2&=\left(x^2y^2-2xy+1\right)+\left(x^2+2xy+y^2\right) \\ &=x^2y^2+x^2+y^2+1 \\ &=x^2\left(y^2+1\right)+\left(y^2+1\right) \\ &=\left(x^2+1\right)\left(y^2+1\right). \end{align*} Clearly, both factors are positive. By the Trivial Inequality, we have \[\left(x^2+1\right)\left(y^2+1\right)\geq\left(0+1\right)\left(0+1\right)=\boxed{1}.\] Note that the least possible value of $(xy-1)^2+(x+y)^2$ occurs at $x=y=0.$

Solution: Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\nabla f(x, y) = \boldsymbol{0}$ . First, find the first partial derivative with respect to x and y and find where they are $0$ \[\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0\] \[\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0\] Thus, there is a local extremum at $(0, 0)$ . Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\implies \boxed{1}$


Answer: 2018.0

What is the least possible value of \[(x+1)(x+2)(x+3)(x+4)+2019\] where $x$ is a real number?
$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$

Solution: Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$ , which can be simplified to $(x^2+5x+5)^2-1+2019$ . Noting that squares are nonnegative, and verifying that $x^2+5x+5=0$ for some real $x$ , the answer is $\boxed{2018}$

Solution:
Let $a=x+\tfrac{5}{2}$ . Then the expression $(x+1)(x+2)(x+3)(x+4)$ becomes $\left(a-\tfrac{3}{2}\right)\left(a-\tfrac{1}{2}\right)\left(a+\tfrac{1}{2}\right)\left(a+\tfrac{3}{2}\right)$
We can now use the difference of two squares to get $\left(a^2-\tfrac{9}{4}\right)\left(a^2-\tfrac{1}{4}\right)$ , and expand this to get $a^4-\tfrac{5}{2}a^2+\tfrac{9}{16}$
Refactor this by completing the square to get $\left(a^2-\tfrac{5}{4}\right)^2-1$ , which has a minimum value of $-1$ . The answer is thus $2019-1=\boxed{2018}$

Solution:
Similar to Solution 1, grouping the first and last terms and the middle terms, we get $(x^2+5x+4)(x^2+5x+6)+2019$
Letting $y=x^2+5x$ , we get the expression $(y+4)(y+6)+2019$ . Now, we can find the critical points of $(y+4)(y+6)$ to minimize the function:
$\frac{d}{dx}(y^2+10y+24)=0$
$2y+10=0$
$2y=-5$
$y=-5,0$
To minimize the result, we use $y=-5$ . Hence, the minimum is $(-5+4)(-5+6)=-1$ , so $-1+2019 = \boxed{2018}$

Solution: The expression is negative when an odd number of the factors are negative. This happens when $-2 < x < -1$ or $-4 < x < -3$ . Plugging in $x = -\frac32$ or $x = -\frac72$ yields $-\frac{15}{16}$ , which is very close to $-1$ . Thus the answer is $-1 + 2019 = \boxed{2018}$

Solution: Answer choices $C$ $D$ , and $E$ are impossible, since $(x+1)(x+2)(x+3)(x+4)$ can be negative (as seen when e.g. $x = -\frac{3}{2}$ ). Plug in $x = -\frac{3}{2}$ to see that it becomes $2019 - \frac{15}{16}$ , so round this to $\boxed{2018}$

Solution:
We can ignore the $2019$ and consider it later, as it is a constant. By difference of squares, we can group this into $\left((x+2.5)^2-0.5^2\right)\left((x+2.5)^2-1.5^2\right)$ . We pull a factor of $4$ into each term to avoid dealing with decimals:
\[\dfrac{\left((2x+5)^2-1\right)\left((2x+5)^2-9\right)}{16}.\]
Now, we let $a=2x+5$ . Our expression becomes:
\[\dfrac{(a-1)(a-9)}{16}=\dfrac{a^2-10a+9}{16}.\]
Taking the derivative, we get $\dfrac{2a-10}{16}=\dfrac{a-5}8.$ This is equal to $0$ when $a=5$ , and plugging in $a=5$ , we get the expression is equal to $-1$ and therefore our answer is $2019-1=\boxed{2018}.$


Answer: 6.0

What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution:
The volume of the cube is $V_{\text{cube}}=6^3=216,$ and the volume of a clay ball is $V_{\text{ball}}=\frac43\cdot\pi\cdot2^3=\frac{32}{3}\pi.$
Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is \[\left\lfloor\frac{V_{\text{cube}}}{V_{\text{ball}}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.\] Approximating with $\pi\approx3.14,$ we have $12<4\pi<13,$ or $\left\lfloor\frac{81}{13}\right\rfloor \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq \left\lfloor\frac{81}{12}\right\rfloor.$ We simplify to get \[6 \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq 6,\] from which $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{6}.$

Solution:
As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is $\left\lfloor\frac{81}{4\pi}\right\rfloor.$
By an underestimation $\pi\approx3,$ we have $4\pi>12,$ or $\frac{81}{4\pi}<6\frac34.$
By an overestimation $\pi\approx\frac{22}{7},$ we have $4\pi<\frac{88}{7},$ or $\frac{81}{4\pi}>6\frac{39}{88}.$
Together, we get \[6 < 6\frac{39}{88} < \frac{81}{4\pi} < 6\frac34 < 7,\] from which $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{6}.$

Solution:
As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is $\left\lfloor\frac{81}{4\pi}\right\rfloor.$
Approximating with $\pi\approx3,$ we have $\frac{81}{4\pi}\approx6\frac34.$ Since $\pi$ is about $5\%$ greater than $3,$ it is safe to claim that $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{6}.$


Answer: 3.0

What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions $y=p(x)$ and $y=q(x)$ , each with leading coefficient 1?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$

Solution: The intersections of the two polynomials, $p(x)$ and $q(x)$ , are precisely the roots of the equation $p(x)=q(x) \rightarrow p(x) - q(x) = 0$ . Since the leading coefficients of both polynomials are $1$ , the degree of $p(x) - q(x) = 0$ is at most three, and the maximum point of intersection is three, because a third degree polynomial can have at most three roots. Thus, the answer is $\boxed{3}$


Answer: 6.0

What is the maximum number of possible points of intersection of a circle and a triangle?
$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

Solution:
Circle-triangle problem.PNG
We can draw a circle and a triangle, such that each side is tangent to the circle. This means that each side would intersect the circle at one point.
You would then have $3$ points, but what if the circle was bigger? Then, each side would intersect the circle at 2 points.
Therefore, $2 \times 3 = \boxed{6}$

Solution: We know that the maximum amount of points that a circle and a line segment can intersect is $2$ . Therefore, because there are $3$ line segments in a triangle, the maximum amount of points of intersection is $2 \times 3 = \boxed{6}$


Answer: 17.0

What is the maximum value of $n$ for which there is a set of distinct positive integers $k_1, k_2, ... k_n$ for which
\[k^2_1 + k^2_2 + ... + k^2_n = 2002?\]
$\text{(A) }14 \qquad \text{(B) }15 \qquad \text{(C) }16 \qquad \text{(D) }17 \qquad \text{(E) }18$

Solution:
Note that $k^2_1 + k^2_2 + ... + k^2_n = 2002 \leq \frac{n(n+1)(2n+1)}{6}$
When $n = 17$ $\frac{n(n+1)(2n+1)}{6} = \frac{(17)(18)(35)}{6} = 1785 < 2002$
When $n = 18$ $\frac{n(n+1)(2n+1)}{6} = 1785 + 18^2 = 2109 > 2002$
Therefore, we know $n \leq 17$
Now we must show that $n = 17$ works. We replace one of $1, 2, ... 17$ with an integer $a > 17$ to account for the amount under $2002$ , which is $2002-1785 = 217$
Essentially, this boils down to writing $217$ as a difference of squares. We know $217 = (7)(31)$ , so we assume there exist positive integers $a$ and $b$ where $a > 17$ and $b \leq 17$ such that $a^2 - b^2 = 217$
We can rewrite this as $(a+b)(a-b) = 217$ , so either $a+b = 217$ and $a-b = 1$ or $a+b = 31$ and $a-b = 7$ . We analyze each case separately.
Case 1: $a+b = 217$ and $a-b = 1$
Solving this system of equations gives $a = 109$ and $b = 108$ . However, $108 > 17$ , so this case does not yield a solution.
Case 2: $a+b = 31$ and $a-b = 7$
Solving this system of equations gives $a = 19$ and $b = 12$ . This satisfies all the requirements of the problem.
The list $1, 2 ... 11, 13, 14 ... 17, 19$ has $17$ terms whose sum of squares equals $2002$ . Therefore, the answer is $\boxed{17}$


Answer: 10.0

What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

Solution: Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the  hour. The central angle formed between $4$ and $5$ is $30$ degrees (since it is 1/12 of a full circle, 360). By $4:20$ , the hour hand would have moved $\frac{1}{3}$ way from 4 to 5 since $\frac{20}{60}$ is reducible to $\frac{1}{3}$ . One third of the way from 4 to 5 is one third of 30 degrees, which is 10 degrees past the 4. Recall that the minute hand is at the 4, so the angle between them is $\boxed{10}$


Answer: 1976.5

Solution:
We can see that $44^2=1936$ which is less than 2020. Therefore, there are $2020-44=1976$ of the $4040$ numbers greater than $2020$ . Also, there are $2020+44=2064$ numbers that are less than or equal to $2020$
Since there are $44$ duplicates/extras, it will shift up our median's placement down $44$ . Had the list of numbers been $1,2,3, \dots, 4040$ , the median of the whole set would be $\dfrac{1+4040}{2}=2020.5$
Thus, our answer is $2020.5-44=\boxed{1976.5}$

Solution:
As we are trying to find the median of a $4040$ -term set, we must find the average of the $2020$ th and $2021$ st terms.
Since $45^2 = 2025$ is slightly greater than $2020$ , we know that the $44$ perfect squares $1^2$ through $44^2$ are less than $2020$ , and the rest are greater. Thus, from the number $1$ to the number $2020$ , there are $2020 + 44 = 2064$ terms. Since $44^2$ is $44 + 45 = 89$ less than $45^2 = 2025$ and $84$ less than $2020$ , we will only need to consider the perfect square terms going down from the $2064$ th term, $2020$ , after going down $84$ terms. Since the $2020$ th and $2021$ st terms are only $44$ and $43$ terms away from the $2064$ th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$ . Averaging the two, we get $\boxed{1976.5}.$

Solution:
We want to know the $2020$ th term and the $2021$ st term to get the median.
We know that $44^2=1936$ . So, numbers $1^2, 2^2, \ldots,44^2$ are in between $1$ and $1936$
So, the sum of $44$ and $1936$ will result in $1980$ , which means that $1936$ is the $1980$ th number.
Also, notice that $45^2=2025$ , which is larger than $2021$
Then the $2020$ th term will be $1936+40 = 1976$ , and similarly the $2021$ th term will be $1977$
Solving for the median of the two numbers, we get $\boxed{1976.5}$

Solution: We note that $44^2 = 1936$ , which is the first square less than $2020$ , which means that there are $44$ additional terms before $2020$ . This makes $2020$ the $2064$ th term. To find the median, we need the $2020$ th and $2021$ st term. We note that every term before $2020$ is one less than the previous term (That is, we subtract $1$ to get the previous term.). If $2020$ is the $2064$ th term, than $2020 - 44$ is the $(2064 - 44)$ th term. So, the $2020$ th term is $1976$ , and the $2021$ st term is $1977$ , and the average of these two terms is the median, or $\boxed{1976.5}$

Solution:
To find the median, we sort the $4040$ numbers in decreasing order, then average the $2020$ th and the $2021$ st numbers of the sorted list.
Since $45^2=2025$ and $44^2=1936,$ the first $2021$ numbers of the sorted list are \[\underbrace{2020^2,2019^2,2018^2,\ldots,46^2,45^2}_{1976\mathrm{ \ numbers}}\phantom{ },\phantom{ }\underbrace{2020,2019,2018,\ldots,1977,1976}_{45\mathrm{ \ numbers}}\phantom{ },\] from which the answer is $\frac{1977+1976}{2}=\boxed{1976.5}.$


Answer: 1976.5

Solution:
We can see that $44^2=1936$ which is less than 2020. Therefore, there are $2020-44=1976$ of the $4040$ numbers greater than $2020$ . Also, there are $2020+44=2064$ numbers that are less than or equal to $2020$
Since there are $44$ duplicates/extras, it will shift up our median's placement down $44$ . Had the list of numbers been $1,2,3, \dots, 4040$ , the median of the whole set would be $\dfrac{1+4040}{2}=2020.5$
Thus, our answer is $2020.5-44=\boxed{1976.5}$

Solution:
As we are trying to find the median of a $4040$ -term set, we must find the average of the $2020$ th and $2021$ st terms.
Since $45^2 = 2025$ is slightly greater than $2020$ , we know that the $44$ perfect squares $1^2$ through $44^2$ are less than $2020$ , and the rest are greater. Thus, from the number $1$ to the number $2020$ , there are $2020 + 44 = 2064$ terms. Since $44^2$ is $44 + 45 = 89$ less than $45^2 = 2025$ and $84$ less than $2020$ , we will only need to consider the perfect square terms going down from the $2064$ th term, $2020$ , after going down $84$ terms. Since the $2020$ th and $2021$ st terms are only $44$ and $43$ terms away from the $2064$ th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$ . Averaging the two, we get $\boxed{1976.5}.$

Solution:
We want to know the $2020$ th term and the $2021$ st term to get the median.
We know that $44^2=1936$ . So, numbers $1^2, 2^2, \ldots,44^2$ are in between $1$ and $1936$
So, the sum of $44$ and $1936$ will result in $1980$ , which means that $1936$ is the $1980$ th number.
Also, notice that $45^2=2025$ , which is larger than $2021$
Then the $2020$ th term will be $1936+40 = 1976$ , and similarly the $2021$ th term will be $1977$
Solving for the median of the two numbers, we get $\boxed{1976.5}$

Solution: We note that $44^2 = 1936$ , which is the first square less than $2020$ , which means that there are $44$ additional terms before $2020$ . This makes $2020$ the $2064$ th term. To find the median, we need the $2020$ th and $2021$ st term. We note that every term before $2020$ is one less than the previous term (That is, we subtract $1$ to get the previous term.). If $2020$ is the $2064$ th term, than $2020 - 44$ is the $(2064 - 44)$ th term. So, the $2020$ th term is $1976$ , and the $2021$ st term is $1977$ , and the average of these two terms is the median, or $\boxed{1976.5}$

Solution:
To find the median, we sort the $4040$ numbers in decreasing order, then average the $2020$ th and the $2021$ st numbers of the sorted list.
Since $45^2=2025$ and $44^2=1936,$ the first $2021$ numbers of the sorted list are \[\underbrace{2020^2,2019^2,2018^2,\ldots,46^2,45^2}_{1976\mathrm{ \ numbers}}\phantom{ },\phantom{ }\underbrace{2020,2019,2018,\ldots,1977,1976}_{45\mathrm{ \ numbers}}\phantom{ },\] from which the answer is $\frac{1977+1976}{2}=\boxed{1976.5}.$


Answer: 26.0

What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$

Solution: We can rewrite the fraction as $\frac{123456789}{2^{22} \cdot 10^4} = \frac{12345.6789}{2^{22}}$ . Since the last digit of the numerator is odd, a $5$ is added to the right if the numerator is divided by $2$ , and this will continuously happen because $5$ , itself, is odd. Indeed, this happens twenty-two times since we divide by $2$ twenty-two times, so we will need $22$ more digits. Hence, the answer is $4 + 22 = \boxed{26}$

Solution: Multiply the numerator and denominator of the fraction by $5^{22}$ (which is the same as multiplying by 1) to give $\frac{5^{22} \cdot 123456789}{10^{26}}$ . Now, instead of thinking about this as a fraction, think of it as the division calculation $(5^{22} \cdot 123456789) \div 10^{26}$ . The dividend is a huge number, but we know it doesn't have any digits to the right of the decimal point. Also, the dividend is not a multiple of 10 (it's not a multiple of 2), so these 26 divisions by 10 will each shift the entire dividend one digit to the right of the decimal point. Thus, $\boxed{26}$ is the minimum number of digits to the right of the decimal point needed.

Solution: The denominator is $10^4 \cdot 2^{22}$ . Each $10$ adds one digit to the right of the decimal, and each additional $2$ adds another digit. The answer is $4 + 22 = \boxed{26}$


Answer: 4.0

What is the minimum number of small squares that must be colored black so that a line of symmetry lies on the diagonal $\overline{BD}$ of square $ABCD$
[asy] defaultpen(linewidth(1)); for ( int x = 0; x < 5; ++x ) {     draw((0,x)--(4,x));     draw((x,0)--(x,4)); }  fill((1,0)--(2,0)--(2,1)--(1,1)--cycle); fill((0,3)--(1,3)--(1,4)--(0,4)--cycle); fill((2,3)--(4,3)--(4,4)--(2,4)--cycle); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle); label("$A$", (0, 4), NW); label("$B$", (4, 4), NE); label("$C$", (4, 0), SE); label("$D$", (0, 0), SW); [/asy]
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Rotating square $ABCD$ counterclockwise $45^\circ$ so that the line of symmetry $BD$ is a vertical line makes it easier to see that $\boxed{4}$ squares need to be colored to match its corresponding square.


Answer: 280.0

What is the minimum possible product of three different numbers of the set $\{-8,-6,-4,0,3,5,7\}$
$\text{(A)}\ -336 \qquad \text{(B)}\ -280 \qquad \text{(C)}\ -210 \qquad \text{(D)}\ -192 \qquad \text{(E)}\ 0$

Solution:
The only way to get a negative product using three numbers is to multiply one negative number and two positives or three negatives. Only two reasonable choices
exist: $(-8)\times(-6)\times(-4) = (-8)\times(24) = -192$ and $(-8)\times5\times7 = (-8)\times35 = -280$ .
The latter is smaller, so $\boxed{280}$


Answer: 49.0

What is the minimum value of $f(x)=\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$

Solution:
If we graph each term separately, we will notice that all of the zeros occur at $\frac{1}{m}$ , where $m$ is any integer from $1$ to $119$ , inclusive: $|mx-1|=0\implies mx=1\implies x=\frac{1}{m}$
The minimum value of $f(x)$ occurs where the absolute value of the sum of the slopes is at a minimum $\ge 0$ , since it is easy to see that the value will be increasing on either side. That means the minimum must happen at some $\frac{1}{m}$
The sum of the slopes at $x = \frac{1}{m}$ is
\begin{align*}&\sum_{i=m+1}^{119}i - \sum_{i=1}^{m}i\\ &=\sum_{i=1}^{119}i - 2\sum_{i=1}^{m}i\\ &=-m^2-m+7140\end{align*}
Now we want to minimize $-m^2-m+7140$ . The zeros occur at $-85$ and $84$ , which means the slope is $0$ where $m = 84, 85$
We can now verify that both $x=\frac{1}{84}$ and $x=\frac{1}{85}$ yield $\boxed{49}$


Answer: 49.0

Solution:
By the triangle inequality, $|x-1|+|2x-1|+|3x-1|+\cdots + |119x-1| \geq |(x-1)+(2x-1)+\cdots+(119x)-1|.$ However, we may change signs of some of these terms to cancel out the $x$ 's.
Since the minimum exists, we want all the $x$ s to cancel out. Thus, we want to find some $n$ such that \[1+2+3+...+n=(n+1)+(n+2)+(n+3)+...+119\] \[\frac{n(n+1)}{2}=\frac{119\cdot120}{2}-\frac{n(n+1)}{2}\] \[n^2+n-7140=0\] \[n=84\]
Then, $x=\frac{1}{n}= \frac{1}{84}$ . The answer(expression's value) is then $84*1+(119-85+1)*(-1)$ , which becomes $84-35=\boxed{49}$


Answer: 150.0

What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?
$\text{(A)}\ 50^\circ \qquad \text{(B)}\ 120^\circ \qquad \text{(C)}\ 135^\circ \qquad \text{(D)}\ 150^\circ \qquad \text{(E)}\ 165^\circ$

Solution: The smaller angle makes up $5/12$ of the circle which is the clock.  A circle is $360^\circ$ , so the measure of the smaller angle is \[\frac{5}{12}\cdot 360^\circ = 150^\circ \rightarrow \boxed{150}\]


Answer: 167.0

What is the number of terms with rational coefficients among the $1001$ terms in the expansion of $\left(x\sqrt[3]{2}+y\sqrt{3}\right)^{1000}?$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 166 \qquad\textbf{(C)}\ 167 \qquad\textbf{(D)}\ 500 \qquad\textbf{(E)}\ 501$

Solution:
By the Binomial Theorem, each term in the expansion is of the form \[\binom{1000}{k}\left(x\sqrt[3]{2}\right)^k\left(y\sqrt{3}\right)^{1000-k}=\binom{1000}{k}2^{\frac k3}3^{\frac{1000-k}{2}}x^k y^{1000-k},\] where $k\in\{0,1,2,\ldots,1000\}.$
This problem is equivalent to counting the values of $k$ such that both $\frac k3$ and $\frac{1000-k}{2}$ are integers. Note that $k$ must be a multiple of $3$ and a multiple of $2,$ so $k$ must be a multiple of $6.$ There are $\boxed{167}$ such values of $k:$ \[6(0), 6(1), 6(2), \ldots, 6(166).\]


Answer: 2.0

What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Solution:
We can rewrite the expression as \[222,222-(22,222+2,222+222+22+2).\]
We note that the units digit of the addition is $0$ because all the units digits of the five numbers are $2$ and $5*2=10$ , which has a units digit of $0$
Now, we have something with a units digit of $0$ subtracted from $222,222$ . The units digit of this expression is obviously $2$ , and we get $\boxed{2}$ as our answer.

Solution:
222,222-22,222 = 200,000
200,000 - 2,222 = 197778
197778 - 222 = 197556
197556 - 22 = 197534
197534 - 2 = 1957532
So our answer is $\boxed{2}$

Solution:
We only care about the unit's digits.
Thus, $2-2$ ends in $0$ $0-2$ ends in $8$ $8-2$ ends in $6$ $6-2$ ends in $4$ , and $4-2$ ends in $\boxed{2}$

Solution: We just take the units digit of each and subtract, or you can do it this way by adding an extra ten to the first number (so we don't get a negative number): \[12-2-(2+2+2+2)=10-8=2\] Thus, we get the answer $\boxed{2}$


Answer: 16.0

What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$ $0\le v\le1,$ and $0\le w\le1$
$\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16$

Solution:
 Notice that we are given a parametric form of the region, and $w$ is used in both $x$ and $y$ . We first fix $u$ and $v$ to $0$ , and graph $(-3w,4w)$ from $0\le w\le1$ . When $w$ is $0$ , we have the point $(0,0)$ , and when $w$ is $1$ , we have the point $(-3,4)$ . We see that since this is a directly proportional function, we can just connect the dots like this:

Now, when we vary $u$ from $0$ to $2$ , this line is translated to the right $2$ units:

We know that any points in the region between the line (or rather segment) and its translation satisfy $w$ and $u$ , so we shade in the region:

We can also shift this quadrilateral one unit up, because of $v$ . Thus, this is our figure:

The length of the boundary is simply $1+2+5+1+2+5$ $5$ can be obtained by Pythagorean theorem since we have side lengths $3$ and $4$ .). This equals $\boxed{16.}$

Solution:
We can find the "boundary points" and work with our intuition to solve the problem. We set each of $u, v, w$ equal to $0, 1$ for a total of $8$ combinations in $u, v, w$ . We now test each one.
Case 1: $u = 0, v = 0, w = 0 \implies (0, 0)$
Case 2: $u = 0, v = 0, w = 1 \implies (-3, 4)$
Case 3: $u = 0, v = 1, w = 0 \implies (0, 1)$
Case 4: $u = 0, v = 1, w = 1 \implies (-3, 5)$
Case 5: $u = 1, v = 0, w = 0 \implies (2, 0)$
Case 6: $u = 1, v = 0, w = 1 \implies (-1, 4)$
Case 7: $u = 1, v = 1, w = 0 \implies (2, 1)$
Case 8: $u = 1, v = 1, w = 1 \implies (-1, 5)$
When graphed on a coordinate plane, the points appear as follows.

Notice how there are two distinct rectangles visible in the figure. This leads us to believe that the region tracks the motion of this region as it travels in space. To understand why this is true, we can imagine a fixed $w$ (as it is present in both the $x$ and $y$ coordinates). Then if we hold one of $u$ or $v$ fixed and let the other vary, we get a straight line parallel to the $x$ or $y$ axis respectively. If we let the other vary, we get the other type of straight line. Together, they form a rectangular region. In addition, $w$ serves as a diagonal translation, so if we now let $w$ vary, it traces out the motion of the rectangle. Keeping this in mind, we connect the dots.

Each of the diagonal sides have length $5$ by the distance formula on $(0,0)$ and $(-3,4)$ (the other diagonal side is congruent), so our total area is $2 + 1 + 5 + 2 + 1 + 5 = \boxed{16}$


Answer: 180.0

Solution: Draw altitudes from $B$ and $C$ to base $AD$ to create a rectangle and two right triangles. The side opposite $BC$ is equal to $50$ . The bases of the right triangles can be found using Pythagorean or special triangles to be $18$ and $7$ . Add it together to get $AD=18+50+7=75$ . The perimeter is $75+30+50+25=\boxed{180}$


Answer: 1003.0

What is the product of $\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2006}{2005}$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1002\qquad\textbf{(C)}\ 1003\qquad\textbf{(D)}\ 2005\qquad\textbf{(E)}\ 2006$

Solution: The numerator in each fraction cancels out with the denominator of the next fraction. There are only two numbers that didn't cancel: $\frac{2006}{2}=\boxed{1003}$


Answer: 81.0

What is the product of all real numbers $x$ such that the distance on the number line between $\log_6x$ and $\log_69$ is twice the distance on the number line between $\log_610$ and $1$
$\textbf{(A) } 10 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 25 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 81$

Solution: First, notice that there must be two such numbers: one greater than $\log_69$ and one less than it. Furthermore, they both have to be the same distance away, namely $2(\log_610 - 1)$ . Let these two numbers be $\log_6a$ and $\log_6b$ . Because they are equidistant from $\log_69$ , we have $\frac{\log_6a + \log_6b}{2} = \log_69$ . Using log properties, this simplifies to $\log_6{\sqrt{ab}} = \log_69$ . We then have $\sqrt{ab} = 9$ , so $ab = \boxed{81}$

Solution:
\[\log_6 9 - \log_6 x_2 = -2d\] Subbing in for $-2d$ and using log rules, \[\log_6 \frac{9}{x_2} = \log_6 \frac{9}{25}\] From this we conclude that \[\frac{9}{x_2} = \frac{9}{25} \implies x_2 = 25\]
Finding the product of the distinct values, $x_1x_2 = \boxed{81}$


Answer: 81.0

Solution:
Let $a = 2 \cdot |\log_6 10 - 1| = |\log_6 9 - \log_6 x|  =  |\log_6 \frac{9}{x}|$
$\pm a = \log_6 \frac{9}{x} \implies 6^{\pm a} = b^{\pm 1} = \frac{9}{x}  \implies x = 9 \cdot b^{\pm 1}$
$9b^1 \cdot 9b^{-1} = \boxed{81}$


Answer: 1.0

Solution: For $\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$ , transform it into $\dfrac{\ln 289+\ln 7}{\ln 7 + \ln x}\cdot \dfrac{\ln 289+\ln 7}{\ln 289 + \ln x}=\dfrac{\ln 289+\ln 7}{\ln 289+\ln 7+\ln x}$ . Replace $\ln x$ with $y$ . Because we want to find the product of all solutions of $x$ , it is equivalent to finding the exponential of the sum of all solutions of $y$ . Change the equation to standard quadratic equation form, the term with 1 power of $y$ is canceled.  By using Vieta, we see that since there does not exist a $by$ term, $\sum y=0$ and $\prod x=e^0=\boxed{1}$

Solution:
\[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
Rearranging it give us:
\[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x\]
\[(\log_{2023}7+\log_{2023}x)(\log_{2023}289+\log_{2023}x)=(\log_{2023}2023+\log_{2023}x)\]
let $\log_{2023}x$ be $a$ , we get
\[(\log_{2023}7+a)(\log_{2023}289+a)=1+a\]
\[a^2+(\log_{2023}7+\log_{2023}289)a+\log_{2023}7 \cdot \log_{2023}289=1+a\]
\[a^2+\log_{2023}7 \cdot \log_{2023}289-1=0\]
by Vieta's Formulas,
\[a_1+a_2=0\]
\[\log_{2023}{x_1}+\log_{2023}{x_2}=0\]
\[\log_{2023}{x_1x_2}=0\]
\[x_1x_2=\boxed{1}\]

Solution: Similar to solution 1, change the bases first \[\frac{\ln 289+\ln 7}{\ln7 + \ln{x}} \cdot \frac{\ln 289+\ln 7}{2\ln17 + \ln{x}} = \frac{\ln 289+\ln 7}{\ln7 + 2\ln17 + \ln{x}}\] Cancel and cross multiply to get \[(\ln7 + 2\ln17)(\ln7 + 2\ln17 + \ln{x}) = (\ln7 + \ln{x})(2\ln17 + \ln{x})\] Simplify to get \[(\ln{x})^2 = 4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2\] \[\ln{x} = \pm \sqrt{4(\ln17)^2 + 2\ln17\ln7 + (\ln7)^2}\] The sum of all possible $\ln{x}$ is 0, thus the product of all solutions of $x$ is $\boxed{1}$

Solution: We take the reciprocal of both sides: \[\frac{1}{\log_{7x}2023}\cdot \frac{1}{\log_{289x}2023}=\frac{1}{\log_{2023x}2023}.\] Using logarithm properties, we have \[\log_{2023}7x\cdot \log_{2023}289x=\log_{2023}2023x.\] Simplify to obtain \[2023x^2=2023x,\] from which we have $x=\boxed{1}$


Answer: 64.0

What is the product of all the roots of the equation \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]
$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$

Solution:
First, square both sides, and isolate the absolute value. \begin{align*} 5|x|+8&=x^2-16\\ 5|x|&=x^2-24\\ |x|&=\frac{x^2-24}{5}. \\ \end{align*} Solve for the absolute value and factor.
Case 1: $x=\frac{x^2-24}{5}$
Multiplying both sides by $5$ gives us \[5x=x^2-24.\] Rearranging and factoring, we have \begin{align*} x^2-5x-24 &=0, \\ (x-8)(x+3) &= 0.\\ \end{align*}
Case 2: $x=\frac{-x^2+24}{5}$
As above, we multiply both sides by $5$ to find \[5x=-x^2+24.\] Rearranging and factoring gives us \begin{align*} x^2+5x-24 &=0, \\ (x+8)(x-3) &= 0. \\ \end{align*}
Combining these cases, we have $x= -8, -3, 3, 8$ . Because our first step of squaring is not reversible, however, we need to check for extraneous solutions. Plug each solution for $x$ back into the original equation to ensure it works. Whether the number is positive or negative does not matter since the absolute value or square will cancel it out anyways.
Trying $|x|=|3|$ , we have \begin{align*} \sqrt{5|3|+8}&=\sqrt{3^2-16}, \\ \sqrt{15+8}&=\sqrt{9-16}, \\ \sqrt{23} &\not= \sqrt{-7}.\\ \end{align*} Therefore, $x = 3$ and $x= -3$ are extraneous.
Checking $|x|=|8|$ , we have \begin{align*} \sqrt{5|8|+8}&=\sqrt{8^2-16}, \\ \sqrt{40+8}&=\sqrt{64-16}, \\ \sqrt{48}&=\sqrt{48}.\\ \end{align*}
The roots of our original equation are $-8$ and $8$ and product is $-8 \times 8 = \boxed{64}$

Solution: Square both sides, to get $5|x| + 8 = x^2-16$ . Rearrange to get $x^2 - 5|x| - 24 = 0$ . Seeing that $x^2 = |x|^2$ , substitute to get $|x|^2 - 5|x| - 24 = 0$ . We see that this is a quadratic in $|x|$ . Factoring, we get $(|x|-8)(|x|+3) = 0$ , so $|x| = \{8,-3\}$ . Since the radicand of the equation can't be negative, the sole solution is $|x| = 8$ . Therefore, $x$ can be $8$ or $-8$ . The product is then $-8 \times 8 = \boxed{64}$

Solution:
First we note that $x \in (-\infty,-4] \cup [4,\infty]$ . This will help us later with finding extraneous solutions.
Next, we have two cases: \[\text{IF } x\leq 4:\] \[\text{ } \sqrt{-5x+8} = \sqrt{x^2-16} \implies x^2+5x-24=0 \rightarrow x = -8,3\] .
We note that $3$ is not in the range of possible $x$ 's and thus is not a solution.
\[\text{IF } x\geq 4:\] \[\text{ } \sqrt{5x+8} = \sqrt{x^2-16} \implies x^2-5x-24=0 \rightarrow x = -3,8\] .
We again not that $-3$ is an extraneous solution.
Thus, we have the two solutions $-8$ and $8$ . Therefore product is $-8 \cdot 8 = \boxed{64}$


Answer: 20.0

What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$

Solution:
If we were to expand by squaring, we would get a quartic polynomial , which isn't always the easiest thing to deal with.
Instead, we substitute $y$ for $x^2+18x+30$ , so that the equation becomes $y=2\sqrt{y+15}$
Now we can square; solving for $y$ , we get $y=10$ or $y=-6$ . The second root is extraneous since $2\sqrt{y+15}$ is always non-negative (and moreover, plugging in $y=-6$ , we get $-6=6$ , which is obviously false). Hence we have $y=10$ as the only solution for $y$ . Substituting $x^2+18x+30$ back in for $y$
Both of the roots of this equation are real, since its discriminant is $18^2 - 4 \cdot 1 \cdot 20 = 244$ , which is positive. Thus by Vieta's formulas , the product of the real roots is simply $\boxed{020}$

Solution:
We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: \[(x^2+ 18x + 45) - 2\sqrt{x^2+18x+45} - 15 = 0.\] Letting $n = \sqrt{x^2+18x+45}$ , we have $n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0$ . Because the square root of a real number can't be negative, the only possible $n$ is $5$
Substituting that in, we have \[\sqrt{x^2+18x+45} = 5 \Longrightarrow x^2 + 18x + 45 = 25 \Longrightarrow x^2+18x+20=0.\]
Reasoning as in Solution 1, the product of the roots is $\boxed{020}$

Solution: Begin by completing the square on both sides of the equation, which gives \[(x+9)^2-51=2\sqrt{(x+3)(x+15)}\] Now by substituting $y=x+9$ , we get $y^2-51=2\sqrt{(y-6)(y+6)}$ , or \[y^4-106y^2+2745=0\] The solutions in $y$ are then \[y=x+9=\pm3\sqrt{5},\pm\sqrt{61}\] Turns out, $\pm3\sqrt{5}$ are a pair of extraneous solutions. Thus, our answer is then \[\left(\sqrt{61}-9\right)\left(-\sqrt{61}-9\right)=81-61=\boxed{020}\] By difference of squares.

Solution: We are given the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] Squaring both sides yields \[(x^2+18x+30)^2=4(x^2+18x+45)\] \[(x^2+18x+30)^2=4(x^2+18x+30+15)\] \[(x^2+18x+30)^2=4(x^2+18x+30)+60\] \[(x^2+18x+30)^2-4(x^2+18x+30)-60=0\] Substituting $y=x^2+18x+30$ yields \[y^2-4y-60=0\] \[(y+6)(y-10)=0\] Thus $y=x^2+18x+30=-6,10$ . However if $y=-6$ , the left side of the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] would be negative while the right side is negative. Thus $y=10$ is the only possible value and we have \[x^2+18x+30=10\] \[x^2+18x+20=0\] Since the discriminant $\sqrt{18^2-4\cdot20}$ is real, both the roots are real. Thus by Vieta's Formulas, the product of the roots is the constant, $\boxed{20}$


Answer: 330.0

What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$

Solution:
To find either the LCM or the GCF of two numbers, always prime factorize first.
The prime factorization of $180 = 3^2 \times  5 \times 2^2$
The prime factorization of $594 = 3^3 \times  11 \times 2$
Then, to find the LCM, we have to find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is $3^3, 5, 11, 2^2$ ). Multiply all of these to get 5940.
For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. $3^2 \times 2$ = 18.
Thus the answer = $\frac{5940}{18}$ $\boxed{330}$

Solution:
We start off with a similar approach as the original solution. From the prime factorizations, the GCF is $18$
It is a well known fact that $\gcd(m,n)\times \operatorname{lcm}(m,n)=|mn|$ . So we have, $18\times \operatorname{lcm} (180,594)=594\times 180$
Dividing by $18$ yields $\operatorname{lcm} (180,594)=594\times 10=5940$
Therefore, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf}(180,594)}=\frac{5940}{18}=\boxed{330}$

Solution:
From Solution 1,
the prime factorization of $180 = 2^2 \cdot 3^2 \cdot 5$ .
The prime factorization of $594 = 2 \cdot 3^3 \cdot 11$ .
Hence, $\operatorname{lcm} (180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11$ , and $\operatorname{gcf} (180,594) = 2 \cdot 3^2$ .
Therefore, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf} (180,594)} = \frac{2^2 \cdot 3^3 \cdot 5 \cdot 11}{2 \cdot 3^2} = 2 \cdot 3 \cdot 5 \cdot 11 = 330 \Longrightarrow \boxed{330}$


Answer: 201.0

What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$

Solution:
Completing the square, then difference of squares:
\begin{align*} 2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\ &= (2^{101} + 1)^2 - 2^{102} + 201\\ &= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201. \end{align*}
Thus, we see that the remainder is surely $\boxed{201}$

Solution:
Let $x=2^{50}$ . We are now looking for the remainder of $\frac{4x^4+202}{2x^2+2x+1}$
We could proceed with polynomial division, but the numerator looks awfully similar to the Sophie Germain Identity ,  which states that \[a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)\]
Let's use the identity, with $a=1$ and $b=x$ , so we have
\[1+4x^4=(1+2x^2+2x)(1+2x^2-2x)\]
Rearranging, we can see that this is exactly what we need:
\[\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1\]
So \[\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1}\]
Since the first half divides cleanly as shown earlier, the remainder must be $\boxed{201}$

Solution:
We let \[x = 2^{50}\] and \[2^{202} + 202 = 4x^{4} + 202\] .
Next we write \[2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1\] .
We know that \[4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)\] by the Sophie Germain identity so to find \[4x^{4} + 202,\] we find that \[4x^{4} + 202 = 4x^{4} + 201 + 1\] which shows that the remainder is $\boxed{201}$

Solution: We let $x=2^{50.5}$ . That means $2^{202}+202=x^{4}+202$ and $2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1$ . Then, we simply do polynomial division, and find that the remainder is $\boxed{201}$

Solution:
Let $n=2^{101}+2^{51}+1$ . Then, mod $n$
$2^{202}+202 \equiv (-2^{51}-1)^2 + 202$
$\equiv 2^{102}+2^{52}+203$
$= 2(n-1)+203 \equiv 201 \pmod{n}$
Thus, the remainder is $\boxed{201}$

Solution:
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:
\[\frac{2^{202}+202}{2^{101}+2^{51}+1}\] \[= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}\] \[= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}\] \[=2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}\] \[=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1}\] \[= 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1}\] \[= 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.\]
Clearly, $201 < 2^{201} + 2^{51} + 1$ , hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is $\boxed{201}$


Answer: 4.0

What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?
$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$

Solution:
The sum of any four consecutive powers of 3 is divisible by $3^0 + 3^1 + 3^2 +3^3 = 40$ and hence is divisible by 8. Therefore
is divisible by 8.  So the required remainder is $3^0 + 3^1 = \boxed{4}$


Answer: 4.0

Solution:
We have $3^2 = 9 \equiv 1 \pmod 8$ . Hence for any $k$ we have $3^{2k}\equiv 1^k = 1 \pmod 8$ , and then $3^{2k+1} = 3\cdot 3^{2k} \equiv 3\cdot 1 = 3  \pmod 8$
Therefore our sum gives the same remainder modulo $8$ as $1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3$ . There are $2010$ terms in the sum, hence there are $2010/2 = 1005$ pairs of $1+3$ , and thus the sum is $1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8$


Answer: 50.0

What is the remainder when $x^{51}+51$ is divided by $x+1$
$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \  } 49 \qquad \mathrm{(D) \  } 50 \qquad \mathrm{(E) \  }51$

Solution: From the Remainder Theorem , the remainder when $x^{51}+51$ is divided by $x+1$ is $(-1)^{51}+51=-1+51=50, \boxed{50}$


Answer: 337.0

What is the smallest integer $n$ , greater than one, for which the root-mean-square of the first $n$ positive integers is an integer?
$\mathbf{Note.}$ The root-mean-square of $n$ numbers $a_1, a_2, \cdots, a_n$ is defined to be \[\left[\frac{a_1^2 + a_2^2 + \cdots + a_n^2}n\right]^{1/2}\]

Solution:
Let's first obtain an algebraic expression for the root mean square of the first $n$ integers, which we denote $I_n$ . By repeatedly using the identity $(x+1)^3 = x^3 + 3x^2 + 3x + 1$ , we can write \[1^3 + 3\cdot 1^2 + 3 \cdot 1 + 1 = 2^3,\] \[1^3 + 3 \cdot(1^2 + 2^2) + 3 \cdot (1 + 2) + 1 + 1  = 3^3,\] and \[1^3 + 3\cdot(1^2 + 2^2 + 3^2) + 3 \cdot (1 + 2 + 3) + 1 + 1 + 1 = 4^3.\] We can continue this pattern indefinitely, and thus for any
positive integer $n$ \[1 + 3\sum_{j=1}^n j^2 + 3 \sum_{j=1}^n j^1 + \sum_{j=1}^n j^0 = (n+1)^3.\] Since $\sum_{j=1}^n j  = n(n+1)/2$ , we obtain \[\sum_{j=1}^n j^2 = \frac{2n^3 + 3n^2 + n}{6}.\] Therefore, \[I_n = \left(\frac{1}{n} \sum_{j=1}^n j^2\right)^{1/2} = \left(\frac{2n^2 + 3n + 1}{6}\right)^{1/2}.\] Requiring that $I_n$ be an integer, we find that \[(2n+1 ) (n+1) = 6k^2,\] where $k$ is an integer. Using the Euclidean algorithm, we see that $\gcd(2n+1, n+1) = \gcd(n+1,n) = 1$ , and so $2n+1$ and $n+1$ share no
factors greater than 1. The equation above thus implies that $2n+1$ and $n+1$ is each proportional to a perfect square. Since $2n+1$ is
odd, there are only two possible cases:
Case 1: $2n+1 = 3 a^2$ and $n+1 = 2b^2$ , where $a$ and $b$ are integers.
Case 2: $2n+1 = a^2$ and $n+1 = 6b^2$
In Case 1, $2n+1 = 4b^2 -1 = 3a^2$ . This means that $(4b^2 -1)/3 = a^2$ for some integers $a$ and $b$ . We proceed by checking whether $(4b^2-1)/3$ is a perfect square for $b=2, 3, 4, \dots$ . (The solution $b=1$ leads to $n=1$ , and we are asked to find a value of $n$ greater than 1.) The smallest positive integer $b$ greater than 1 for
which $(4b^2-1)/3$ is a perfect square is $b=13$ , which results in $n=337$
In Case 2, $2n+1 = 12b^2 - 1 = a^2$ . Note that $a^2 = 2n+1$ is an odd square, and hence is congruent to $1 \pmod 4$ . But $12b^2 -1 \equiv 3 \pmod 4$ for any $b$ , so Case 2 has no solutions.
Alternatively, one can proceed by checking whether $12b^2 -1$ is a perfect square for $b=1, 2 ,3 ,\dots$ . We find that $12b^2 -1$ is not a perfect square for $b = 1,2, 3, ..., 7, 8$ , and $n= 383$ when $b=8$ . Thus the smallest positive integers $a$ and $b$ for which $12b^2- 1 = a^2$ result in a value of $n$ exceeding the value found in Case 1, which was 337.
In summary, the smallest value of $n$ greater than 1 for which $I_n$ is an integer is $\boxed{337}$


Answer: 970.0

What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$
$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$

Solution: $(\sqrt{3}+\sqrt{2})^6=(5+2\sqrt{6})^3=(5+2\sqrt{6})(49+20\sqrt{6})=(485+198\sqrt{6})$ Then, find that $\sqrt{6}$ is about $2.449$ . Finally, multiply and add to find that the smallest integer higher is $\boxed{970}$

Solution: Let's evaluate $(\sqrt{3}+\sqrt{2})^6 + (\sqrt{3}-\sqrt{2})^6$ . We see that all the irrational terms cancel. Then, using binomial theorem, we evaluate all the rational terms in the first expression to get 485. Then, the sum of the rational parts of the 2nd term will be 485 as well. Then, we get a total of 970 and since $(\sqrt{3}-\sqrt{2})^6) < 1$ , the greatest integer greater than our original expression is $\boxed{970}$


Answer: 2.0

What is the smallest integral value of $k$ such that \[2x(kx-4)-x^2+6=0\] has no real roots?
$\mathrm{(A)\ } -1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \  } 3 \qquad \mathrm{(D) \  } 4 \qquad \mathrm{(E) \  }5$

Solution: Expanding, we have $2kx^2-8x-x^2+6=0$ , or $(2k-1)x^2-8x+6=0$ . For this quadratic not to have real roots, it must have a negative discriminant. Therefore, $(-8)^2-4(2k-1)(6)<0\implies 64-48k+24<0\implies k>\frac{11}{6}$ . From here, we can easily see that the smallest integral value of $k$ is $2, \boxed{2}$


Answer: 2501.0

What is the smallest positive integer $n$ such that $\sqrt{n}-\sqrt{n-1}<.01$
$\textbf{(A) }2499\qquad \textbf{(B) }2500\qquad \textbf{(C) }2501\qquad \textbf{(D) }10,000\qquad  \textbf{(E) }\text{There is no such integer}$

Solution: Adding $\sqrt{n - 1}$ to both sides, we get \[\sqrt{n} < \sqrt{n - 1} + 0.01.\] Squaring both sides, we get \[n < n - 1 + 0.02 \sqrt{n - 1} + 0.0001,\] which simplifies to \[0.9999 < 0.02 \sqrt{n - 1},\] or \[\sqrt{n - 1} > 49.995.\] Squaring both sides again, we get \[n - 1 > 2499.500025,\] so $n > 2500.500025$ . The smallest positive integer $n$ that satisfies this inequality is $\boxed{2501}$


Answer: 495.0

What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?

Solution:
Let the desired integer be $n$ . From the information given, it can be determined that, for positive integers $a, \ b, \ c$
$n = 9a + 36 = 10b + 45 = 11c + 55$
This can be rewritten as the following congruences:
$n \equiv 0 \pmod{9}$
$n \equiv 5 \pmod{10}$
$n \equiv 0 \pmod{11}$
Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is $\boxed{495}$

Solution: Let $n$ be the desired integer. From the given information, we have \begin{align*}9x &= a \\ 11y &= a \\ 10z + 5 &= a, \end{align*} here, $x,$ and $y$ are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have $z$ as the 4th term of the sequence. Since, $a$ is a multiple of $9$ and $11,$ it is also a multiple of $\text{lcm}[9,11]=99.$ Hence, $a=99m,$ for some $m.$ So, we have $10z + 5 = 99m.$ It follows that $99(5) = \boxed{495}$ is the smallest integer that can be represented in such a way.

Solution: By the method in Solution 1, we find that the number $n$ can be written as $9a+36=10b+45=11c+55$ for some integers $a,b,c$ . From this, we can see that $n$ must be divisible by 9, 5, and 11. This means $n$ must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that $n$ cannot be divisible by 10, so $n$ must equal $\boxed{495}$ . Solution by Zeroman.

Solution: First note that the integer clearly must be divisible by $9$ and $11$ since we can use the "let the middle number be x" trick. Let the number be $99k$ for some integer $k.$ Now let the $10$ numbers be $x,x+1, \cdots x+9.$ We have $10x+45 = 99k.$ Taking mod $5$ yields $k \equiv 0 \pmod{5}.$ Since $k$ is positive, we take $k=5$ thus obtaining $99 \cdot 5 = \boxed{495}$ as our answer.


Answer: 3127.0

What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?
$\textbf{(A)}\hspace{.05in}3127\qquad\textbf{(B)}\hspace{.05in}3133\qquad\textbf{(C)}\hspace{.05in}3137\qquad\textbf{(D)}\hspace{.05in}3139\qquad\textbf{(E)}\hspace{.05in}3149$

Solution:
The problem states that the answer cannot be a perfect square or have prime factors less than $50$ . Therefore, the answer will be the product of at least two different primes greater than $50$ . The two smallest primes greater than $50$ are $53$ and $59$ . Multiplying these two primes, we obtain the number $3127$ , which is also the smallest number on the list of answer choices.
So we are done, and the answer is $\boxed{3127}$


Answer: 180.0

What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?

Solution:
We use the fact that the number of divisors of a number $n = p_1^{e_1}p_2^{e_2} \cdots p_k^{e_k}$ is $(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$ . If a number has $18 = 2 \cdot 3 \cdot 3$ factors, then it can have at most $3$ distinct primes in its factorization.
Dividing the greatest power of $2$ from $n$ , we have an odd integer with six positive divisors, which indicates that it either is ( $6 = 2 \cdot 3$ ) a prime raised to the $5$ th power, or two primes, one of which is squared. The smallest example of the former is $3^5 = 243$ , while the smallest example of the latter is $3^2 \cdot 5 = 45$
Suppose we now divide all of the odd factors from $n$ ; then we require a power of $2$ with $\frac{18}{6} = 3$ factors, namely $2^{3-1} = 4$ . Thus, our answer is $2^2 \cdot 3^2 \cdot 5 = \boxed{180}$

Solution:
Somewhat similar to the first solution, we see that the number $n$ has two even factors for every odd factor.  Thus, if $x$ is an odd factor of $n$ , then $2x$ and $4x$ must be the two corresponding even factors.  So, the prime factorization of $n$ is $2^2 3^a 5^b 7^c...$ for some set of integers $a, b, c, ...$
Since there are $18$ factors of $n$ , we can write:
$(2+1)(a+1)(b+1)(c+1)... = 18$
$(a+1)(b+1)(c+1)... = 6$
Since $6$ only has factors from the set $1, 2, 3, 6$ , either $a=5$ and all other variables are $0$ , or $a=3$ and $b=2$ , with again all other variables equal to $0$ .  This gives the two numbers $2^2 \cdot 3^5$ and $2^2 \cdot 3^2 \cdot 5$ .  The latter number is smaller, and is equal to $\boxed{180}$


Answer: 9.0

What is the smallest positive odd integer $n$ such that the product $2^{1/7}2^{3/7}\cdots2^{(2n+1)/7}$ is greater than $1000$ ?
(In the product the denominators of the exponents are all sevens, and the numerators are the successive odd integers from $1$ to $2n+1$ .)
$\textbf{(A) }7\qquad \textbf{(B) }9\qquad \textbf{(C) }11\qquad \textbf{(D) }17\qquad  \textbf{(E) }19$

Solution:
Combine the terms in the product to get $2^{\frac{1+3+5+ \dots +(2n-1)+(2n+1)}{7}}$
The exponent can be simplified to \[\frac{1+3+5+ \dots +(2n-1)+(2n+1)}{7} \Rightarrow \frac{\frac{n(1+(2n+1))}{2}}{7} \Rightarrow \frac{n^2}{7}.\]
We want this inequality to be true with the smallest positive odd integer value of $n$ \[2^{\frac{n^2}{7}} > 1000.\]
Now, let's test the answer choices. For $n=7$ , we have $2^{49/7}=2^{7}<1000$ . For $n=9$ , we have $2^{81/7}>2^{10}>1000$
So our answer is $\boxed{9}$ . ~ jiang147369


Answer: 5.0

What is the smallest possible average of four distinct positive even integers?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$

Solution: In order to get the smallest possible average, we want the 4 even numbers to be as small as possible. The first 4 positive even numbers are 2, 4, 6, and 8. Their average is $\frac{2+4+6+8}{4}=\boxed{5}$


Answer: 36.0

What is the smallest result that can be obtained from the following process?
$\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 56$

Solution:
Since we want the smallest possible result, and we are only adding and multiplying positive numbers over $1$ , we can "prune" the set to the three smallest numbers $\{3,5,7\}$ .  Using bigger numbers will create bigger sums and bigger products.
From there, compute the $3$ ways you can do the two operations:
$(3+5)7 = 8\cdot 7 = 56$
$(3 + 7)5 = 10\cdot 5 = 50$
$(7+5)3 = 12\cdot 3 = 36$
The smallest number is 36, giving an answer of $\boxed{36}$


Answer: 1047.0

Solution:
Let the two three-digit numbers be $\overline{abc}$ and $\overline{def}$ .  Their sum is equal to $100(a+d)+10(b+e)+(c+f)$
To minimize this, we need to minimize the contribution of the $100$ factor, so we let $a=4$ and $d=5$ .  Similarly, we let $b=6$ $e=7$ , and then $c=8$ and $f=9$ .  The sum is \[100(9)+10(13)+(17)=1047 \rightarrow \boxed{1047}\]


Answer: 48.0

What is the smallest whole number larger than the perimeter of any triangle with a side of length $5$ and a side of length $19$
$\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57$

Solution: We know from the Triangle Inequality that the last side, $s$ , fulfills $s<5+19=24$ . Adding $5+19$ to both sides of the inequality, we get $s+5+19<48$ , and because $s+5+19$ is the perimeter of our triangle, $\boxed{48}$ is our answer.


Answer: 12.0

What is the sum of all integer solutions to $1<(x-2)^2<25$
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25$

Solution:
$(x-2)^2$ = perfect square.
1 < perfect square < 25
Perfect square can equal: 4, 9, or 16
Solve for $x$
$(x-2)^2=4$
$x=4,0$
and
$(x-2)^2=9$
$x=5,-1$
and
$(x-2)^2=16$
$x=6,-2$
The sum of all integer solutions is
$4+5+6+0+(-1)+(-2)=\boxed{12}$


Answer: 10.0

What is the sum of all possible values of $k$ for which the polynomials $x^2 - 3x + 2$ and $x^2 - 5x + k$ have a root in common?
$\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10$

Solution: We factor $x^2-3x+2$ into $(x-1)(x-2)$ . Thus, either $1$ or $2$ is a root of $x^2-5x+k$ . If $1$ is a root, then $1^2-5\cdot1+k=0$ , so $k=4$ . If $2$ is a root, then $2^2-5\cdot2+k=0$ , so $k=6$ . The sum of all possible values of $k$ is $\boxed{10}$


Answer: 380.0

What is the sum of all possible values of $t$ between $0$ and $360$ such that the triangle in the coordinate plane whose vertices are \[(\cos 40^\circ,\sin 40^\circ), (\cos 60^\circ,\sin 60^\circ), \text{ and } (\cos t^\circ,\sin t^\circ)\] is isosceles?
$\textbf{(A)} \: 100 \qquad\textbf{(B)} \: 150 \qquad\textbf{(C)} \: 330 \qquad\textbf{(D)} \: 360 \qquad\textbf{(E)} \: 380$

Solution:
Let $A = (\cos 40^{\circ}, \sin 40^{\circ}), B = (\cos 60^{\circ}, \sin 60^{\circ}),$ and $C = (\cos t^{\circ}, \sin t^{\circ}).$ We apply casework to the legs of isosceles $\triangle ABC:$
Together, the sum of all such possible values of $t$ is $20+80+50+230=\boxed{380}.$


Answer: 18.0

What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25$

Solution:
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
Case 1:
The equation yields $x^2-12x+34=2$ , which is equal to $(x-4)(x-8)=0$ . Therefore, the two values for the positive case is $4$ and $8$
Case 2:
Similarly, taking the nonpositive case for the value inside the absolute value notation yields $x^2-12x+34=-2$ . Factoring and simplifying gives $(x-6)^2=0$ , so the only value for this case is $6$
Summing all the values results in $4+8+6=\boxed{18}$

Solution:
We have the equations $x^2-12x+32=0$ and $x^2-12x+36=0$
Notice that the second is a perfect square with a double root at $x=6$ , and the first has two distinct real roots. By Vieta's, the sum of the roots of the first equation is $-(-12)$ or $12$ $12+6=\boxed{18}$

Solution:
Completing the square gives \begin{align*} \left|(x-6)^2-2\right|&=2 \\ (x-6)^2-2&=\pm2. \hspace{15mm}(\bigstar) \end{align*} Note that the graph of $y=(x-6)^2-2$ is an upward parabola with the vertex $(6,-2)$ and the axis of symmetry $x=6;$ the graphs of $y=\pm2$ are horizontal lines.
We apply casework to $(\bigstar):$
Finally, the sum of all solutions is $12+6=\boxed{18}.$


Answer: 5.0

What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?
$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}$

Solution:
The mean is $\frac{4+6+8+17+x}{5}=\frac{35+x}{5}$
There are three possibilities for the median: it is either $6$ $8$ , or $x$
Let's start with $6$
$\frac{35+x}{5}=6$ has solution $x=-5$ , and the sequence is $-5, 4, 6, 8, 17$ , which does have median $6$ , so this is a valid solution.
Now let the median be $8$
$\frac{35+x}{5}=8$ gives $x=5$ , so the sequence is $4, 5, 6, 8, 17$ , which has median $6$ , so this is not valid.
Finally we let the median be $x$
$\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75$ , and the sequence is $4, 6, 8, 8.75, 17$ , which has median $8$ . This case is therefore again not valid.
Hence the only possible value of $x$ is $\boxed{5}.$


Answer: 5.0

Solution:
The mean is $\frac{4+6+8+17+x}{5}=\frac{35+x}{5}$
There are three possibilities for the median: it is either $6$ $8$ , or $x$
Let's start with $6$
$\frac{35+x}{5}=6$ has solution $x=-5$ , and the sequence is $-5, 4, 6, 8, 17$ , which does have median $6$ , so this is a valid solution.
Now let the median be $8$
$\frac{35+x}{5}=8$ gives $x=5$ , so the sequence is $4, 5, 6, 8, 17$ , which has median $6$ , so this is not valid.
Finally we let the median be $x$
$\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75$ , and the sequence is $4, 6, 8, 8.75, 17$ , which has median $8$ . This case is therefore again not valid.
Hence the only possible value of $x$ is $\boxed{5}.$


Answer: 92.0

What is the sum of all the solutions of $x = \left|2x-|60-2x|\right|$
$\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 92 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 124$

Solution:
We evaluate this in cases:
Case 1 $x<30$
When $x<30$ we are going to have $60-2x>0$ .  When $x>0$ we are going to have $|x|>0\implies x>0$ and when $-x>0$ we are going to have $|x|>0\implies -x>0$ .  Therefore we have $x=|2x-(60-2x)|$ $x=|2x-60+2x|\implies x=|4x-60|$
Subcase 1 $30>x>15$
When $30>x>15$ we are going to have $4x-60>0$ . When this happens, we can express $|4x-60|$ as $4x-60$ .
Therefore we get $x=4x-60\implies -3x=-60\implies x=20$ .  We check if $x=20$ is in the domain of the numbers that we put into this subcase, and it is, since $30>20>15$ . Therefore $20$ is one possible solution.
Subcase 2 $x<15$
When $x<15$ we are going to have $4x-60<0$ , therefore $|4x-60|$ can be expressed in the form $60-4x$ .
We have the equation $x=60-4x\implies 5x=60\implies x=12$ .  Since $12$ is less than $15$ $12$ is another possible solution. $x=|2x-|60-2x||$
Case 2 $x>30$
When $x>30$ $60-2x<0$ . When $x<0$ we can express this in the form $-x$ .  Therefore we have $-(60-2x)=2x-60$ .  This makes sure that this is positive, since we just took the negative of a negative to get a positive.  Therefore we have $x=|2x-(2x-60)|$
$x=|2x-2x+60|$
$x=|60|$
$x=60$
We have now evaluated all the cases, and found the solution to be $\{60,12,20\}$ which have a sum of $\boxed{92}$

Solution:
From the equation $x = \left|2x-|60-2x|\right|$ , we have $x = 2x-|60-2x|$ , or $-x = 2x-|60-2x|$ . Therefore, $x=|60-2x|$ , or $3x=|60-2x|$ . From here we have four possible cases:
1. $x=60-2x$ ; this simplifies to $3x=60$ , so $x=20$
2. $-x=60-2x$ ; this simplifies to $x=60$
3. $3x=60-2x$ ; this simplifies to $5x=60$ , so $x=12$
4. $-3x=60-2x$ ; this simplifies to $-x=60$ , so $x=-60$ . However, this solution is extraneous because the absolute value of $2x-|60-2x|$ cannot be negative.
The sum of all of the solutions of $x$ is $20+60+12=\boxed{92}$


Answer: 7.0

What is the sum of the digits of the decimal form of the product $2^{1999}\cdot 5^{2001}$
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10$

Solution: $2^{1999}\cdot5^{2001}=2^{1999}\cdot5^{1999}\cdot5^{2}=25\cdot10^{1999}$ , a number with the digits "25" followed by 1999 zeros. The sum of the digits in the decimal form would be $2+5=7$ , thus making the answer $\boxed{7}$


Answer: 81.0

What is the sum of the digits of the square of $\text 111111111$
$\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81$

Solution: Using the standard multiplication algorithm, $111,111,111^2=12,345,678,987,654,321,$ whose digit sum is $\boxed{81.}$

Solution:
We note that
$1^2 = 1$
$11^2 = 121$
$111^2 = 12321$
and $1,111^2 = 1234321$
We can clearly see the pattern: If $X$ is $111\cdots111$ , with $n$ ones (and for the sake of simplicity, assume that $n<10$ ), then the sum of the digits of $X^2$ is
$1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1$
$=(1+2+3\cdots n)+(1+2+3+\cdots n-1)$
$=\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}$
$=\dfrac{n(n+1+n-1)}{2}=\dfrac{2n^2}{2}=n^2.$
Aha! We know that $111,111,111$ has $9$ digits, so its digit sum is $9^2=\boxed{81}$

Solution:
We see that $111^2$ can be written as $111(100+10+1)=11100+1110+111=12321$
We can apply this strategy to find $111,111,111^2$ , as seen below.
$111111111^2=111111111(100000000+10000000\cdots+10+1)$
$=11111111100000000+1111111110000000+\cdots+111111111$
$=12,345,678,987,654,321$
The digit sum is thus $1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{81}$


Answer: 12.0

What is the sum of the distinct prime integer divisors of $2016$
$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$

Solution: The prime factorization is $2016=2^5\times3^2\times7$ .  Since the problem is only asking us for the distinct prime factors, we have $2,3,7$ .  Their desired sum is then $\boxed{12}$

Solution: We notice that $9 \mid 2016$ , since $2+0+1+6 = 9$ , and $9 \mid 9$ . We can divide $2016$ by $9$ to get $224$ . This is divisible by $4$ , as $4 \mid 24$ . Dividing $224$ by $4$ , we have $56$ . This is clearly divisible by $7$ , leaving $8$ . We have $2016 = 9\cdot 4\cdot 7\cdot 8$ . We know that $4$ and $8$ are both multiples of $2$ $9$ is $3^2$ , and $7$ is prime. This means that the distinct prime factors are $2,3,$ and $7$ . Their sum is $\boxed{12}$


Answer: 8.0

What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

Solution: Looking at the prime numbers under $12$ , we see that there are $\left\lfloor\frac{12}{2}\right\rfloor+\left\lfloor\frac{12}{2^2}\right\rfloor+\left\lfloor\frac{12}{2^3}\right\rfloor=6+3+1=10$ factors of $2$ $\left\lfloor\frac{12}{3}\right\rfloor+\left\lfloor\frac{12}{3^2}\right\rfloor=4+1=5$ factors of $3$ , and $\left\lfloor\frac{12}{5}\right\rfloor=2$ factors of $5$ . All greater primes are represented once or none in $12!$ , so they cannot be part of the square. Since we are looking for a perfect square, the exponents on its prime factors must be even, so we can only use $4$ of the $5$ factors of $3$ . Therefore, the prime factorization of the square is $2^{10}\cdot3^4\cdot5^2$ . To find the square root of this, we halve the exponents, leaving $2^5\cdot3^2\cdot5$ . The sum of the exponents is $\boxed{8}$


Answer: 7.5

What is the sum of the mean, median, and mode of the numbers $2,3,0,3,1,4,0,3$
$\textbf{(A)}\ 6.5 \qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 7.5\qquad\textbf{(D)}\ 8.5\qquad\textbf{(E)}\ 9$

Solution: Putting the numbers in numerical order we get the list $0,0,1,2,3,3,3,4.$ The mode is $3.$ The median is $\frac{2+3}{2}=2.5.$ The average is $\frac{0+0+1+2+3+3+3+4}{8}=\frac{16}{8}=2.$ The sum of all three is $3+2.5+2=\boxed{7.5}$


Answer: 77.0

What is the sum of the prime factors of $2010$
$\textbf{(A)}\ 67\qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210$

Solution: First, we must find the prime factorization of $2010$ $2010=2\cdot 3 \cdot 5 \cdot 67$ . We add the factors up to get $\boxed{77}$


Answer: 1.0

What is the sum of the reciprocals of the roots of the equation $\frac{2003}{2004}x+1+\frac{1}{x}=0$
$\mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003}$

Solution:
Multiplying both sides by $x$
$\frac{2003}{2004}x^{2}+1x+1=0$
Let the roots be $a$ and $b$
The problem is asking for $\frac{1}{a}+\frac{1}{b}= \frac{a+b}{ab}$
By Vieta's formulas
$a+b=(-1)^{1}\frac{1}{\frac{2003}{2004}}=-\frac{2004}{2003}$
$ab=(-1)^{2}\frac{1}{\frac{2003}{2004}}=\frac{2004}{2003}$
So the answer is $\frac{a+b}{ab}=\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-1 \Rightarrow\boxed{1}$

Solution:
Dividing both sides by $x$
$\frac{2003}{2004}+\frac 1x+\frac{1}{x^2}=0$
we see by Vieta's formulas that the sum of the roots is $-1  \Rightarrow\boxed{1}$

Solution:
Re-stating and multiplying $\frac{1}{x}$ by $\frac{2004}{2004}$ , we have $\frac{2003x}{2004} + \frac{2004}{2004x} + 1 = 0$
Simplifying, we have $\frac{2003}{2004}+\frac{\frac{2004}{x}}{2004}=-1.$
Putting this together, we have $\frac{2003+\frac{2004}{x}}{2004}=-1.$
Simplifying, we have $2004x + \frac{2004}{x} = -2004.$
Multiplying both sides by $x,$ we have $2003x^2+2004=-2004x.$
Moving $-2004x$ to the left side, we have $2003x^2 + 2004x + 2004 = 0.$
Let's go back and see what we want.
We want the sum of the reciprocals of the roots, in which if the roots were to be called $r_1$ and $r_2,$ we would want $\frac{1}{r_1}+\frac{1}{r_2}=\frac{r_1 + r_2}{r_1 \cdot r_2}$
We can use Vieta's formulas to solve the sum of the roots and the product of the roots.
We have that the sum of the two roots is $\frac{-b}{a}$ where the quadratic is $ax^2+bx+c=0.$
In this case, $a=2003$ and $b=2004$
Therefore, $r_1+r_2=-\frac{2004}{2003}.$
We have that the product of the two roots is $\frac{c}{a}$ where the quadratic is $ax^2+bx+c=0.$
In this case, $c=2004$ and $a=2003.$
Therefore, $r_1 \cdot r_2 = \frac{2004}{2003}.$
Now that we have the sum and product of the roots, we can substitute back into the expression of what we want.
We have $\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-\frac{2004}{2003} \cdot \frac{2003}{2004} = \boxed{1}.$


Answer: 337.0

What is the sum of the solutions to the equation $\sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}$

Solution:
Let $y = \sqrt[4]{x}$ . Then we have $y(7 - y) = 12$ , or, by simplifying, \[y^2 - 7y + 12 = (y - 3)(y - 4) = 0.\]
This means that $\sqrt[4]{x} = y = 3$ or $4$
Thus the sum of the possible solutions for $x$ is $4^4 + 3^4 = \boxed{337}$


Answer: 4.0

What is the tens digit in the sum $7!+8!+9!+...+2006!$
$\textbf{(A) } 1\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 6\qquad \textbf{(E) } 9$

Solution:
Since $10!$ is divisible by $100$ , any factorial greater than $10!$ is also divisible by $100$ . The last two digits of all factorials greater than $10!$ are $00$ , so the last two digits of $10!+11!+...+2006!$ are $00$ .
(*)
So all that is needed is the tens digit of the sum $7!+8!+9!$
$7!+8!+9!=5040+40320+362880=408240$
So the tens digit is $\boxed{4}$


Answer: 0.0

What is the tens digit of $2015^{2016}-2017?$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8$

Solution:
Since we only need the tens digits, we only need to care about the multiplication of tens and ones. (If you want to use mathematical terms then we only need to look at the exponents in $mod 100$ .) We will use the " $\equiv$ " sign to denote congruence in modulus, basically taking the last two digits and ignoring everything else.
\[\begin{split} 15^{1}\equiv15 \\ 15^{2}\equiv25 \end{split}\] From here we only need to multiply $15\cdot25$ and we can ignore the hundreds digits. \[\begin{split} 15^{3}\equiv75 \\ 15^{4}\equiv25 \\ 15^{5}\equiv75 \end{split}\] Notice that for every $x\neq1$ $15^{x}\equiv25$ if $x$ is even, and $15^{x}\equiv75$ if $x$ is odd. Since $2015^{2016}$ has an even exponent, we conclude that the last two digits will be $25$ , and subtracting $25-17=0 \Longrightarrow \boxed{0}$ .
~JH. L


Answer: 0.0

Solution:
Notice that, for $n\ge 2$ $2015^n\equiv 15^n$ is congruent to $25\pmod{100}$ when $n$ is even and $75\pmod{100}$ when $n$ is odd. (Check for yourself).  Since $2016$ is even, $2015^{2016} \equiv 25\pmod{100}$ and $2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}$
So the answer is $\textbf{(A)}\ \boxed{0}$


Answer: 4.0

What is the tens digit of $7^{2011}$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$

Solution: Since we want the tens digit, we can find the last two digits of $7^{2011}$ .  We can do this by using modular arithmetic. \[7^1\equiv 07 \pmod{100}.\] \[7^2\equiv 49 \pmod{100}.\] \[7^3\equiv 43 \pmod{100}.\] \[7^4\equiv 01 \pmod{100}.\] We can write $7^{2011}$ as $(7^4)^{502}\times 7^3$ .  Using this, we can say: \[7^{2011}\equiv (7^4)^{502}\times 7^3\equiv 7^3\equiv 343\equiv 43\pmod{100}.\] From the above, we can conclude that the last two digits of $7^{2011}$ are 43.  Since they have asked us to find the tens digit, our answer is $\boxed{4}$

Solution:
We can use patterns to figure out the answer.
7 to the power of 2 is 49. So, the tens digit is 4.
7 to the power of 3 is 343. So, the tens digit is 4.
7 to the power of 4 is 2401.So, the tens digit is 0.
7 to the power of 5 is 16807. So, the tens digit is 0.
By now, we can notice the pattern. The tens digit for 7 to the power of 2 is 4, then 4, then 0, then 0. It keeps on going, 2 fours, and then 2 zeros in 4 numbers. If we round up for 2011/4, we get 503. 503 * 4 is 2012. So 7 to the 2012 power has a tens digit of 0, since 2012 is a mutiple of 4, and 7 to the power of 4 has a tens digit of 0. We have to subtract a power from 7 to the 2012 power, so the tens digit goes back from 0 to 4 because if we subtract a power from 7 to the power of 4, we have 7 to the power of 3, which has a tens digit of 4. Hence the answer is $\boxed{4}$


Answer: 7.0

What is the units digit of $13^{2003}$
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9$

Solution:
$13^{2003}\equiv 3^{2003}\pmod{10}$
Since $3^4=81\equiv1\pmod{10}$
$3^{2003}=(3^{4})^{500}\cdot3^{3}\equiv1^{500}\cdot27\equiv7\pmod{10}$
Therefore, the units digit is $7 \Rightarrow\boxed{7}$

Solution:
Since we are looking for the units digit of $13^{2003}$ , we only have to focus on the units digit of the base (13) as none of the other digits of the base affect the units digit of the resulting value.
By testing the first few values or through previous knowledge, you might see that the units digit of exponents with base 3 follow this pattern: \[3^1=3\] \[3^2=9\] \[3^3=27\] \[3^4=81,\] giving us the rotation $3-9-7-1.$
As this cycle resets every time the index increases by 4, we know that this cycle ends on 2000, and starts once again on 2001. As our expression is raised to the power of 2003, we know that the units digit of our expression must end with the third term of our pattern: $7$
Therefore, the units digit of our expression is $7 \Rightarrow\boxed{7}$