Answer: 683.0

Solution:
Consider the elements of $U$ modulo $3.$
Ignore the $0$ 's because we're gonna multiply $\binom{6}{0}+..+\binom{6}{6}=2^6$ at the end. Let $a$ be the $1's$ and $b$ be the $2's$ . The key here is that $2 \equiv -1 \pmod{3}$ so the difference between the number of $a$ and $b$ is a multiple of $3$
1. Counted twice because $a$ and $b$ can be switched:
$6a$
$6a,3b$
$5a,2b$
$4a,b$
$3a$
2. Counted once:
$6a,6b$ ... $0a,0b$
By Vandermonde's Identity on the second case, this is $2^6 \left( 2\left(1+20+90+90+20\right) + \binom{12}{6} \right)\implies \boxed{683}$

Solution:
Elements $0 \pmod{3}$ can either be included or excluded, for a total of $2^6$ . Then, like the previous solution, let $a$ be the number of elements $1 \pmod{3}$ and $b$ be the number of elements $2 \pmod{3}$ . Then, $a + 2b \equiv 0 \pmod{3} \implies a - b \equiv 0 \pmod{3}$ . Since $0 \le a, b \le 6$ , there are 3 cases:
$\>\>\>\> 1. \> a - b = 0$
Then, we have, \[\binom{6}{0}\binom{6}{0} + \binom{6}{1}\binom{6}{1} + \cdots + \binom{6}{6} \binom{6}{6}\]
Using the substitution $\binom{6}{k} = \binom{6}{6-k}$ on every second binomial coefficient, it is clear that the bottom numbers sum to $6$ , and the ones above sum to $12$ . Apply Vandermonde's, we obtain $\binom{12}{6}$
$\>\>\>\> 2. \> a - b = \pm 3$
For $a - b = 3$ , using the same substitution and applying Vandermonde's, we get: \[\binom{6}{3} \binom{6}{0} + \binom{6}{4} \binom{6}{1} + \cdots + \binom{6}{6} \binom{6}{3} = \binom{12}{3}\]
Then, for $a - b = -3$ , it is completely symmetric with $a - b = 3$ . We get $\binom{12}{3}$
$\>\>\>\> 3. \> a - b = \pm 6$
We have $\binom{6}{6} \binom{6}{0} + \binom{6}{0} \binom{6}{6} = 2$
We get $\frac{2^6 (\binom{12}{6} + 2 \binom{12}{3} + 2)}{2^{18}} = \frac{683}{2^{11}} \implies m = \boxed{683}$

Solution:
Note that in general, the answer will be around $1/3$ of $2^{18}$ . We will use this to our advantage. Partition $U$ into disjoint subsets $\{1,2,3\}, \{4,5,6\}, \cdots, \{16,17,18\}$ . Within each of these subsets, there are 3 possible remainders $\mod 3$ depending on what elements we choose to include into $T$ : (Using set $\{1,2,3\}$ for demonstration purposes)
Remainder of zero: Include $\{3\}, \{1,2\}, \{\}, \{1,2,3\}$
Remainder of one: Include $\{1\}, \{1,3\}$
Remainder of two: Include $\{2\}, \{2,3\}$
Suppose all subsets are of the form $\{\}, \{1,2,3\}$ . Then, since both choices are $0 \pmod{3}$ , we can choose either one for all six subsets, giving us $2^6$ combinations.
On the other hand, suppose there exists a subset $V$ that isn't of the form $\{\}, \{1,2,3\}$ . Then, for $V$ , it is equally likely that a remainder of zero, one, or two is chosen, since each remainder has $2$ ways to be achieved, i.e. there is a $1/3$ chance for each. Consider the sum of the other subsets $T \setminus V$ . The sum is either $0, 1, 2 \pmod{3}$ . Whatever that remainder $r$ might be, we can always choose $3-r$ as the remainder for our set $V$ , giving us a total of $r + 3 - r \equiv 0 \pmod{3}$ . The probability we choose remainder $3-r$ is $1/3$ as previously mentioned. So, we get $\frac{1}{3}(2^{18} - 2^6)$ total combinations.
Therefore, we get $\frac{2^6 + \frac{1}{3}(2^{18} - 2^6)}{2^{18}} = \frac{683}{2^{11}} \implies m = \boxed{683}$

Solution:
Rewrite the set after mod 3 as above
1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0
All 0s can be omitted
Case 1
No 1 No 2 $1$
Case 2 $222$ $20$
Case 3 $222222$ $1$
Case 4 $12$ $6*6=36$
Case 5 $12222$ $6*15=90$
Case 6 $1122$ $15*15=225$
Case 7 $1122222$ $15*6=90$
Case 8 $111$ $20$
Case 9 $111222$ $20*20=400$
Case 10 $111222222$ $20$
Case 11 $11112$ $15*6=90$
Case 12 $11112222$ $15*15=225$
Case 13 $1111122$ $6*15=90$
Case 14 $1111122222$ $6*6=36$
Case 15 $111111$ $1$
Case 16 $111111222$ $20$
Case 17 $111111222222$ $1$
Total $1+20+1+36+90+225+90+20+400+20+90+225+90+36+1+20+1=484+360+450+72=1366$
$P=\frac{1366}{2^{12}}=\frac{683}{2^{11}}$
ANS= $\boxed{683}$

Solution:
Notice that six numbers are $0\pmod3$ , six are $1\pmod3$ , and six are $2\pmod3$ .  Having numbers $0\pmod3$ will not change the remainder when $s(T)$ is divided by $3$ , so we can choose any number of these in our subset.  We ignore these for now.  The number of numbers that are $1\pmod3$ , minus the number of numbers that are $2\pmod3$ , must be a multiple of $3$ , possibly zero or negative.  We can now split into cases based on how many numbers that are $1\pmod3$ are in the set.
CASE 1- $0$ $3$ , or $6$ integers:  There can be $0$ $3$ , or $6$ integers that are $2\pmod3$ .  We can choose these in \[\left(\binom60+\binom63+\binom66\right)\cdot\left(\binom60+\binom63+\binom66\right)=(1+20+1)^2=484.\]
CASE 2- $1$ or $4$ integers:  There can be $2$ or $5$ integers that are $2\pmod3$ .  We can choose these in \[\left(\binom61+\binom64\right)\cdot\left(\binom62+\binom65\right)=(6+15)^2=441.\]
CASE 3- $2$ or $5$ integers:  There can be $1$ or $4$ integers that are $2\pmod3$ .  We can choose these in \[\left(\binom62+\binom65\right)\cdot\left(\binom61+\binom64\right)=(15+6)^2=441.\]
Adding these up, we get that there are $1366$ ways to choose the numbers such that their sum is a multiple of three.  Putting back in the possibility that there can be multiples of $3$ in our set, we have that there are \[1366\cdot\left(\binom60+\binom61+\binom62+\binom63+\binom64+\binom65+\binom66\right)=1366\cdot2^6\] subsets $T$ with a sum that is a multiple of $3$ .  Since there are $2^{18}$ total subsets, the probability is \[\frac{1366\cdot2^6}{2^{18}}=\frac{683}{2^{11}},\] so the answer is $\boxed{683}$

Solution: We use generating functions. Each element of $U$ has two choices that occur with equal probability--either it is in the set or out of the set. Therefore, given $n\in U$ , the probability generating function is \[\frac{1}{2}+\frac{1}{2}x^n.\] Therefore, in the generating function \[\frac{1}{2^{18}}(1+x)(1+x^2)(1+x^3)\cdots (1+x^{18}),\] the coefficient of $x^k$ represents the probability of obtaining a sum of $k$ . We wish to find the sum of the coefficients of all terms of the form $x^{3k}$ (where $k \in \mathbb{Z}_{\geq0}$ ). If $\omega=e^{2\pi i/3}$ is a cube root of unity, then it is well know that for a polynomial $P(x)$ \[\frac{P(1)+P(\omega)+P(\omega^2)}{3}\] will yield the sum of the coefficients of the terms of the form $x^{3k}$ . (This is known as the third Roots of Unity Filter.) Then we find \begin{align*} \frac{1}{2^{18}}(1+1)(1+1^2)(1+1^3)\cdots (1+1^{18})&=1\\\frac{1}{2^{18}}(1+\omega)(1+\omega^2)(1+\omega^3)\cdots (1+\omega^{18})&=\frac{1}{2^{12}}\\\frac{1}{2^{18}}(1+\omega^2)(1+\omega^4)(1+\omega^6)\cdots (1+\omega^{36})&=\frac{1}{2^{12}}. \end{align*} To evaluate the last two products, we utilized the facts that $\omega^3=1$ and $1+\omega+\omega^2=0$ . Therefore, the desired probability is \[\frac{1+1/2^{12}+1/2^{12}}{3}=\frac{683}{2^{11}}.\] Thus the answer is $\boxed{683}$

Solution:
Define a set that "works" to be a set for which the sum of the terms is $0$ mod $3$ . The given set mod $3$ is \[\{1,2,0,1,2,0,1,2,0,1,2,0,1,2,0,1,2,0\}.\] Let $w(N)$ be the number of subsets of the first $N$ terms of this set that "work."
Consider just the first $3$ terms: \[\{1,2,0\}.\] There are $2^3=8$ total subsets, and $w(3)=4$ (the subsets are $\emptyset, \{0\}, \{1,2\},$ and $\{1,2,0\}$ ). Now consider the first $6$ terms: \[\{1,2,0,1,2,0\}.\] To find $w(6)$ , we set aside the last $3$ terms as a "reserve" that we can pair with subsets of the first $3$ terms (which we looked at earlier).
First, all $2^3$ subsets of the first $3$ terms can either be paired with a $1$ or a $2$ (or nothing) from the "reserve" terms so that they "work," creating $2^3$ subsets that "work" already. But for each of these, we have the option to add a $0$ from the reserve, so we now have $2\cdot2^3$ subsets that "work." For each of the $w(3)=4$ subsets of the first $3$ terms that "work", we can also add on a $\{1,2\}$ or a $\{1,2,0\}$ from the reserves, creating $2w(3)$ new subsets that "work." And that covers all of them. With all of this information, we can write $w(6)$ as \[w(6)=2\cdot2^3+2w(3)=2(2^3+w(3)).\] The same process can be used to calculate $w(9)$ then $w(12)$ etc. so we can generalize it to \[w(N)=2(2^{N-3}+w(N-3)).\] Thus, we calculate $w(18)$ with this formula: \[w(18)=2( 2^{15} + 2( 2^{12} + 2( 2^9 + 2( 2^6 + 2( 2^3 + 4 ) ) ) ) )=87424.\] Because there are $2^{18}$ total possible subsets, the desired probability is \[\frac{w(18)}{2^{18}}=\frac{87424}{2^{18}}=\frac{683}{2048},\] so the answer is $\boxed{683}.$

Solution:
Try smaller cases and find a pattern. Using similar casework as in Solution 1, we can easily find the desired probability if $U$ is of a small size.
If $U = \{ 1,2,0\} \pmod 3$ , then $4$ out of $8$ subsets work, for a probability of $\tfrac12$
If $U = \{ 1,2,0,1,2,0\} \pmod 3$ , then $24$ out of $64$ subsets work, for a probability of $\tfrac38$
If $U = \{ 1,2,0,1,2,0,1,2,0\} \pmod 3$ , then $176$ out of $512$ subsets work, for a probability of $\tfrac{11}{32}$
Let $a_n$ be the numerator of the desired probability if $U$ is of size $3n$ . Then $a_1 = 1, a_2 = 3,$ and $a_3 = 11$ . Noticing that the denominators are multiplied by $4$ each time, we can conclude that the pattern of the numerators is $a_n = 4a_{n-1} - 1$ , so $a_6 = \boxed{683}$

Solution:
The total number of subsets is simply $2^{18}.$ Now we need to find the number of subsets that have a sum divisible by $3.$
Ignore the 6 numbers in the list that are divisible by 3. We look only at the number of subsets of $\{1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17\}$ then multiply by $2^6$ at the end. This is because adding a multiple of $3$ to the sum will not change whether it is divisible by $3$ or not and for each of the $6$ multiples, we have two options of whether to add it to the subset or not.
Now, let $f(3k)$ be the number of subsets of $\{1, 2, 4, 5, \dots, 3k-2, 3k-1\}$ that have a sum divisible by $3$ . There are $2k$ numbers in the set. Let us look at the first $2k - 2$ numbers: all subsets of the first $2k - 2$ numbers will have a residue $0, 1,$ or $2$ mod $3.$ If it is $0,$ add both $3k-2$ and $3k-1$ to the subset. If it is $1$ add just $3k-1$ to the set, and if it is $2$ add just $3k-2.$ We don't have to compute all three cases separately; since there is a 1 to 1 correspondence, we only need the sum of the three cases (which we know is $2^{2k-2}$ ).
Now, this counts all the subsets except for those that include neither $3k-1$ nor $3k-2.$ However, this is just $f(3k - 3).$ Thus, $f(3k) = 2^{2k-2} + f(3k-3)$ and the base case is $f(0) = 1.$ We have,
$f(18) = 2^{10} + f(15) = 2^{10} + 2^{8} + f(12) = \dots = 2^{10} + 2^8 + 2^6 + 2^4 + 2^2 + 2^0 + f(0).$ $=2^{10} + 2^8 + \dots + 2^2 + 2.$
Multiplying this by $2^6$ we have, \[\frac{2^7(1 + 2^1 + 2^3 + 2^5 + 2^7 + 2^9)}{2^{18}} .\] The $2^7$ cancels out with the denominator. However, the second term in the product is obviously odd and so does not cancel further with the denominator, which is just a power of $2.$ Thus, we want to find $1 + 2 + 2^3 + 2^5 + 2^7 + 2^9,$ which equals $\boxed{683}.$

Solution:
The total number of subsets is $\sum_{i=0}^{18}\tbinom{18}{i}=2^{18}$ . If $s(T)\equiv 0\bmod{3}$ , the sum of the elements divides 3. We can rewrite the set as 6 0s, 6 1s, and 6 2s. We can ignore the zeros for now, since they won't influence the sum so we focus on each configuration of the 6 1s and 6 2s such that the sum is divisible by 3 and then multiply by $\sum_{i=0}^{6}\tbinom{6}{i}=2^6$ at the end. We proceed with casework on the number of values that are equivalent to $2\pmod{3}$ as follows:
Case 1: There are zero elements:
Then, there are only configurations of the values congruent to 1 mod 3. There are $\tbinom{6}{0}+\tbinom{6}{3}+\tbinom{6}{6}$ ways to assign values in the set that are congruent to 1 mod 3 such that the sum of those values divides 3. Therefore there are $22$ configurations for this case.
Case 2: There is one element:
There are $6$ ways to choose which element is included in the subset and $\tbinom{6}{1}+\tbinom{6}{4}$ ways to assign values to the numbers congruent to 1 mod 3 in the set. Therefore there are $6\cdot21=126$ configurations for this case.
Case 3: There are two elements:
There are $\tbinom{6}{2}=12$ ways to choose the 2 elements in the set that are congruent to 2 mod 3 and $\tbinom{6}{2}+\tbinom{6}{5}$ possible ways to assign values to the numbers congruent to 1 mod 3. Therefore there are $15\cdot21=315$ configurations for this case.
Case 4: There are three elements:
There are $\tbinom{6}{3}=20$ ways to choose three elements in the set that are congruent to 2 mod 3. There are $\tbinom{6}{0}+\tbinom{6}{3}+\tbinom{6}{6}$ ways to assign values to the numbers congruent to 1 mod 3. Therefore, there are $20\cdot22=440$ configurations for this case.
Case 5, Case 6, and Case 7 are symmetric to Case 3, Case 2, and Case 1 respectively so we can multiply by 2 for those cases.
Putting all the cases together we obtain $2(22+126+315) + 440=1366$ . Multiplying by $2^6$ gives $2^6\cdot1366=2^7\cdot683$ . Since $n=2^{18}$ $\frac{m}{n}=\frac{683}{2^{11}}$ . Therefore, $m=\boxed{683}$


Answer: 404.0

For every triple $(a,b,c)$ of non-zero real numbers, form the number $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$ .
The set of all numbers formed is
$\textbf{(A)}\ {0} \qquad \textbf{(B)}\ \{-4,0,4\} \qquad \textbf{(C)}\ \{-4,-2,0,2,4\} \qquad \textbf{(D)}\ \{-4,-2,2,4\}\qquad \textbf{(E)}\ \text{none of these}$

Solution:
Solution by e_power_pi_times_i
$\dfrac{x}{|x|} = 1$ or $-1$ depending whether $x$ is positive or negative. If $a$ $b$ , and $c$ are positive, then the entire thing amounts to $4$ . If one of the three is negative and the other two positive, the answer is $0$ . If two of the three is negative and one is positive, the answer is $0$ . If all three are negative, the answer is $-4$ . Therefore the set is $\boxed{4,0,4}$


Answer: 1000.0

For his birthday, Bert gets a box that holds 125 jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans did Carrie get?
$\text{(A)}\ 250\qquad\text{(B)}\ 500\qquad\text{(C)}\ 625\qquad\text{(D)}\ 750\qquad\text{(E)}\ 1000$

Solution:
Since the volume ratio is equal to the sides ratio cubed, then the ratio of the larger box's volume to the smaller one is 2 cubed.
$2^3=8$
Now multiply 125 (the number of jellybeans that Bert's box can hold) by 8.
$8\cdot125= \boxed{1000}$


Answer: 20.0

For how many $n$ in $\{1, 2, 3, ..., 100 \}$ is the tens digit of $n^2$ odd?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50$

Solution: Let $n=10a+b$ . So $n^2=(10a+b)^2=100a^2+20ab+b^2$ . The term $100a^2$ only contributes digits starting at the hundreds place, so this does not affect whether the tens digit is odd.  The term $20ab$ only contributes digits starting at the tens place, and the tens digit contributed will be the ones digit of $2ab$ which is even.  So we see this term also does not affect whether the tens digit is odd.  This means only $b^2$ can affect whether the tens digit is odd.  We can quickly check $1^2=1, \dots, 9^2=81$ and discover only for $b=4$ or $b=6$ does $b^2$ have an odd tens digit. The total number of positive integers less than or equal to $100$ that have $4$ or $6$ as the units digit is \[10\times 2=\boxed{20.}\]


Answer: 9.0

For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

Solution:
Note that \[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.\] Since this expression is an integer, we need:
Taking the intersection gives $-5\leq n\leq3.$ So, there are $3-(-5)+1=\boxed{9}$ integer values of $n.$

Solution: Note that $4000\cdot \left(\frac{2}{5}\right)^n$ will be an integer if the denominator is a factor of $4000$ . We also know that the denominator will always be a power of $5$ for positive values and a power of $2$ for all negative values. So we can proceed to divide $4000$ by $5^n$ for each increasing positive value of $n$ until we get a non-factor of $4000$ and also divide $4000$ by $2^{-n}$ for each decreasing negative value of $n$ . For positive values we get $n= 1, 2, 3$ and for negative values we get $n= -1, -2, -3, -4, -5$ . Also keep in mind that the expression will be an integer for $n=0$ , which gives us a total of $\boxed{9}$ for $n.$

Solution:
The values for $n$ are $-5, -4, -3, -2, -1, 0, 1, 2,$ and $3.$
The corresponding values for $4000\cdot \left(\frac{2}{5}\right)^n$ are $390625, 156250, 62500, 25000, 10000, 4000, 1600, 640,$ and $256,$ respectively.
In total, there are $\boxed{9}$ values for $n.$


Answer: 9.0

Solution:
Note that \[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.\] Since this expression is an integer, we need:
Taking the intersection gives $-5\leq n\leq3.$ So, there are $3-(-5)+1=\boxed{9}$ integer values of $n.$

Solution: Note that $4000\cdot \left(\frac{2}{5}\right)^n$ will be an integer if the denominator is a factor of $4000$ . We also know that the denominator will always be a power of $5$ for positive values and a power of $2$ for all negative values. So we can proceed to divide $4000$ by $5^n$ for each increasing positive value of $n$ until we get a non-factor of $4000$ and also divide $4000$ by $2^{-n}$ for each decreasing negative value of $n$ . For positive values we get $n= 1, 2, 3$ and for negative values we get $n= -1, -2, -3, -4, -5$ . Also keep in mind that the expression will be an integer for $n=0$ , which gives us a total of $\boxed{9}$ for $n.$

Solution:
The values for $n$ are $-5, -4, -3, -2, -1, 0, 1, 2,$ and $3.$
The corresponding values for $4000\cdot \left(\frac{2}{5}\right)^n$ are $390625, 156250, 62500, 25000, 10000, 4000, 1600, 640,$ and $256,$ respectively.
In total, there are $\boxed{9}$ values for $n.$


Answer: 2.0

For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$

Solution:
The image below shows the two curves for $k=4$ . The blue curve is $x^2+y^2=k^2$ , which is clearly a circle with radius $k$ , and the red curve is a part of the curve $xy=k$

In the special case $k=0$ the blue curve is just the point $(0,0)$ , and as $0\cdot 0=0$ , this point is on the red curve as well, hence they intersect.
The case $k<0$ is symmetric to $k>0$ : the blue curve remains the same and the red curve is flipped according to the $x$ axis. Hence we just need to focus on $k>0$
Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as $x$ approaches 0, $y$ approaches $\infty$ . Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most $k$
At this point we can guess that on the red curve the point where $x=y$ is always closest to the origin, and skip the rest of this solution.
For an exact solution, fix $k$ and consider any point $(x,y)$ on the red curve. Its distance from the origin is $\sqrt{ x^2 + (k/x)^2 }$ . To minimize this distance, it is enough to minimize $x^2 + (k/x)^2$ . By the Arithmetic Mean-Geometric Mean Inequality we get that this value is at least $2k$ , and that equality holds whenever $x^2 = (k/x)^2$ , i.e., $x=\pm\sqrt k$
Now recall that the red curve intersects the blue one if and only if its closest point is at most $k$ from the origin. We just computed that the distance between the origin and the closest point on the red curve is $\sqrt{2k}$ . Therefore, we want to find all positive integers $k$ such that $\sqrt{2k} > k$
Clearly the only such integer is $k=1$ , hence the two curves are only disjoint for $k=1$ and $k=-1$ .
This is a total of $\boxed{2}$ values.

Solution:
From the graph shown above, we see that there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the graph of the circle. This should be enough to conclude that the hyperbola does not intersect the circle.
Therefore, for each value of k, we only need to check said value to determine intersection. Let said point, closest to the circle have coordinates $(x, k/x)$ derived from the equation. Then, all coordinates that satisfy $\sqrt{ x^2+ (k/x)^2 } \leq k$ intersect the circle.
Squaring, we find $x^2+(k/x)^2 \leq k^2.$ After multiplying through by $x^2$ and rearranging, we find $x^4-x^2k^2+k^2 \leq 0$ .
We see this is a quadratic in $x^2$ and consider taking the determinant, which tells us that solutions are real when, after factoring: $k^2(k^2-4) \geq 0$ We plot this inequality on the number line to find it is satisfied for all values except: $(-1, 0, 1)$ .
We then eliminate 0 because it is extraneous as both $xy=0$ and $x^2+y^2=0$ are points which coincide.
Therefore, there are a total of $\boxed{2}$ values.

Solution:
Since $xy=k$ , multiply the equation by 2 on both sides to get $2xy=2k$ . Now we can add the two equations to get $(x+y)^2=k^2+2k$ , for which the only value of $k$ that does not satisfy the equation is $-1$ , as that makes the RHS negative. Similarly, if we subtract the two equations, we obtain $(x-y)^2=k^2-2k$ , for which the only value of $k$ that does not satisfy the equation is $1$ , for the same reason above.
Thus, the only values are $k = 1, -1$ , giving us a total of $\boxed{2}$ values.

Solution: Multiply $k=xy$ by and substitute it into $k^2=x^2+y^2$ . Then, $k=\frac{x^2+y^2}{xy}$ . Recognize it? It's also $k=\frac{x}{y}+\frac{y}{x}$ . The minimum of this function (more accurately the minimum absolute value of the function) is k=2, -2 (when x=y or x=-y). As long as k>2 or k<-2, the function is valid. As such, $k\neq1,-1 \implies \boxed{2}$ . Elegant, huh?

Solution: Assume that $k\ge 0$ since if $k$ works then $-k$ also works. Let $x = ka$ and $y = kb$ . Then the given equations become $ab = \frac{1}{k}$ and $a^2 + b^2 = 1$ , which we don't want to intersect. The points that are the closest on this graph are $(1/\sqrt{2}, 1/\sqrt{2})$ and $(1/\sqrt{k})(1/\sqrt{k})$ . We don't want the hyperbola to intersect or go inside the circle, so we require $1/\sqrt{k} > 1/\sqrt{2}$ , so $k< 2\implies k = 0, 1$ . Also, by symmetry $k = -1$ also works.Obviously $k = 0$ doesn't work, so we can discard that, leaving $\boxed{2}$ integral solutions for $k$


Answer: 2.0

Solution:
To know when the system does not intersect, we first need to find when they do intersect. To do so, we substitute $xy=k$ into $x^{2}+y^{2}=k^{2}$ giving:
$x^2+y^2=(xy)^2$
By Simon's Favorite Factoring Trick, we get $1=(1-x)(1+x)(1-y)(1+y)$
Notice that other than for $x$ and/or $y$ equal to $-1$ and/or $1$ because for any $x$ or $y$ we can get some coordinates that work even if they're fractions which through $xy=k$ it can become an integer as $\dfrac{1}{2}\times2=1$
This means we have to exclude $xy=k=1$ or $-1$ . This is $\boxed{2}$ values that do not work.

Solution:
The first equation is describing a circle with radius $k$ centered at $(0,0)$ , so we can rewrite it in terms of a new variable $\theta$
\[(k\cos(\theta))^2+(k\sin(\theta))^2=k^2.\]
A quick check shows that this identity holds since $\sin^2+\cos^2=1$ . Now, with an alternate expression for $x$ and $y$ , we can plug this into the second equation to get:
\[(k\sin(\theta))(k\cos(\theta))=k\rightarrow k^2\sin(\theta)\cos(\theta)=k.\]
Since the two equations does intersect when $k=0$ at $(x,y)=(0,0)$ , we do not need to consider the $k=0$ case, so we can safely divide both sides by $k^2$ to get:
\[\sin(\theta)\cos(\theta)=\frac{1}{k}.\]
We can see that the intersections of these two equations will remain unchanged, only now that the solutions will be in terms of $k$ and $\theta$ . Because the range of both $\sin$ and $\cos$ are limited to $[0,1]$ , the range of $\sin(\theta)\cos(\theta)$ will not exceed $1$ . However, the only way that the product could equal 1 is if both $\sin(\theta)$ and $\cos(\theta)$ to equal 1 at the same time, so since the maxima of the functions don't match up, this is an impossible case. Therefore, this equation will only have valid solutions when $k>1$ , not at $k=1$ . A quick sanity check with $\theta=\frac{\pi}{4}$ shows that $\frac{1}{k}=\frac{1}{2}$ and all values less than $\frac{1}{2}$ is indeed contained in the range.
For all $k<0$ , symmetry about the $y$ -axis will give an identical argument with the equation $\sin(\theta)\cos(\theta)=-\frac{1}{k}$ , with $k<-1$ producing valid solutions for our system. Then, the only two values of $k$ such that the system of equations won't intersect are $1$ and $-1$ , a total of $\boxed{2}$ values.


Answer: 34.0

For how many integers $n$ between $1$ and $50$ , inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$ .)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

Solution:
The main insight is that
\[\frac{(n^2)!}{(n!)^{n+1}}\]
is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$ . Thus,
\[\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}\]
is an integer if $n^2 \mid n!$ , or in other words, if $\frac{(n-1)!}{n}$ , is an integer. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson's Theorem . There are $15$ primes between $1$ and $50$ , inclusive, so there are $15 + 1 = 16$ terms for which
\[\frac{(n^2-1)!}{(n!)^{n}}\]
is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of $2$ in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$ , as there are more factors of p in the denominator than the numerator. Thus all $16$ values of n make the expression not an integer and the answer is $50-16=\boxed{34}$

Solution:
We can use the P-Adic Valuation (more info could be found here: Mathematicial notation ) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by $v_p (n)$ and is defined as the greatest power of some prime 'p' that divides n. For example, $v_2 (6)=1$ or $v_7 (245)=2$ .) Using Legendre's formula, we know that :
\[v_p (n!)= \sum_{i=1}^\infty   \lfloor \frac {n}{p^i} \rfloor\]
Seeing factorials involved in the problem, this prompts us to use Legendre's Formula where n is a power of a prime.
We also know that , $v_p (m^n) = n \cdot v_p (m)$ .
Knowing that $a\mid b$ if $v_p (a) \le v_p (b)$ , we have that :
\[n \cdot v_p (n!) \le v_p ((n^2 -1 )!)\] and we must find all n for which this is true.
If we plug in $n=p$ , by Legendre's we get two equations:
\[v_p ((n^2 -1)!) = \sum_{i=1}^\infty  \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1\]
And we also get :
\[v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty   \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p\]
But we are asked to prove that $n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1$ which is false for all 'n' where n is prime.
Now we try the same for $n=p^2$ , where p is a prime. By Legendre we arrive at:
\[v_p ((p^4 -1)!) = p^3 + p^2 + p -3\]
(as $v_p(p^4!) = p^3 + p^2 + p + 1$ and $p^4$ contains 4 factors of $p$ ) and
\[p^2 \cdot v_p (p^2 !) = p^3 + p^2\]
Then we get:
\[p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3\] Which is true for all primes except for 2, so $2^2 = 4$ doesn't work. It can easily be verified that for all $n=p^i$ where $i$ is an integer greater than 2, satisfies the inequality : \[n \cdot v_p (n!) \le v_p ((n^2 -1 )!).\]
Therefore, there are 16 values that don't work and $50-16 = \boxed{34}$ values that work.


Answer: 34.0

Solution: Notice all $15$ primes don't work as there are $n$ factors of $n$ in the denominator and $n-1$ factors of $n$ in the numerator. Further experimentation finds that $n=4$ does not work as there are 11 factors of 2 in the numerator and 12 in the denominator. We also find that it seems that all other values of $n$ work. So we get $50-15-1=\boxed{34}$ and that happens to be right.


Answer: 34.0

Solution:
The main insight is that
\[\frac{(n^2)!}{(n!)^{n+1}}\]
is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$ . Thus,
\[\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}\]
is an integer if $n^2 \mid n!$ , or in other words, if $\frac{(n-1)!}{n}$ , is an integer. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson's Theorem . There are $15$ primes between $1$ and $50$ , inclusive, so there are $15 + 1 = 16$ terms for which
\[\frac{(n^2-1)!}{(n!)^{n}}\]
is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of $2$ in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$ , as there are more factors of p in the denominator than the numerator. Thus all $16$ values of n make the expression not an integer and the answer is $50-16=\boxed{34}$

Solution:
We can use the P-Adic Valuation (more info could be found here: Mathematicial notation ) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by $v_p (n)$ and is defined as the greatest power of some prime 'p' that divides n. For example, $v_2 (6)=1$ or $v_7 (245)=2$ .) Using Legendre's formula, we know that :
\[v_p (n!)= \sum_{i=1}^\infty   \lfloor \frac {n}{p^i} \rfloor\]
Seeing factorials involved in the problem, this prompts us to use Legendre's Formula where n is a power of a prime.
We also know that , $v_p (m^n) = n \cdot v_p (m)$ .
Knowing that $a\mid b$ if $v_p (a) \le v_p (b)$ , we have that :
\[n \cdot v_p (n!) \le v_p ((n^2 -1 )!)\] and we must find all n for which this is true.
If we plug in $n=p$ , by Legendre's we get two equations:
\[v_p ((n^2 -1)!) = \sum_{i=1}^\infty  \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1\]
And we also get :
\[v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty   \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p\]
But we are asked to prove that $n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1$ which is false for all 'n' where n is prime.
Now we try the same for $n=p^2$ , where p is a prime. By Legendre we arrive at:
\[v_p ((p^4 -1)!) = p^3 + p^2 + p -3\]
(as $v_p(p^4!) = p^3 + p^2 + p + 1$ and $p^4$ contains 4 factors of $p$ ) and
\[p^2 \cdot v_p (p^2 !) = p^3 + p^2\]
Then we get:
\[p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3\] Which is true for all primes except for 2, so $2^2 = 4$ doesn't work. It can easily be verified that for all $n=p^i$ where $i$ is an integer greater than 2, satisfies the inequality : \[n \cdot v_p (n!) \le v_p ((n^2 -1 )!).\]
Therefore, there are 16 values that don't work and $50-16 = \boxed{34}$ values that work.


Answer: 34.0

Solution: Notice all $15$ primes don't work as there are $n$ factors of $n$ in the denominator and $n-1$ factors of $n$ in the numerator. Further experimentation finds that $n=4$ does not work as there are 11 factors of 2 in the numerator and 12 in the denominator. We also find that it seems that all other values of $n$ work. So we get $50-15-1=\boxed{34}$ and that happens to be right.


Answer: 9.0

For how many integers $n$ between 1 and 100 does $x^2+x-n$ factor into the product of two linear factors with integer coefficients?
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$

Solution:
For $x^2+x-n$ to factor into a product of two linear factors, we must have $x^2+x-n = (x + a)(x + b)$ , where $a$ and $b$ are integers.
By expansion of the product of the linear factors and comparison to the original quadratic,
$ab = -n$
$a + b = 1$
The only way for this to work if $n$ is a positive integer is if $a = -b +1$
Here are the possible pairs:
This gives us 9 integers for $n$ $\boxed{9}$

Solution:
For $x^2+x-n$ to factor into a product of two linear factors, the discriminant must be a perfect square. The discriminant equals $\sqrt{1^2-4(1)(-n)}$ or $\sqrt{4n+1}$
Since $n$ must be a positive integer, $4n+1$ must be odd because if it were even, $4n$ would have to be both odd and divisible by 4, which is a contradiction. Therefore, $n$ must be even.
The maximum value of $4n+1$ is 401, and the minimum is 5. Therefore, we must find the number of odd perfect squares between 5 and 401.
There are 9: $9,25,49,81,121,169,225,289,$ and $361$ . Therefore, the answer is $\boxed{9}$


Answer: 901.0

For how many integers $n$ does the expression \[\sqrt{\frac{\log (n^2) - (\log n)^2}{\log n - 3}}\] represent a real number, where log denotes the base $10$ logarithm?
$\textbf{(A) }900 \qquad \textbf{(B) }2\qquad \textbf{(C) }902 \qquad \textbf{(D) } 2  \qquad \textbf{(E) }901$

Solution:
We have \begin{align*} \sqrt{\frac{\log \left( n^2 \right) - \left( \log n \right)^2}{\log n - 3}} & = \sqrt{\frac{2 \log n - \left( \log n \right)^2}{\log n - 3}} \\ & = \sqrt{\frac{\left( \log n \right) \left( 2 - \log n\right)}{\log n - 3}} . \end{align*}
Because $n$ is an integer and $\log n$ is well defined, $n$ must be a positive integer.
Case 1: $n = 1$ or $10^2$
The above expression is 0. So these are valid solutions.
Case 2: $n \neq 1, 10^2$
Thus, $\log n > 0$ and $2 - \log n \neq 0$ .
To make the above expression real, we must have $2 < \log n < 3$ .
Thus, $100 < n < 1000$ .
Thus, $101 \leq n \leq 999$ .
Hence, the number of solutions in this case is 899.
Putting all cases together, the total number of solutions is $\boxed{901}$

Solution:
Notice $\log(n^2)$ can be written as $2\log(n)$ . Setting $a=\log(n)$ , the equation becomes $\sqrt{\frac{2a-a^2}{a-3}}$ which can be written as $\sqrt{\frac{a(2-a)}{a-3}}$
Case 1: $a \ge 3$ The expression is undefined when $a=3$ . For $a>3$ , it is trivial to see that the denominator is positive and the numerator is negative, thus resulting in no real solutions.
Case 2: $2 \le a<3$ For $a=2$ , the numerator is zero, giving us a valid solution. When $a>2$ , both the denominator and numerator are negative so all real values of a in this interval is a solution to the equation. All integers of n that makes this true are between $10^2$ and $10^3-1$ . There are 900 solutions here.
Case 3: $0<a<2$ The numerator will be positive but the denominator is negative, no real solutions exist.
Case 4: $a=0$ The expression evaluates to zero, $1$ valid solution exists.
Case 5: $a<0$ All values for $a<0$ requires $0<n<1$ , no integer solutions exist.
Adding up the cases: $900+1=\boxed{901}$


Answer: 4.0

For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$

Solution: Let $x^2 = \frac{n}{20-n}$ , with $x \ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$ ). Then rewriting, $n = \frac{20x^2}{x^2 + 1}$ . Since $\text{gcd}(x^2, x^2 + 1) = 1$ , it follows that $x^2 + 1$ divides $20$ . Listing the factors of $20$ , we find that $x = 0, 1, 2 , 3$ are the only $\boxed{4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ).

Solution:
For $n<0$ and $n>20$ the fraction is negative, for $n=20$ it is not defined, and for $n\in\{1,\dots,9\}$ it is between 0 and 1.
Thus we only need to examine $n=0$ and $n\in\{10,\dots,19\}$
For $n=0$ and $n=10$ we obviously get the squares $0$ and $1$ respectively.
For prime $n$ the fraction will not be an integer, as the denominator will not contain the prime in the numerator.
This leaves $n\in\{12,14,15,16,18\}$ , and a quick substitution shows that out of these only $n=16$ and $n=18$ yield a square. Therefore, there are only $\boxed{4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ).

Solution: If $\frac{n}{20-n} = k^2 \ge 0$ , then $n \ge 0$ and $20-n > 0$ , otherwise $\frac{n}{20-n}$ will be negative. Thus $0 \le n \le 19$ and \[0 = \frac{0}{20-(0)} \le \frac{n}{20-n} \le \frac{19}{20-(19)} = 19\] Checking all $k$ for which $0 \le k^2 \le 19$ , we have $0$ $1$ $2$ $3$ as the possibilities. $\boxed{4}$

Solution: For all integers x, $x^2$ is always a positive integer. So solve for $\frac{n}{20-n} = 0$ , getting $n=0$ and $\frac{n}{20-n} = 1$ , getting $n =10$ . For all values n less than 0 and greater than 20, the value $\frac{n}{20-n}$ is negative, so now try values of n between 10 and 20. Quick substitution finds $0$ $10$ $16$ , and $18$ which yields $x=0$ $x=1$ $x=2$ , and $x=3$ respectively. 4 values, or $\boxed{4}$

Solution:
Simon's Favourite Factoring Trick
Since \frac{n}{20-n} is an integer k, w multiple both sides by 20-n. This gives us $n$ $20k^2$ $nk^2$ . We subtract $20k^2$ on both sides, then add $nk^2$ on both sides as a prerequisite for using Simon's Favorite Factoring Trick. We have (k^2+1)(n-20)=20. We then consider the different factors of 20 that k^2+1 can be. It could be $1$ $2$ $4$ $5$ $10$ , and $20$ . After checking case by case, we then are able to identify that there are 4 such k values that also yield an integer n value, meaning that there are 4 values, so the correct answer is $\boxed{4}$ ~CharmaineMa07292010


Answer: 4.0

Solution: Let $x^2 = \frac{n}{20-n}$ , with $x \ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$ ). Then rewriting, $n = \frac{20x^2}{x^2 + 1}$ . Since $\text{gcd}(x^2, x^2 + 1) = 1$ , it follows that $x^2 + 1$ divides $20$ . Listing the factors of $20$ , we find that $x = 0, 1, 2 , 3$ are the only $\boxed{4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ).

Solution:
For $n<0$ and $n>20$ the fraction is negative, for $n=20$ it is not defined, and for $n\in\{1,\dots,9\}$ it is between 0 and 1.
Thus we only need to examine $n=0$ and $n\in\{10,\dots,19\}$
For $n=0$ and $n=10$ we obviously get the squares $0$ and $1$ respectively.
For prime $n$ the fraction will not be an integer, as the denominator will not contain the prime in the numerator.
This leaves $n\in\{12,14,15,16,18\}$ , and a quick substitution shows that out of these only $n=16$ and $n=18$ yield a square. Therefore, there are only $\boxed{4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ).

Solution: If $\frac{n}{20-n} = k^2 \ge 0$ , then $n \ge 0$ and $20-n > 0$ , otherwise $\frac{n}{20-n}$ will be negative. Thus $0 \le n \le 19$ and \[0 = \frac{0}{20-(0)} \le \frac{n}{20-n} \le \frac{19}{20-(19)} = 19\] Checking all $k$ for which $0 \le k^2 \le 19$ , we have $0$ $1$ $2$ $3$ as the possibilities. $\boxed{4}$

Solution: For all integers x, $x^2$ is always a positive integer. So solve for $\frac{n}{20-n} = 0$ , getting $n=0$ and $\frac{n}{20-n} = 1$ , getting $n =10$ . For all values n less than 0 and greater than 20, the value $\frac{n}{20-n}$ is negative, so now try values of n between 10 and 20. Quick substitution finds $0$ $10$ $16$ , and $18$ which yields $x=0$ $x=1$ $x=2$ , and $x=3$ respectively. 4 values, or $\boxed{4}$

Solution:
Simon's Favourite Factoring Trick
Since \frac{n}{20-n} is an integer k, w multiple both sides by 20-n. This gives us $n$ $20k^2$ $nk^2$ . We subtract $20k^2$ on both sides, then add $nk^2$ on both sides as a prerequisite for using Simon's Favorite Factoring Trick. We have (k^2+1)(n-20)=20. We then consider the different factors of 20 that k^2+1 can be. It could be $1$ $2$ $4$ $5$ $10$ , and $20$ . After checking case by case, we then are able to identify that there are 4 such k values that also yield an integer n value, meaning that there are 4 values, so the correct answer is $\boxed{4}$ ~CharmaineMa07292010


Answer: 4.0

For how many integers $x$ does a triangle with side lengths $10, 24$ and $x$ have all its angles acute?
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{more than } 7$

Solution:
We first notice that the sides $10$ and $24$ , can be part of $2$ different right triangles, one with sides $10,24,26$ , and the other with a leg
somewhere between $21$ and $22$ . We now notice that if $x$ is less than or equal to $21$ , one of the angles is obtuse, and that the same is the same for
any value of $x$ above $26$ . Thus the only integer values of $x$ that fit the conditions, are $x=22, 23, 24, \text{and }25.$ So, the answer is $\boxed{4}$


Answer: 12.0

For how many integers $x$ is the number $x^4-51x^2+50$ negative?
$\textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16$

Solution:
First, note that $50+1=51$ , which motivates us to factor the polynomial as $(x^2-50)(x^2-1)$ . Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so $x^2-50<0<x^2-1$ . Solving this inequality, we find $1<x^2<50$ . There are  exactly $12$ integers $x$ that satisfy this inequality, $\pm \{2,3,4,5,6,7\}$
Thus our answer is $\boxed{12}$

Solution: Since the $x^4-51x^2$ part of $x^4-51x^2+50$ has to be less than $-50$ (because we want $x^4-51x^2+50$ to be negative), we have the inequality $x^4-51x^2<-50 \rightarrow x^2(x^2-51) <-50$ $x^2$ has to be positive, so $(x^2-51)$ is negative. Then we have $x^2<51$ . We know that if we find a positive number that works, it's parallel negative will work. Therefore, we just have to find how many positive numbers work, then multiply that by $2$ . If we try $1$ , we get $1^4-51(1)^4+50 = -50+50 = 0$ , and $0$ therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above $2$ that satisfy $x^2<51$ work, that is the set { ${2,3,4,5,6,7}$ }. That equates to $6$ numbers. Since each numbers' negative counterparts work, $6\cdot2=\boxed{12}$

Solution:
Let $x^{2}=u$ . Then the expression becomes $u^{2}-51u+50$ which can be factored as $\left(u-1\right)\left(u-50\right)$ . Since the expression is negative, one of $\left(u-1\right)$ and $\left(u-50\right)$ need to be negative. $u-1>u-50$ , so $u-1>0$ and $u-50<0$ , which yields the inequality $50>u>1$ . Remember, since $u=x^{2}$ where $x$ is an integer, this means that $u$ is a perfect square between $1$ and $50$ . There are $6$ values of $u$ that satisfy this constraint, namely $4$ $9$ $16$ $25$ $36$ , and $49$ . Solving each of these values for $x$ yields $12$ values (as $x$ can be negative or positive) $\Longrightarrow \boxed{12}$ .
~JH. L


Answer: 59.0

For how many integral values of $x$ can a triangle of positive area be formed having side lengths $\log_{2} x, \log_{4} x, 3$
$\textbf{(A) } 57 \qquad\textbf{(B) } 59 \qquad\textbf{(C) } 61 \qquad\textbf{(D) } 62 \qquad\textbf{(E) } 63$

Solution:
For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need \[\log_2{x} + \log_4{x} > 3\] \[\log_2{x} + 3 > \log_4{x}\] \[\log_4{x} + 3 > \log_2{x}.\] The second inequality is redundant, as it's always less restrictive than the last inequality.
Let's raise the first inequality to the power of $4$ . This gives $4^{\log_2{x}} \cdot 4^{\log_4{x}} > 64 \Rightarrow \left(2^2\right)^{\log_2{x}} \cdot x > 64 \Rightarrow x^2 \cdot x > 64$ . Thus, $x > 4$
Doing the same for the second inequality gives $4^{\log_4{x}} \cdot 64 > 4^{\log_2{x}} \Rightarrow 64x > x^2 \Rightarrow x < 64$ (where we are allowed to divide both sides by $x$ since $x$ must be positive in order for the logarithms given in the problem statement to even have real values).
Combining our results, $x$ is an integer strictly between $4$ and $64$ , so the number of possible values of $x$ is $64 - 4 - 1 = \boxed{59}$

Solution: Using the triangle inequality, you get $\log_2{x}+\log_4{x} > 3$ . Solving for $x$ , you get $x > 4$ . Now we need an upper-bound for $x$ and since we're dealing with bases of $2$ and $4$ , we're looking for answer choices close to a power of $2$ and $4$ . All the answer choices seem to be around $64$ , and plugging that into the inequality $3+\log_4{x} > \log_2{x}$ we see $64$ is the correct number. Now we have $64 > x > 4$ and the number of integers in between is $64-4-1 = \boxed{59}$ --OGBooger


Answer: 3.0

For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

Solution: This question is simply asking how many of the listed quadrilaterals are cyclic (since the point equidistant from all four vertices would be the center of the circumscribed circle). A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer is $\boxed{3}$

Solution: We can use a process of elimination. Going down the list, we can see a square obviously works. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point that is equidistant, but isosceles trapezoids do have such a point, so the answer is $\boxed{3}$

Solution: The perpendicular bisector of a line segment is the locus of all points that are equidistant from the endpoints. The question then boils down to finding the shapes where the perpendicular bisectors of the sides all intersect at a point. This is true for a square, rectangle, and isosceles trapezoid, so the answer is $\boxed{3}$


Answer: 5.0

For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 4$

Solution:
Denote three roots as $r_1 < r_2 < r_3$ .
Following from Vieta's formula, $r_1r_2r_3 = -6$
Case 1: All roots are negative.
We have the following solution: $\left( -3, -2, -1 \right)$
Case 2: One root is negative and two roots are positive.
We have the following solutions: $\left( -3, 1, 2 \right)$ $\left( -2, 1, 3 \right)$ $\left( -1, 2, 3 \right)$ $\left( -1, 1, 6 \right)$
Putting all cases together, the total number of solutions is $\boxed{5}$


Answer: 6.0

For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$

Solution:
A quadratic equation does not have two distinct real solutions if and only if the discriminant is nonpositive. We conclude that:
Squaring the first inequality, we get $b^4\leq 16c^2.$ Multiplying the second inequality by $16,$ we get $16c^2\leq 64b.$ Combining these results, we get \[b^4\leq 16c^2\leq 64b.\] We apply casework to the value of $b:$
Together, there are $\boxed{6}$ ordered pairs $(b,c),$ namely $(1,1),(1,2),(2,1),(2,2),(3,3),$ and $(4,4).$

Solution:
Similar to Solution 1, use the discriminant to get $b^2\leq 4c$ and $c^2\leq 4b$ . These can be rearranged to $c\geq \frac{1}{4}b^2$ and $b\geq \frac{1}{4}c^2$ . Now, we can roughly graph these two inequalities, letting one of them be the $x$ axis and the other be $y$ .
The graph of solutions should be above the parabola and under its inverse, meaning we want points on the graph or in the first area enclosed by the two graphs:  We are looking for lattice points (since $b$ and $c$ are positive integers), of which we can count $\boxed{6}$

Solution:
We need to solve the following system of inequalities: \[ \left\{ \begin{array}{ll} b^2 - 4 c \leq 0 \\ c^2 - 4 b \leq 0 \end{array} \right.. \] Feasible solutions are in the region formed between two parabolas $b^2 - 4 c = 0$ and $c^2 - 4 b = 0$
Define $f \left( b \right) = \frac{b^2}{4}$ and $g \left( b \right) = 2 \sqrt{b}$ .
Therefore, all feasible solutions are in the region formed between the graphs of these two functions.
For $b = 1$ , we have $f(b) = \frac{1}{4}$ and $g(b) = 2$ .
Hence, the feasible $c$ are $1, 2$
For $b = 2$ , we have $f(b) = 1$ and $g(b) = 2 \sqrt{2}$ .
Hence, the feasible $c$ are $1, 2$
For $b = 3$ , we have $f(b) = \frac{9}{4}$ and $g(b) = 2 \sqrt{3}$ .
Hence, the feasible $c$ is $3$
For $b = 4$ , we have $f(b) = 4$ and $g(b) = 4$ .
Hence, the feasible $c$ is $4$
For $b > 4$ , we have $f(b) > g(b)$ . Hence, there is no feasible $c$
Putting all cases together, the correct answer is $\boxed{6}$

Solution:
A quadratic equation $Ax^2+Bx+C=0$ has one real solution if and only if $\sqrt{B^2-4AC}=0.$ Similarly, it has imaginary solutions if and only if $\sqrt{B^2-4AC}<0.$ We proceed as following:
We want both $x^2+bx+c$ to be $1$ value or imaginary and $x^2+cx+b$ to be $1$ value or imaginary. $x^2+4x+4$ is one such case since $\sqrt {b^2-4ac}$ is $0.$ Also, $x^2+3x+3, x^2+2x+2, x^2+x+1$ are always imaginary for both $b$ and $c.$ We also have $x^2+x+2$ along with $x^2+2x+1$ since the latter has one solution, while the first one is imaginary. Therefore, we have $\boxed{6}$ total ordered pairs of integers.

Solution: We see that $b^2 \leq 4c$ and $c^2 \leq 4b.$ WLOG, assume that $b \geq c.$ Then we have that $b^2 \leq 4c \leq 4b$ , so $b^2 \leq 4b$ and therefore $b \leq 4$ , also meaning that $c \leq 4.$ This means that we only need to try 16 cases. Now we can get rid of the assumption that $b \geq c$ , because we want ordered pairs. For $b = 1$ and $b = 2$ $c = 1$ and $c = 2$ work. When $b = 3$ $c$ can only be $3$ , and when $b = 4$ , only $c = 4$ works, for a total of $\boxed{6}$ ordered pairs of integers.

Solution:
We need both $b^2\leq 4c$ and $c^2\leq 4b$
If $b=c$ then the above become $b^2\leq 4b\iff b\leq 4$ , so we have four solutions $(k,k)$ , where $k=1$ $2$ $3$ $4$
If $b<c$ then we only need $c^2\leq 4b$ since it implies $b^2< 4c$ . Now $c^2\leq 4b\leq 4(c-1) \implies (c-2)^2\leq 0 \implies c=2$ , so $b=1$ . We plug $b=1$ $c=2$ back into $c^2\leq 4b$ and it works. So there is another solution $(1,2)$
By symmetry, if $b>c$ then $(b,c)=(2,1)$
Therefore the total number of solutions is $\boxed{6}$

Solution: Since $b^{2} - 4c \le 0$ and $c^{2} - 4b \le 0$ , adding the two together yields $b^{2} + c^{2} \le 4(c+b)$ . Obviously, this is not true if either $b$ or $c$ get too large, and they are equal when $b = c = 4$ , so the greatest pair is $(4,4)$ and both numbers must be lesser for further pairs. For there to be two distinct real solutions, we can test all these pairs where $(b,c)$ are less than 4 (except for the already valid solution) on the original quadratics, and we find the working pairs are $(1,1)$ $(2,1)$ $(2,1)$ $(2,2)$ $(3,3)$ $(4,4)$ meaning there are $\boxed{6}$ pairs.


Answer: 6.0

Solution:
A quadratic equation does not have two distinct real solutions if and only if the discriminant is nonpositive. We conclude that:
Squaring the first inequality, we get $b^4\leq 16c^2.$ Multiplying the second inequality by $16,$ we get $16c^2\leq 64b.$ Combining these results, we get \[b^4\leq 16c^2\leq 64b.\] We apply casework to the value of $b:$
Together, there are $\boxed{6}$ ordered pairs $(b,c),$ namely $(1,1),(1,2),(2,1),(2,2),(3,3),$ and $(4,4).$

Solution:
Similar to Solution 1, use the discriminant to get $b^2\leq 4c$ and $c^2\leq 4b$ . These can be rearranged to $c\geq \frac{1}{4}b^2$ and $b\geq \frac{1}{4}c^2$ . Now, we can roughly graph these two inequalities, letting one of them be the $x$ axis and the other be $y$ .
The graph of solutions should be above the parabola and under its inverse, meaning we want points on the graph or in the first area enclosed by the two graphs:  We are looking for lattice points (since $b$ and $c$ are positive integers), of which we can count $\boxed{6}$

Solution:
We need to solve the following system of inequalities: \[ \left\{ \begin{array}{ll} b^2 - 4 c \leq 0 \\ c^2 - 4 b \leq 0 \end{array} \right.. \] Feasible solutions are in the region formed between two parabolas $b^2 - 4 c = 0$ and $c^2 - 4 b = 0$
Define $f \left( b \right) = \frac{b^2}{4}$ and $g \left( b \right) = 2 \sqrt{b}$ .
Therefore, all feasible solutions are in the region formed between the graphs of these two functions.
For $b = 1$ , we have $f(b) = \frac{1}{4}$ and $g(b) = 2$ .
Hence, the feasible $c$ are $1, 2$
For $b = 2$ , we have $f(b) = 1$ and $g(b) = 2 \sqrt{2}$ .
Hence, the feasible $c$ are $1, 2$
For $b = 3$ , we have $f(b) = \frac{9}{4}$ and $g(b) = 2 \sqrt{3}$ .
Hence, the feasible $c$ is $3$
For $b = 4$ , we have $f(b) = 4$ and $g(b) = 4$ .
Hence, the feasible $c$ is $4$
For $b > 4$ , we have $f(b) > g(b)$ . Hence, there is no feasible $c$
Putting all cases together, the correct answer is $\boxed{6}$

Solution:
A quadratic equation $Ax^2+Bx+C=0$ has one real solution if and only if $\sqrt{B^2-4AC}=0.$ Similarly, it has imaginary solutions if and only if $\sqrt{B^2-4AC}<0.$ We proceed as following:
We want both $x^2+bx+c$ to be $1$ value or imaginary and $x^2+cx+b$ to be $1$ value or imaginary. $x^2+4x+4$ is one such case since $\sqrt {b^2-4ac}$ is $0.$ Also, $x^2+3x+3, x^2+2x+2, x^2+x+1$ are always imaginary for both $b$ and $c.$ We also have $x^2+x+2$ along with $x^2+2x+1$ since the latter has one solution, while the first one is imaginary. Therefore, we have $\boxed{6}$ total ordered pairs of integers.

Solution: We see that $b^2 \leq 4c$ and $c^2 \leq 4b.$ WLOG, assume that $b \geq c.$ Then we have that $b^2 \leq 4c \leq 4b$ , so $b^2 \leq 4b$ and therefore $b \leq 4$ , also meaning that $c \leq 4.$ This means that we only need to try 16 cases. Now we can get rid of the assumption that $b \geq c$ , because we want ordered pairs. For $b = 1$ and $b = 2$ $c = 1$ and $c = 2$ work. When $b = 3$ $c$ can only be $3$ , and when $b = 4$ , only $c = 4$ works, for a total of $\boxed{6}$ ordered pairs of integers.

Solution:
We need both $b^2\leq 4c$ and $c^2\leq 4b$
If $b=c$ then the above become $b^2\leq 4b\iff b\leq 4$ , so we have four solutions $(k,k)$ , where $k=1$ $2$ $3$ $4$
If $b<c$ then we only need $c^2\leq 4b$ since it implies $b^2< 4c$ . Now $c^2\leq 4b\leq 4(c-1) \implies (c-2)^2\leq 0 \implies c=2$ , so $b=1$ . We plug $b=1$ $c=2$ back into $c^2\leq 4b$ and it works. So there is another solution $(1,2)$
By symmetry, if $b>c$ then $(b,c)=(2,1)$
Therefore the total number of solutions is $\boxed{6}$

Solution: Since $b^{2} - 4c \le 0$ and $c^{2} - 4b \le 0$ , adding the two together yields $b^{2} + c^{2} \le 4(c+b)$ . Obviously, this is not true if either $b$ or $c$ get too large, and they are equal when $b = c = 4$ , so the greatest pair is $(4,4)$ and both numbers must be lesser for further pairs. For there to be two distinct real solutions, we can test all these pairs where $(b,c)$ are less than 4 (except for the already valid solution) on the original quadratics, and we find the working pairs are $(1,1)$ $(2,1)$ $(2,1)$ $(2,2)$ $(3,3)$ $(4,4)$ meaning there are $\boxed{6}$ pairs.


Answer: 997.0

For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$

Solution:
\begin{eqnarray*} \frac{x+y}{2} &=& \sqrt{xy} + 2\\ x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\ \sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}
Because $y > x$ , we only consider $+2$
For simplicity, we can count how many valid pairs of $(\sqrt{x},\sqrt{y})$ that satisfy our equation.
The maximum that $\sqrt{y}$ can be is $\sqrt{10^6} - 1 = 999$ because $\sqrt{y}$ must be an integer (this is because $\sqrt{y} - \sqrt{x} = 2$ , an integer). Then $\sqrt{x} = 997$ , and we continue this downward until $\sqrt{y} = 3$ , in which case $\sqrt{x} = 1$ . The number of pairs of $(\sqrt{x},\sqrt{y})$ , and so $(x,y)$ is then $\boxed{997}$

Solution:
Let $a^2$ $x$ and $b^2$ $y$ , where $a$ and $b$ are positive.
Then \[\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2\] \[a^2 + b^2 = 2ab + 4\] \[(a-b)^2 = 4\] \[(a-b) = \pm 2\]
This makes counting a lot easier since now we just have to find all pairs $(a,b)$ that differ by 2.
Because $\sqrt{10^6} = 10^3$ , then we can use all positive integers less than 1000 for $a$ and $b$
We know that because $x < y$ , we get $a < b$
We can count even and odd pairs separately to make things easier*:
Odd: \[(1,3) , (3,5) , (5,7)  .  .  .  (997,999)\]
Even: \[(2,4) , (4,6) , (6,8)  .  .  .  (996,998)\]
This makes 499 odd pairs and 498 even pairs, for a total of $\boxed{997}$ pairs.

Solution:
Since the arithmetic mean is 2 more than the geometric mean, $\frac{x+y}{2} = 2 + \sqrt{xy}$ . We can multiply by 2 to get $x + y = 4 + 2\sqrt{xy}$ . Subtracting 4 and squaring gives \[((x+y)-4)^2 = 4xy\] \[((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy\] \[x^2 - 2xy + y^2 + 16 - 8x - 8y = 0\]
Notice that $((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y$ , so the problem asks for solutions of \[(x-y-4)^2 = 16y\] Since the left hand side is a perfect square, and 16 is a perfect square, $y$ must also be a perfect square. Since $0 < y < (1000)^2$ $y$ must be from $1^2$ to $999^2$ , giving at most 999 options for $y$
However if $y = 1^2$ , you get $(x-5)^2 = 16$ , which has solutions $x = 9$ and $x = 1$ . Both of those solutions are not less than $y$ , so $y$ cannot be equal to 1. If $y = 2^2 = 4$ , you get $(x - 8)^2 = 64$ , which has 2 solutions, $x = 16$ , and $x = 0$ . 16 is not less than 4, and $x$ cannot be 0, so $y$ cannot be 4. However, for all other $y$ , you get exactly 1 solution for $x$ , and that gives a total of $999 - 2 = \boxed{997}$ pairs.

Solution:
Rearranging our conditions to
\[x^2-2xy+y^2+16-8x-8y=0 \implies\] \[(y-x)^2=8(x+y-2).\]
Thus, $4|y-x.$
Now, let $y = 4k+x.$ Plugging this back into our expression, we get
\[(k-1)^2=x-1.\]
There, a unique value of $x, y$ is formed for every value of $k$ . However, we must have
\[y<10^6 \implies (k+1)^2< 10^6-1\]
and
\[x=(k-1)^2+1>0.\]
Therefore, there are only $\boxed{997}$ pairs of $(x,y).$


Answer: 49.0

For how many ordered pairs of positive integers $(x,y)$ is $x + 2y = 100$
$\text {(A)} 33 \qquad \text {(B)} 49 \qquad \text {(C)} 50 \qquad \text {(D)} 99 \qquad \text {(E)}100$

Solution: If $x$ and $2y$ must each be positive integers, then we can say that $x$ is at least 1 and $2y$ is at least 1. From there, we want to find out how many ways there are to distribute the other 98 ones (the smallest positive integer addends of 100). 98 identical objects can be distributed to two distinct bins in 99 ways (think stars and bars), yet this 99 is an overcount. Because $y$ must be an integer, $2y$ must be even; thus only $\left\lfloor \frac{99}{2} \right\rfloor = \boxed{49}$ ways exist to distribute these ones.


Answer: 85.0

For how many ordered pairs of positive integers $(x,y),$ with $y<x\le 100,$ are both $\frac xy$ and $\frac{x+1}{y+1}$ integers?

Solution:
Since $y|x$ $y+1|x+1$ , then $\text{gcd}\,(y,x)=y$ (the bars indicate divisibility ) and $\text{gcd}\,(y+1,x+1)=y+1$ . By the Euclidean algorithm , these can be rewritten respectively as $\text{gcd}\,(y,x-y)=y$ and $\text{gcd}\, (y+1,x-y)=y+1$ , which implies that both $y,y+1 | x-y$ . Also, as $\text{gcd}\,(y,y+1) = 1$ , it follows that $y(y+1)|x-y$
Thus, for a given value of $y$ , we need the number of multiples of $y(y+1)$ from $0$ to $100-y$ (as $x \le 100$ ). It follows that there are $\left\lfloor\frac{100-y}{y(y+1)} \right\rfloor$ satisfactory positive integers for all integers $y \le 100$ . The answer is
\[\sum_{y=1}^{99} \left\lfloor\frac{100-y}{y(y+1)} \right\rfloor = 49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.\]

Solution:
We know that $x \equiv 0 \mod y$ and $x+1 \equiv 0 \mod y+1$
Write $x$ as $ky$ for some integer $k$ . Then, $ky+1 \equiv 0\mod y+1$ . We can add $k$ to each side in order to factor out a $y+1$ . So, $ky+k+1 \equiv k \mod y+1$ or $k(y+1)+1 \equiv k \mod y+1$ . We know that $k(y+1) \equiv 0 \mod y+1$ . We finally achieve the congruence $1-k \equiv 0 \mod y+1$
We can now write $k$ as $(y+1)a+1$ . Plugging this back in, if we have a value for $y$ , then $x = ky = ((y+1)a+1)y = y(y+1)a+y$ . We only have to check values of $y$ when $y(y+1)<100$ . This yields the equations $x = 2a+1, 6a+2, 12a+3, 20a+4, 30a+5, 42a+6, 56a+7, 72a+8, 90a+9$
Finding all possible values of $a$ such that $y<x<100$ , we get $49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.$

Solution:
We use casework:
For $y=1$ , we have $3,5,\cdots ,99$ , or $49$ cases.
For $y=2$ , we have $8,14,\cdots ,98$ , or $16$ cases.
For $y=3$ , we have $15,27,\cdots ,99$ , or $8$ cases.
For $y=4$ , we have $24,44\cdots ,84$ , or $4$ cases.
For $y=5$ , we have $35,65,95$ , or $3$ cases.
For $y=6$ , we have $48,90$ , or $2$ cases.
For $y=7$ , we have $63$ , or $1$ case.
For $y=8$ , we have $80$ , or $1$ case.
For $y=9$ , we have $99$ , or $1$ case.
Adding, we get our final result of $\boxed{085}.$


Answer: 225.0

For how many ordered triples $(x,y,z)$ of nonnegative integers less than $20$ are there exactly two distinct elements in the set $\{i^x, (1+i)^y, z\}$ , where $i=\sqrt{-1}$
$\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235$

Solution:
We have either $i^{x}=(1+i)^{y}\neq z$ $i^{x}=z\neq(1+i)^{y}$ , or $(1+i)^{y}=z\neq i^x$
$i^{x}=(1+i)^{y}$ only occurs when it is $1$ $(1+i)^{y}=1$ has only one solution, namely, $y=0$ $i^{x}=1$ has five solutions between 0 and 19, $x=0, x=4, x=8, x=12$ , and $x=16$ $z\neq 1$ has nineteen integer solutions between zero and nineteen. So for $i^{x}=(1+i)^{y}\neq z$ , we have $5\cdot 1\cdot 19=95$ ordered triples.
For $i^{x}=z\neq(1+i)^{y}$ , again this only occurs at $1$ $(1+i)^{y}\neq 1$ has nineteen solutions, $i^{x}=1$ has five solutions, and $z=1$ has one solution, so again we have $5\cdot 1\cdot 19=95$ ordered triples.
For $(1+i)^{y}=z\neq i^x$ , this occurs at $1$ and $16$ $(1+i)^{y}=1$ and $z=1$ both have one solution while $i^{x}\neq 1$ has fifteen solutions. $(1+i)^{y}=16$ and $z=16$ both have one solution, namely, $y=8$ and $z=16$ , while $i^{x}\neq 16$ has twenty solutions ( $i^x$ only cycles as $1, i, -1, -i$ ). So we have $15\cdot 1\cdot 1+20\cdot 1\cdot 1=35$ ordered triples.
In total we have ${95+95+35=\boxed{225}$ ordered triples


Answer: 156.0

For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added?

Solution:
For one such pair of consecutive integers, let the smaller integer be $\underline{1ABC},$ where $A,B,$ and $C$ are digits from $0$ through $9.$
We wish to count the ordered triples $(A,B,C).$ By casework, we consider all possible forms of the larger integer, as shown below. \[\begin{array}{c|c|c|c|c|c|c} & & & & & & \\ [-2.5ex] \textbf{Case} & \textbf{Thousands} & \textbf{Hundreds} & \textbf{Tens} & \textbf{Ones} & \textbf{Conditions for No Carrying} & \boldsymbol{\#}\textbf{ of Ordered Triples} \\ [0.5ex] \hline & & & & & & \\ [-2ex] 0\leq C\leq 8 & 1 & A & B & C+1 & 0\leq A,B,C\leq 4 & 5^3 \\ 0\leq B\leq 8; \ C=9 & 1 & A & B+1 & 0 & 0\leq A,B\leq 4; \ C=9 & 5^2 \\ 0\leq A\leq 8; \ B=C=9 & 1 & A+1 & 0 & 0 & 0\leq A\leq 4; \ B=C=9 & 5 \\ A=B=C=9 & 2 & 0 & 0 & 0 & A=B=C=9 & 1 \end{array}\] Together, the answer is $5^3+5^2+5+1=\boxed{156}.$

Solution:
Consider what carrying means: If carrying is needed to add two numbers with digits $abcd$ and $efgh$ , then $h+d\ge 10$ or $c+g\ge 10$ or $b+f\ge 10$ . 6.  Consider $c \in \{0, 1, 2, 3, 4\}$ $1abc + 1ab(c+1)$ has no carry if $a, b \in \{0, 1, 2, 3, 4\}$ . This gives $5^3=125$ possible solutions.
With $c \in \{5, 6, 7, 8\}$ , there obviously must be a carry. Consider $c = 9$ $a, b \in \{0, 1, 2, 3, 4\}$ have no carry. This gives $5^2=25$ possible solutions. Considering $b = 9$ $a \in \{0, 1, 2, 3, 4, 9\}$ have no carry. Thus, the solution is $125 + 25 + 6=\boxed{156}$

Solution:
Consider the ordered pair $(1abc , 1abc - 1)$ where $a,b$ and $c$ are digits. We are trying to find all ordered pairs where $(1abc) + (1abc - 1)$ does not require carrying. For the addition to require no carrying, $2a,2b < 10$ , so $a,b < 5$ unless $1abc$ ends in $00$ , which we will address later.  Clearly, if $c \in \{0, 1, 2, 3, 4 ,5\}$ , then adding $(1abc) + (1abc - 1)$ will require no carrying. We have $5$ possibilities for the value of $a$ $5$ for $b$ , and $6$ for $c$ , giving a total of $(5)(5)(6) = 150$ , but we are not done yet.
We now have to consider the cases where $b,c = 0$ , specifically when $1abc \in \{1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000\}$ . We can see that $1100, 1200, 1300, 1400, 1500$ , and $2000$ all work, giving a grand total of $150 + 6 = \boxed{156}$ ordered pairs.


Answer: 7.0

For how many positive integer values of $N$ is the expression $\dfrac{36}{N+2}$ an integer?
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

Solution: We should list all the positive divisors of $36$ and count them. By trial and error, the divisors of $36$ are found to be $1,2,3,4,6,9,12,18,36$ , for a total of $9$ . However, $1$ and $2$ can't be equal to $N+2$ for a POSITIVE integer N, so the number of possibilities is $\boxed{7}$

Solution: To find the number of positive divisors of $36$ , first prime factorize $36$ to get $2^2*3^2$ . Then add $1$ to the power of both $2$ and $3$ to get $3$ . Multiply $3*3$ to get $9$ . Since the problem is asking only for positive integer values of N, subtract $2$ from $9$ (since $2-2$ and $1-2$ result in integers that are not positive) to get $\boxed{7}$


Answer: 12.0

For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 27\qquad \textbf{(D)}\ 33\qquad \textbf{(E)}\ 34$

Solution: Instead of finding n, we find $x=\frac{n}{3}$ . We want $x$ and $9x$ to be three-digit whole numbers. The smallest three-digit whole number is $100$ , so that is our minimum value for $x$ , since if $x \in \mathbb{Z^+}$ , then $9x \in \mathbb{Z^+}$ . The largest three-digit whole number divisible by $9$ is $999$ , so our maximum value for $x$ is $\frac{999}{9}=111$ . There are $12$ whole numbers in the closed set $\left[100,111\right]$ , so the answer is $\boxed{12}$

Solution:
We can set the following inequalities up to satisfy the conditions given by the question, $100 \leq \frac{n}{3} \leq 999$ ,
and $100 \leq 3n \leq 999$ .
Once we simplify these and combine the restrictions, we get the inequality, $300 \leq n \leq 333$ .
Now we have to find all multiples of 3 in this range for $\frac{n}{3}$ to be an integer. We can compute this by setting $\frac{n} {3}=x$ , where $x \in \mathbb{Z^+}$ . Substituting $x$ for $n$ in the previous inequality, we get, $100 \leq x \leq 111$ , and there are $111-100+1$ integers in this range giving us the answer, $\boxed{12}$

Solution: So we know the largest $3$ digit number is $999$ and the lowest is $100$ . This means $\dfrac{n}{3} \ge 100 \rightarrow n \ge 300$ but $3n \le 999 \rightarrow n \le 333$ . So we have the set ${300, 301, 302, \cdots, 330, 331, 332, 333}$ for $n$ . Now we have to find the multiples of $3$ suitable for $n$ , or else $\dfrac{n}{3}$ will be a decimal. Only numbers ${300, 303, \cdots, 333}$ are counted. We can  divide by $3$ to make the difference $1$ again, getting ${100, 101 \cdots , 111}$ . Due to it being inclusive, we have $111-100+1 =\boxed{12}$


Answer: 22.0

For how many positive integers $n \le 1000$ is \[\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor\] not divisible by $3$ ? (Recall that $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ .)
$\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26$

Solution:
Clearly, $n=1$ fails. Except for the special case of $n=1$ \[\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor\] equals either $0$ or $1$ . If it equals $0$ , this implies that $\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor$ , so their sum is clearly a multiple of $3$ , so this will always fail. If it equals $1$ , the sum of the three floor terms is $3 \left\lfloor \frac{999}{n} \right\rfloor \pm 1$ , so it is never a multiple of $3$ . Thus, we are looking for all $n \neq 1$ such that \[\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor = 1.\] This implies that either \[\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor,\] or \[\left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor.\]
Let's analyze the first equation of these two. This equation is equivalent to the statement that there is a positive integer $a$ such that \[\frac{998}{n} < a \leq \frac{999}{n} \implies 998 < an \leq 999 \implies an = 999 \implies a = \frac{999}{n} \implies n | 999.*\] Analogously, the second equation implies that \[n | 1000.\] So our only $n$ that satisfy this condition are $n \neq 1$ that divide $999$ or $1000$ . Using the method to find the number of divisors of a number, we see that $999$ has $8$ divisors and $1000$ has $16$ divisors. Their only common factor is $1$ , so there are $8+16-1 = 23$ positive integers that divide either $999$ or $1000$ . Since the integer $1$ is a special case and does not count, we must subtract this from our $23$ , so our final answer is $23-1 = \boxed{22}.$

Solution:
Counting down $n$ from $1000$ $999$ $998$ ... we notice that $\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor$ is only not divisible by $3$ when n is a divisor of only $1000$ or $999 (1000, 999, 500, 333$ ...). Notice how the factors of $998: 1, 2, 499$ , and $998$ , do not work.
The prime factorization of $999$ is $3^3\cdot37$ , so $4\cdot2=8$ factors in total.
The prime factorization of $1000$ is $2^3\cdot5^3$ , so $4\cdot4=16$ factors in total.
However, $1$ obviously does not work, so we have to subtract $2$ $1$ is counted twice) from the total. $8 + 16 - 2$ $\boxed{22}$

Solution:
First, we notice the following lemma:
$\textbf{Lemma}$ : For $N, n \in \mathbb{N}$ $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1$ if $n \mid N$ ; and $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor$ if $n \nmid N.$
$\textbf{Proof}$ : Let $A = kn + r$ , with $0 \leq r < n$ . If $n \mid N$ , then $r = 0$ . Hence $\left\lfloor \frac{N}{n} \right\rfloor = k$ $\left\lfloor \frac{N-1}{n} \right\rfloor = \left\lfloor \frac{(k-1)n+n-1}{n} \right\rfloor = k-1 + \left\lfloor \frac{n-1}{n} \right\rfloor = k-1$ , and $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1.$
If $n \nmid N$ , then $1 \leq r < n$ . Hence $\left\lfloor \frac{N}{n} \right\rfloor = k$ $\left\lfloor \frac{N-1}{n} \right\rfloor = k + \left\lfloor \frac{r-1}{n} \right\rfloor = k$ , and $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor.$
From the lemma and the given equation, we have four possible cases: \[\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor - 1 \qquad (1)\] \[\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (2)\] \[\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (3)\] \[\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (4)\]
Note that cases (2) and (3) are the cases in which the term, $\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor,$ is not divisible by $3$ . So we only need to count the number of $n$ 's for which cases (2) and (3) stand.
Case (2): By the lemma, we have $n \mid 1000$ and $n \nmid 999.$ Hence $n$ can be any factor of $1000$ except for $n = 1$ . Since $1000 = 2^3 * 5^3,$ there are $(3+1)(3+1) - 1 = 15$ possible values of $n$ for this case.
Case (3): By the lemma, we have $n \mid 999$ and $n \nmid 998.$ Hence $n$ can be any factor of $999$ except for $n = 1$ . Since $999 = 3^3 * 37^1,$ there are $(3+1)(1+1) - 1 = 7$ possible values of $n$ for this case.
So in total, we have total of $15+7=\boxed{22}$ possible $n$ 's.

Solution:
Note that $\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor$ is a multiple of $3$ if $\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor$ lies between two consecutive multiples of $n$
$\textbf{Explanation:}$
Let's assume that the above expression does indeed lie betweent two consecutive multiples of $n$ . This implies that $n$ does not divide either $998, 999$ or $1000$ , meaning that when divided by $n$ , none of the quotients are whole. In turn, this also means that they will all have the same whole number part.
If $\frac{998}{n}, \frac{999}{n},$ and $\frac{1000}{n}$ were to have a different whole number part, then $n$ would have to lie within $998$ and $1000$ . If this is confusing, think of why $\frac{6}{5}, \frac{7}{5}$ and $\frac{8}{5}$ have the same whole number part ( $1$ in this case). Here, $n=5$ . What happens to the whole number part when $n=7$
Since $\frac{998}{n}, \frac{999}{n},$ and $\frac{1000}{n}$ have identical whole number parts, their floors are identical, since the floor of a number is equal to its whole number part, discarding its fractional component.
Let $k$ be the number we get when we floor $\frac{998}{n}, \frac{999}{n},$ or $\frac{1000}{n}$ if the three of those numbers lie between two consecutive multiples of $n$ . Adding them up, we get $3k$ (due to their floors being the same), which is a mulutiple of $3$ . So no matter what, we cannot have $\frac{998}{n}, \frac{999}{n},$ and $\frac{1000}{n}$ lie between two consecutive multiples of $n$
What does this mean? It means that there must be some multiple of $n$ within this expression (with some restrictions, as we'll see), in order to prevent the violation of the main restriction. We can proceed with casework now.
$\textbf{Case 1}$
Let's assume that the multiple of $n$ is located at $\frac{998}{n}$ . Luckily, the only prime factors of $998$ are $2$ and $499$
We can observe that $499$ doesn't work, since $\frac{998}{499}$ and $\frac{1000}{499}$ will both round down to $2$ when divided by it. However, $2$ does work, since it divides both $998$ and $1000$ . The floor of $\frac{999}{2}$ is $499$ , , meaning that when we evaluate the given expression for $n=2$ , we will get $2\cdot{499}+500$ which isn't a multiple of $3$
This case works because we have a multiple of $n$ at the end of the expression, as well as the beginning, so the whole number parts of $998, 999$ and $1000$ when divided by $n$ are not all the same, due to $n$ dividing two of these numbers.
The restriction of $n$ dividing more than one number within the expression is only valid when we're testing the first number in the given expression (otherwise, the other floors wouldn't round down to the same value as the first).
Case $1$ is complete, and we've found that only $2$ works.
$\textbf{Case 2}$
Now, let's assume that the multiple of $n$ is located at $\frac{999}{n}$ . In this case, if $n$ divides $999$ , it doesn't divide $998$ (since two multiples of a number greater than $1$ are never consecutive), nor does it divide $1000$ , for the same reason.
The prime factorization of $999$ is $3^3\cdot37$ , and thus has $(3+1)(1+1)=8$ divisors.
When testing a few values of $n$ initially, we observed that $n=1$ causes the expression to be divisible by $3$ . Subtracting $1$ , we see that there are $7$ values of $n$ that work for this case.
$\textbf{Case 3}$
Finally, let's assume that the multiple of $n$ is located at $\frac{1000}{n}.$
Our goal is to have the other multiple of $n$ below whatever $\frac{998}{n}$ is, so we won't have identical floors throughout the expression.
Once again, any factor of $1000$ , except for $2$ is relatively prime to both $998$ and $999$ , so when we floor those two numbers, we get an integer that isn't $500$
$1000$ factors as $2^3\cdot{5^3}$ , meaning that it has $(3+1)(3+1)=16$ divisors. $1$ and $2$ either don't work or have already been counted, so there are $14$ valid values of $n$ for this case.
$\textbf{Ending}$
Adding these three cases, we get $1+7+14= \boxed{22}$ values of $n$


Answer: 5.0

For how many positive integers $n$ does $1+2+...+n$ evenly divide from $6n$
$\textbf{(A) } 3\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 9\qquad \textbf{(E) } 11$

Solution:
If $1+2+...+n$ evenly divides $6n$ , then $\frac{6n}{1+2+...+n}$ is an integer
Since $1+2+...+n = \frac{n(n+1)}{2}$ we may substitute the RHS in the above fraction . So the problem asks us for how many positive integers $n$ is $\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}$ an integer, or equivalently when $k(n+1) = 12$ for a positive integer $k$
$\frac{12}{n+1}$ is an integer when $n+1$ is a factor of $12$
The factors of $12$ are $1, 2, 3, 4, 6,$ and $12$ , so the possible values of $n$ are $0, 1, 2, 3, 5,$ and $11$
But since $0$ isn't a positive integer, only $1, 2, 3, 5,$ and $11$ are the possible values of $n$ . Therefore the number of possible values of $n$ is $\boxed{5}$


Answer: 7.0

For how many positive integers $n$ is $\frac{n}{30-n}$ also a positive integer?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution: We know that $n \le 30$ or else $30-n$ will be negative, resulting in a negative fraction.  We also know that $n \ge 15$ or else the fraction's denominator will exceed its numerator making the fraction unable to equal a positive integer value.  Substituting all values $n$ from $15$ to $30$ gives us integer values for $n=15, 20, 24, 25, 27, 28, 29$ .  Counting them up, we have $\boxed{7}$ possible values for $n$

Solution: Let $\frac{n}{30-n}=m$ , where $m \in \mathbb{N}$ . Solving for $n$ , we find that $n=\frac{30m}{m+1}$ . Because $m$ and $m+1$ are relatively prime, $m+1|30$ . Our answer is the number of proper divisors of $2^13^15^1$ , which is $(1+1)(1+1)(1+1)-1 = \boxed{7}$


Answer: 3.0

For how many positive integers $n$ is $n^3 - 8n^2 + 20n - 13$ a prime number?
$\text{(A) 1} \qquad \text{(B) 2} \qquad \text{(C) 3} \qquad \text{(D) 4} \qquad \text{(E) more than 4}$

Solution:
Since this is a number theory question, it is clear that the main challenge here is factoring the given cubic. In general, the rational root theorem will be very useful for these situations.
The rational root theorem states that all rational roots of $n^3 - 8n^2 + 20n - 13$ will be among $1, 13, -1$ , and $-13$ . Evaluating the cubic at these values will give $n = 1$ as a root. Doing some synthetic division gives \[n^3 - 8n^2 + 20n - 13 = (n-1)(n^2 - 7n + 13)\]
Since $n > 0$ $n-1$ must be nonnegative. Since $(n-1)(n^2 - 7n + 13)$ evaluates to a prime, it is clear that exactly one of $n-1$ and $n^2 - 7n - 13$ is $1$ . We proceed by splitting the problem into 2 cases.
Case 1: $n-1 = 1$ It is clear that $n = 2$ .  We have $2^2 - 7(2) + 13 = 3$ , so this case yields $n = 2$ as a solution.
Case 2: $n^2 - 7n + 13 = 1$ Solving for $n$ gives $n^2 - 7n + 12 = 0$ or $(n-3)(n-4) = 0$ . Therefore, $n = 3$ or $n = 4$ .
Since both $3-1 = 2$ and $4-1 = 3$ are prime, both $n = 3$ and $n = 4$ work, yielding 2 solutions.
Putting everything together, the answer is $1 + 2 = \boxed{3}$


Answer: 250.0

For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$

Solution:
We know by De Moivre's Theorem that $(\cos t + i \sin t)^n = \cos nt + i \sin nt$ for all real numbers $t$ and all integers $n$ .  So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.
Recall the trigonometric identities $\cos \left(\frac{\pi}2 - u\right) = \sin u$ and $\sin \left(\frac{\pi}2 - u\right) = \cos u$ hold for all real $u$ .  If our original equation holds for all $t$ , it must certainly hold for $t = \frac{\pi}2 - u$ .  Thus, the question is equivalent to asking for how many positive integers $n \leq 1000$ we have that $\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)$ holds for all real $u$
$\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu$ .  We know that two complex numbers are equal if and only if both their real part and imaginary part are equal.  Thus, we need to find all $n$ such that $\cos n u = \sin n\left(\frac\pi2 - u\right)$ and $\sin nu = \cos n\left(\frac\pi2 - u\right)$ hold for all real $u$
$\sin x = \cos y$ if and only if either $x + y = \frac \pi 2 + 2\pi \cdot k$ or $x - y = \frac\pi2 + 2\pi\cdot k$ for some integer $k$ .  So from the equality of the real parts we need either $nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$ , in which case $n = 1 + 4k$ , or we need $-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$ , in which case $n$ will depend on $u$ and so the equation will not hold for all real values of $u$ .  Checking $n = 1 + 4k$ in the equation for the imaginary parts, we see that it works there as well, so exactly those values of $n$ congruent to $1 \pmod 4$ work.  There are $\boxed{250}$ of them in the given range.

Solution: This problem begs us to use the familiar identity $e^{it} = \cos(t) + i \sin(t)$ . Notice, $\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}$ since $\sin(-t) = -\sin(t)$ . Using this, $(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)$ is recast as $(i e^{-it})^n = i e^{-itn}$ . Hence we must have $i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}$ . Thus since $1000$ is a multiple of $4$ exactly one quarter of the residues are congruent to $1$ hence we have $\boxed{250}$

Solution:
We can rewrite $\sin(t)$ as $\cos\left(\frac{\pi}{2}-t\right)$ and $\cos(t)$ as $\sin\left(\frac{\pi}{2}-t\right)$ . This means that $\sin t + i\cos t = e^{i\left(\frac{\pi}{2}-t\right)}=\frac{e^{\frac{\pi i}{2}}}{e^{it}}$ . This theorem also tells us that $e^{\frac{\pi i}{2}}=i$ , so $\sin t + i\cos t = \frac{i}{e^{it}}$ . By the same line of reasoning, we have $\sin nt + i\cos nt = \frac{i}{e^{int}}$
For the statement in the question to be true, we must have $\left(\frac{i}{e^{it}}\right)^n=\frac{i}{e^{int}}$ . The left hand side simplifies to $\frac{i^n}{e^{int}}$ . We cancel the denominators and find that the only thing that needs to be true is that $i^n=i$ . This is true if $n\equiv1\pmod{4}$ , and there are $\boxed{250}$ such numbers between $1$ and $1000$ . Solution by Zeroman

Solution:
We are using degrees in this solution instead of radians. I just process stuff better that way.
We can see that the LHS is $cis(n(90^{\circ}-t))$ , and the RHS is $cis(90^{\circ}-nt)$ So, $n(90-t) \equiv 90-nt \mod 360$ Expanding and canceling the nt terms, we will get $90n \equiv 90 \mod 360$ . Canceling gets $n \equiv 1 \mod 4$ , and thus there are $\boxed{250}$ values of n.

Solution: Let $t=0$ . Then, we have $i^n=i$ which means $n\equiv 1\pmod{4}$ . Thus, the answer is $\boxed{250}$

Solution: We factor out $i^n$ from $(\sin t + i \cos t)^n = i^n (\cos(t) - i \sin t)= i^n(\cos(nt) - i\sin nt).$ We know the final expression must be the same as $\sin nt + i \cos nt$ so we must have $i^n(\cos(nt) - i\sin nt) = \sin nt + i \cos nt$ in which testing yields $n \equiv 1 \pmod{4}$ is the only mod that works, so we have a total of $1000 \cdot\frac{1}{4} = \boxed{250}$ integers $n$ that work.

Solution: Note that this looks like de Moivre's except switched around. Using de Moivre's as motivation we try to convert the given expression into de Moivre's. Note that $\sin t = \cos(90 - t)$ and $\cos t = \sin(90 - t)$ . So we rewrite the expression and setting it equal to the given expression in the problem, we get $\cos(90 - nt) + i\sin(90 - nt) = \cos(90n - nt) + i\sin(90n - nt)$ . Now we can just look at the imaginary parts. Doing so and simplifying, we see that $1 + 4k= n$ . From this we see that $n \equiv 1\pmod{4}$ . So there are $\boxed{250}$ solutions.

Solution:
$(\sin\theta + i\cos\theta)^{n} = i^{n} (\cos\theta - i\sin\theta)^n$
Hence the required condition is just $i^{n} = i$ which is true for exactly 1 in 4 consecutive numbers. Thus $\boxed{250}$


Answer: 16.0

For how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1 + 2 + \cdots + n?$
$\textbf{(A) } 8 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 21$

Solution:
Since $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ , the condition is equivalent to having an integer value for $\frac{n!} {\frac{n(n+1)}{2}}$ . This reduces, when $n\ge 1$ , to having an integer value for $\frac{2(n-1)!}{n+1}$ . This fraction is an
integer unless $n+1$ is an odd prime. There are $8$ odd primes less than or equal to $24$ , so there
are $24 - 8 =  \boxed{16}$ numbers less than or equal to $24$ that satisfy the condition.


Answer: 18.0

For how many positive integers $x$ is $\log_{10}(x-40) + \log_{10}(60-x) < 2$
$\textbf{(A) }10\qquad \textbf{(B) }18\qquad \textbf{(C) }19\qquad \textbf{(D) }20\qquad \textbf{(E) }\text{infinitely many}\qquad$

Solution: The domain of the LHS implies that \[40<x<60\] Begin from the left hand side \[\log_{10}[(x-40)(60-x)]<2\] \[(x-40)(60-x)<100\] \[-x^2+100x-2500<0\] \[(x-50)^2>0\] \[x \not = 50\] Hence, we have integers from 41 to 49 and 51 to 59. There are $\boxed{18}$ integers.


Answer: 10.0

For how many real numbers $a$ does the quadratic equation $x^2 + ax + 6a=0$ have only integer roots for $x$

Solution:
By Vieta's formulas $x_1 + x_2 = -a$ where $x_1, x_2$ are the roots of the quadratic, and since $x_1,x_2$ are integers, $a$ must be an integer. Applying the quadratic formula
\[x = \frac{-a \pm \sqrt{a^2 - 24a}}{2}\]
Since $-a$ is an integer, we need $\sqrt{a^2-24a}$ to be an integer (let this be $b$ ): $b^2 = a^2 - 24a$ Completing the square , we get
\[(a - 12)^2 = b^2 + 144\]
Which implies that $b^2 + 144$ is a perfect square also (let this be $c^2$ ). Then
\[c^2 - b^2 = 144 \Longrightarrow (c+b)(c-b) = 144\]
The pairs of factors of $144$ are $(\pm1,\pm144),( \pm 2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)$ ; since $c$ is the average of each respective pair and is also an integer, the pairs that work must have the same parity . Thus we get $\boxed{10}$ pairs (counting positive and negative) of factors that work, and substituting them backwards show that they all work.

Solution:
Let $x^2 + ax + 6a = (x - s)(x - r)$ . Vieta's yields $s + r = - a, sr = 6a$ \begin{eqnarray*}sr + 6s + 6r &=& 0\\ sr + 6s + 6r + 36 &=& 36\\ (s + 6)(r + 6) &=& 36 \end{eqnarray*}
Without loss of generality let $r \le s$
The possible values of $(r + 6,s + 6)$ are: $( - 36, - 1),( - 18, - 2),( - 12, - 3),( - 9, - 4),( - 6, - 6),(1,36),(2,18),(3,12),(4,9),(6,6)$ $\Rightarrow \boxed{10} a$


Answer: 11.0

For how many real values of $x$ is $\sqrt{120-\sqrt{x}}$ an integer?
$\textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) }  11$

Solution:
For $\sqrt{120-\sqrt{x}}$ to be an integer, $120-\sqrt{x}$ must be a perfect square.
Since $\sqrt{x}$ can't be negative, $120-\sqrt{x} \leq 120$
The perfect squares that are less than or equal to $120$ are $\{0,1,4,9,16,25,36,49,64,81,100\}$ , so there are $11$ values for $120-\sqrt{x}$
Since every value of $120-\sqrt{x}$ gives one and only one possible value for $x$ , the number of values of $x$ is $\boxed{11}$


Answer: 11.0

Solution:
For $\sqrt{120-\sqrt{x}}$ to be an integer, $120-\sqrt{x}$ must be a perfect square.
Since $\sqrt{x}$ can't be negative, $120-\sqrt{x} \leq 120$
The perfect squares that are less than or equal to $120$ are $\{0,1,4,9,16,25,36,49,64,81,100\}$ , so there are $11$ values for $120-\sqrt{x}$
Since every value of $120-\sqrt{x}$ gives one and only one possible value for $x$ , the number of values of $x$ is $\boxed{11}$


Answer: 6.0

For how many three-digit whole numbers does the sum of the digits equal $25$
$\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$

Solution:
Because $8+8+8=24$ , it follows that one of the digits must be a $9$ . The other two digits them have a sum of $25-9=16$ . The groups of digits that produce a sum of $25$ are $799, 889$ and can be arranged as follows
\[799,979,997,889,898,988\]
The number of configurations is $\boxed{6}$


Answer: 40.0

For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$

Solution:
$2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ . We wish to figure out the number of ways to distribute these prime factors amongst 3 different integers, without over counting triples which are simply permutations of one another.
We can account for permutations by assuming WLOG that $a$ contains the prime factor 2. Thus, there are $3^4$ ways to position the other 4 prime numbers. Note that, with the exception of when all of the prime factors belong to $a$ , we have over counted each case twice, as for when we put certain prime factors into $b$ and the rest into $c$ , we count the exact same case when we put those prime factors which were in $b$ into $c$
Thus, our total number of cases is $\frac{3^4 - 1}{2} = 40 \Rightarrow \boxed{40}.$


Answer: 40.0

Solution:
The prime factorization of $2310$ is $2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11.$ Therefore, we have the equation \[abc = 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11,\] where $a, b, c$ must be distinct positive integers and order does not matter. There are $3$ ways to assign each prime number on the right-hand side to one of the variables $a, b, c,$ which gives $3^5 = 243$ solutions for $(a, b, c).$ However, three of these solutions have two $1$ s and one $2310,$ which contradicts the fact that $a, b, c$ must be distinct. Because each prime factor appears only once, all other solutions have $a, b, c$ distinct. Correcting for this, we get $243 - 3 = 240$ ordered triples $(a, b, c)$ where $a, b, c$ are all distinct.
Finally, since order does not matter, we must divide by $3!,$ the number of ways to order $a, b, c.$ This gives the final answer, \[\frac{240}{3!} = \frac{240}{6} = \boxed{40}.\]


Answer: 2.0

For how many values of $a$ is it true that the line $y = x + a$ passes through the
vertex of the parabola $y = x^2 + a^2$
$\mathrm{(A)}\ 0      \qquad \mathrm{(B)}\ 1      \qquad \mathrm{(C)}\ 2      \qquad \mathrm{(D)}\ 10      \qquad \mathrm{(E)}\ \text{infinitely many}$

Solution: We see that the vertex of the quadratic function $y = x^2 + a^2$ is $(0,\,a^2)$ . The y-intercept of the line $y = x + a$ is $(0,\,a)$ . We want to find the values (if any) such that $a=a^2$ . Solving for $a$ , the only values that satisfy this are $0$ and $1$ , so the answer is $\boxed{2}$


Answer: 25.0

For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6$ $8^8$ , and $k$

Solution:
It is evident that $k$ has only 2s and 3s in its prime factorization, or $k = 2^a3^b$
The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. $[6^6,8^8] = 2^{24}3^6$ . Therefore $12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}$ , and $b = 12$ .  Since $0 \le a \le 24$ , there are $\boxed{25}$ values of $k$


Answer: 4.0

For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$

Solution:
First of all, we have to agree on the range of $\sin^{-1}$ and $\cos^{-1}$ . This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: $\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2$ and $\forall x: 0\leq \cos^{-1}(x) \leq \pi$
Hence we get that $\forall x\in[0,\pi]: \cos^{ - 1}(\cos x) = x$ , thus our equation simplifies to $\sin^{ - 1}(\sin 6x) = x$
Consider the function $f(x) = \sin^{ - 1}(\sin 6x)  - x$ . We are looking for roots of $f$ on $[0,\pi]$
By analyzing properties of $\sin$ and $\sin^{-1}$ (or by computing the derivative of $f$ ) one can discover the following properties of $f$
For $x=\pi/6$ we have $f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0$ . Hence $f$ has exactly one root on $(0,\pi/6)$
For $x=2\pi/6$ we have $f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0$ . Hence $f$ is negative on the entire interval $[\pi/6,2\pi/6]$
Now note that $\forall t: \sin^{-1}(t) \leq \pi/2$ . Hence for $x > 3\pi/6$ we have $f(x) < 0$ , and we can easily check that $f(3\pi/6)<0$ as well.
Thus the only unknown part of $f$ is the interval $(2\pi/6,3\pi/6)$ . On this interval, $f$ is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval.
To prove that there are two roots, it is enough to find any $x$ from this interval such that $f(x)>0$
A good guess is its midpoint, $x=5\pi/12$ , where the function $\sin^{-1}(\sin 6x)$ has its local maximum. We can evaluate: $f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0$
Summary: The function $f$ has $\boxed{4}$ roots on $[0,\pi]$ : the first one is $0$ , the second one is in $(0,\pi/6)$ , and the last two are in $(2\pi/6,3\pi/6)$

Solution:
Since $0\leq \cos^{-1}(a) \leq \pi$ for all $a$ , the equation reduces to $\sin^{-1}(\sin(6x)) = x$ . Since $-\pi/2 \leq \sin^{-1}(a) \leq \pi/2$ for all $a$ $0 \leq x \leq \pi/2$ . To make the problem easier, we will measure angles in degrees. We will consider each sixth of the interval $[0, 90]$
For $0 \leq x \leq 15$ $6x$ is in the first quadrant. Thus, $\sin^{-1}(\sin(6x)) = 6x$ . Setting this equal to $x$ yields the solution $x = 0$
For $15 \leq x \leq 30$ $6x$ is in the second quadrant. Thus, $\sin^{-1}(\sin(6x)) = 180 - 6x$ . This yields the solution $x = \frac{180}7$
For $30 \leq x \leq 45$ $6x$ is in the third quadrant. Thus, $\sin^{-1}(\sin(6x)) = 180 - 6x$ . As $\frac{180}{7}$ is not on the interval $30 \leq x \leq 45$ , this yields no solution.
For $45 \leq x \leq 60$ $6x$ is in the fourth quadrant. Thus, $\sin^{-1}(\sin(6x)) = 6x - 360$ . As $72$ is not on the interval $45 \leq x \leq 60$ , this yields no solution.
For $60 \leq x \leq 75$ $6x$ is in the first quadrant plus a full revolution. Thus, $\sin^{-1}(\sin(6x)) = 6x - 360$ . This yields the solution $x = 72$
For $75 \leq x \leq 90$ $6x$ is in the second quadrant plus a full revolution. Thus $\sin^{-1}(\sin(6x)) = 540 - 6x$ . This yields the solution $x = \frac{540}7$
There are $\boxed{4}$ solutions, $x=0$ $x=\pi/7$ $x=2\pi/5$ , and $x=3\pi/7$

Solution:
Algebraically, the inverse function of a function should just cancel out, leaving $6x=x$ . However, upon inspection we find that the graphs of these "inverse function of the function" equations are also periodic, like their normal trig function counterparts, due to the fact that inverse trig functions will never return any angle value higher than $2\pi$ . But instead of a smooth wave, these graphs are made up of zigzags with slope $1$ and $-1$ . Trying a few values, we see that
$y = \arcsin{\sin{x}}$
peaks at $\frac{\pi}{2}$ and
$y = \arccos{\cos{x}}$
peaks at $\pi$
But we want the graph of $y = \arcsin{(\sin6{x})}$ , which has a period of $\frac{\pi}{3}$ instead of $2\pi$ . So this means the interval $[0, \pi]$ will show $3$ periods instead of $\frac{1}{2}$ of a period. Visually it would be lines with slopes $6$ and $-6$ . Using the graph paper given to us, we plot out the two equations according to the above and we see that they intersect $4$ times $\Rightarrow \boxed{4}$

Solution:
The most conventional domain and range for $\cos^{-1} x$ is $x \in [-1, 1]$ and $\cos^{-1} x \in [0, \pi]$ . For $\cos x$ $x \in [0, \pi]$ and $\cos x \in [-1, 1]$ . Since the domain of $\cos x$ is equal to the range of $\cos^{-1} x$ , and the range of $\cos x$ is equal to the domain of $\cos^{-1} x$ $\cos^{-1} (\cos x) = x$
Meaning that $\sin^{ - 1}(\sin 6x) = x$ $\sin x = \sin (6x)$
By graphing $\sin x$ and $\sin 6x$ together, it can be seen that there are 7 intersections for $x \in [0, \pi]$
However, the range of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . Therefore $x \in [0, \frac{\pi}{2}]$ for $x = \sin^{ - 1}(\sin 6x)$ . There are $\boxed{4}]$