Answer: 19.0

How many integer values of $x$ satisfy $|x|<3\pi$
$\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20$

Solution: Since $3\pi\approx9.42$ , we multiply $9$ by $2$ for the integers from $1$ to $9$ and the integers from $-1$ to $-9$ and add $1$ to account for $0$ to get $\boxed{19}$

Solution: $|x|<3\pi$ $\iff$ $-3\pi<x<3\pi$ . Since $\pi$ is approximately $3.14$ $3\pi$ is approximately $9.42$ . We are trying to solve for $-9.42<x<9.42$ , where $x\in\mathbb{Z}$ . Hence, $-9.42<x<9.42$ $\implies$ $-9\leq x\leq9$ , for $x\in\mathbb{Z}$ . The number of integer values of $x$ is $9-(-9)+1=19$ . Therefore, the answer is $\boxed{19}$

Solution:
$3\pi \approx 9.4.$ There are two cases here.
When $x>0, |x|>0,$ and $x = |x|.$ So then $x<9.4$
When $x<0, |x|>0,$ and $x = -|x|.$ So then $-x<9.4$ . Dividing by $-1$ and flipping the sign, we get $x>-9.4.$
From case 1 and 2, we know that $-9.4 < x < 9.4$ . Since $x$ is an integer, we must have $x$ between $-9$ and $9$ . There are a total of \[9-(-9) + 1 = \boxed{19}.\] PureSwag

Solution: Looking at the problem, we see that instead of directly saying $x$ , we see that it is $|x|.$ That means all the possible values of $x$ in this case are positive and negative. Rounding $\pi$ to $3$ we get $3(3)=9.$ There are $9$ positive solutions and $9$ negative solutions: $9+9=18.$ But what about zero? Even though zero is neither negative nor positive, but we still need to add it into the solution. Hence, the answer is $9+9+1=18+1=\boxed{19}.$

Solution: There are an odd number of integer solutions $x$ to this inequality since if any non-zero integer $x$ satisfies this inequality, then so does $-x,$ and we must also account for $0,$ which gives us the desired. Then, the answer is either $\textbf{(A)}$ or $\textbf{(D)},$ and since $3 \pi > 3 \cdot 3 > 9,$ the answer is at least $9 \cdot 2 + 1 = 19,$ yielding $\boxed{19}.$


Answer: 19.0

Solution: Since $3\pi\approx9.42$ , we multiply $9$ by $2$ for the integers from $1$ to $9$ and the integers from $-1$ to $-9$ and add $1$ to account for $0$ to get $\boxed{19}$

Solution: $|x|<3\pi$ $\iff$ $-3\pi<x<3\pi$ . Since $\pi$ is approximately $3.14$ $3\pi$ is approximately $9.42$ . We are trying to solve for $-9.42<x<9.42$ , where $x\in\mathbb{Z}$ . Hence, $-9.42<x<9.42$ $\implies$ $-9\leq x\leq9$ , for $x\in\mathbb{Z}$ . The number of integer values of $x$ is $9-(-9)+1=19$ . Therefore, the answer is $\boxed{19}$

Solution:
$3\pi \approx 9.4.$ There are two cases here.
When $x>0, |x|>0,$ and $x = |x|.$ So then $x<9.4$
When $x<0, |x|>0,$ and $x = -|x|.$ So then $-x<9.4$ . Dividing by $-1$ and flipping the sign, we get $x>-9.4.$
From case 1 and 2, we know that $-9.4 < x < 9.4$ . Since $x$ is an integer, we must have $x$ between $-9$ and $9$ . There are a total of \[9-(-9) + 1 = \boxed{19}.\] PureSwag

Solution: Looking at the problem, we see that instead of directly saying $x$ , we see that it is $|x|.$ That means all the possible values of $x$ in this case are positive and negative. Rounding $\pi$ to $3$ we get $3(3)=9.$ There are $9$ positive solutions and $9$ negative solutions: $9+9=18.$ But what about zero? Even though zero is neither negative nor positive, but we still need to add it into the solution. Hence, the answer is $9+9+1=18+1=\boxed{19}.$

Solution: There are an odd number of integer solutions $x$ to this inequality since if any non-zero integer $x$ satisfies this inequality, then so does $-x,$ and we must also account for $0,$ which gives us the desired. Then, the answer is either $\textbf{(A)}$ or $\textbf{(D)},$ and since $3 \pi > 3 \cdot 3 > 9,$ the answer is at least $9 \cdot 2 + 1 = 19,$ yielding $\boxed{19}.$


Answer: 15.0

How many integers $N$ less than $1000$ can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$

Solution:
Let the first odd integer be $2n+1$ $n\geq 0$ . Then the final odd integer is $2n+1 + 2(j-1) = 2(n+j) - 1$ . The odd integers form an arithmetic sequence with sum $N = j\left(\frac{(2n+1) + (2(n+j)-1)}{2}\right) = j(2n+j)$ . Thus, $j$ is a factor of $N$
Since $n\geq 0$ , it follows that $2n+j \geq j$ and $j\leq \sqrt{N}$
Since there are exactly $5$ values of $j$ that satisfy the equation, there must be either $9$ or $10$ factors of $N$ . This means $N=p_1^2p_2^2$ or $N=p_1p_2^4$ . Unfortunately, we cannot simply observe prime factorizations of $N$ because the factor $(2n+j)$ does not cover all integers for any given value of $j$
Instead we do some casework:
\[(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)\]
\[N = 4k(n+k) \Longrightarrow \frac{N}{4} = k(n+k)\]
\[\frac{N}{4} = (2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13), (3^4\cdot2)\]
The total number of integers $N$ is $5 + 10 = \boxed{15}$

Solution:
Let the largest odd number below the sequence be the $q$ th positive odd number, and the largest odd number in the sequence be the $p$ th positive odd number. Therefore, the sum is $p^2-q^2=(p+q)(p-q)$ by sum of consecutive odd numbers. Note that $p+q$ and $p-q$ have the same parity, and $q$ can equal $0$ . We then perform casework based on the parity of $p-q$
If $p-q$ is odd, then $p^2-q^2$ must always be odd.  Therefore, to have 5 pairs of odd factors, we must have either $9$ (in which case the number is a perfect square) or $10$ factors. Considering the upper bound, the only way this can happen is $p_1^4\cdot{p_2}$ or $p_1^2\cdot{p_2^2}$ . N must then be one of \[(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)\] So, there are $5$ solutions when $(p+q)(p-q)$ is odd.
If $p-q$ is even, then $(p+q)(p-q)$ must have at least two factors of $2$ , so we can rewrite the expression as $4(k)(k-q)$ where $k=\frac{p+q}{2}$ . We can disregard the $4$ by dividing by $4$ and restricting our upper bound to $250$ . Since $k$ and $q$ don't have to be the same parity, we can include all cases less than 250 that have 9 or 10 factors. We then have \[(2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13), (3^4\cdot2)\] as the possibilities.
Therefore, there are $10+5=\boxed{015}$ possibilities for $p^2-q^2=N$ ~sigma


Answer: 2.0

How many integers $n \geq 2$ are there such that whenever $z_1, z_2, ..., z_n$ are complex numbers such that
\[|z_1| = |z_2| = ... = |z_n| = 1 \text{    and    } z_1 + z_2 + ... + z_n = 0,\] then the numbers $z_1, z_2, ..., z_n$ are equally spaced on the unit circle in the complex plane?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5$

Solution:
For $n=2$ , we see that if $z_{1}+z_{2}=0$ , then $z_{1}=-z_{2}$ , so they are evenly spaced along the unit circle.
For $n=3$ , WLOG, we can set $z_{1}=1$ . Notice that now Re $(z_{2}+z_{3})=-1$ and Im $\{z_{2}\}$ $-$ Im $\{z_{3}\}$ . This forces $z_{2}$ and $z_{3}$ to be equal to $e^{i\frac{2\pi}{3}}$ and $e^{-i\frac{2\pi}{3}}$ , meaning that all three are equally spaced along the unit circle.
We can now show that we can construct complex numbers when $n\geq 4$ that do not satisfy the conditions in the problem.
Suppose that the condition in the problem holds for some $n=k$ . We can now add two points $z_{k+1}$ and $z_{k+2}$ anywhere on the unit circle such that $z_{k+1}=-z_{k+2}$ , which will break the condition. Now that we have shown that $n=2$ and $n=3$ works, by this construction, any $n\geq 4$ does not work, making the answer $\boxed{2}$


Answer: 4.0

How many integers $x$ satisfy the equation \[\left(x^2-x-1\right)^{x+2} = 1?\]
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 5 \qquad \mathrm{(E) \  }\text{none of these}$

Solution:
We recall that for real numbers $a$ and $b$ , there are exactly $3$ ways in which we can have $a^b = 1$ , namely $a = 1$ $b = 0$ and $a \neq 0$ ; or $a = -1$ and $b$ is an even integer.
The first case therefore gives \begin{align*}x^2-x-1 = 1 &\iff x^2-x-2 = 0 \\&\iff (x-2)(x+1) = 0 \\&\iff x = 2 \text{ or } x = -1.\end{align*}
Similarly, the second case gives $x+2 = 0$ , i.e. $x = -2$ , and this indeed gives $x^2-x-1 = 4+2-1 = 5 \neq 0$ , so $x = -2$ is a further valid solution.
Lastly, for the third case, we have \begin{align*}x^2-x-1 = -1 &\iff x^2-x = 0 \\&\iff x(x-1) = 0 \\&\iff x = 0 \text{ or } x = 1,\end{align*} but $x = 1$ would give $x+2 = 3$ , which is odd, whereas $x = 0$ gives $x+2 = 2$ , which is even. Therefore, this case gives only one further solution, namely $x = 0$
Accordingly, the possible values of $x = -2$ $-1$ $0$ , or $2$ , yielding a total of $\boxed{4}$ solutions.


Answer: 226.0

How many integers between $100$ and $999$ , inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\textbf{(A) } 226 \qquad \textbf{(B) } 243 \qquad \textbf{(C) } 270 \qquad \textbf{(D) } 469 \qquad \textbf{(E) } 486$

Solution:
There are 81 multiples of 11 between $100$ and $999$ inclusive. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Switching shows we have overcounted by a factor of 2, so assign $6 \div 2 = 3$ permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have $0$ as a digit. Since $0$ cannot be the digit of the hundreds place, we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243) to get 226. $\boxed{226}$

Solution:
We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:
$\textbf{Case 1:}$ All three digits are the same.
By inspection, we find that there are no multiples of $11$ here.
$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.
$\textbf{Case 2a:}$ There are $8$ multiples of $11$ without a zero that have this property: $121$ $242$ $363$ $484$ $616$ $737$ $858$ $979$ .
Each contributes $3$ valid permutations, so there are $8 \cdot 3 = 24$ permutations in this subcase.
$\textbf{Case 2b:}$ There are $9$ multiples of $11$ with a zero that have this property: $110$ $220$ $330$ $440$ $550$ $660$ $770$ $880$ $990$ .
Each one contributes $2$ valid permutations (the first digit can't be zero), so there are $9 \cdot 2 = 18$ permutations in this subcase.
$\textbf{Case 3:}$ All the digits are different.
Since there are $\frac{990-110}{11}+1 = 81$ multiples of $11$ between $100$ and $999$ , there are $81-8-9 = 64$ multiples of $11$ remaining in this case. However, $8$ of them contain a zero, namely $209$ $308$ $407$ $506$ $605$ $704$ $803$ , and $902$ . Each of those multiples of $11$ contributes $2 \cdot 2=4$ valid permutations, but we overcounted by a factor of $2$ ; every permutation of $209$ , for example, is also a permutation of $902$ . Therefore, there are $8 \cdot 4 / 2 = 16$ . Therefore, there are $64-8=56$ remaining multiples of $11$ without a $0$ in this case. Each one contributes $3! = 6$ valid permutations, but once again, we overcounted by a factor of $2$ (note that if a number ABC is a multiple of $11$ , then so is CBA). Therefore, there are $56 \cdot 6 / 2 = 168$ valid permutations in this subcase.
Adding up all the permutations from all the cases, we have $24+18+16+168 = \boxed{226}$

Solution:
We can first overcount and then subtract.
We know that there are $81$ multiples of $11$
We can then multiply by $6$ for each permutation of these multiples. (Yet some multiples do not have six distinct permutations.)
Now divide by $2$ , because if a number $abc$ with digits $a$ $b$ , and $c$ is a multiple of $11$ , then $cba$ is also a multiple of $11$ so we have counted the same permutations twice.
Basically, each multiple of $11$ has its own $3$ permutations (say $abc$ has $abc$ $acb$ and $bac$ whereas $cba$ has $cba$ $cab$ and $bca$ ). We know that each multiple of $11$ has at least $3$ permutations because it cannot have $3$ repeating digits.
Hence we have $243$ permutations without subtracting for overcounting.
Now note that we overcounted cases in which we have $0$ 's at the start of each number. So, in theory, we could just answer $A$ and then move on.
If we want to solve it, then we continue.
We overcounted cases where the middle digit of the number is $0$ and the last digit is $0$
Note that we assigned each multiple of $11$ three permutations.
The last digit is $0$ gives $9$ possibilities where we overcounted by $1$ permutation for each of $110, 220, ... , 990$
The middle digit is $0$ gives $8$ possibilities where we overcount by $1$ $605, 704, 803, 902$ and $506, 407, 308, 209$
Subtracting $17$ gives $\boxed{226}$

Solution:
Taken from solution three, we notice that there are a total of $81$ multiples of $11$ between $100$ and $999$ , and each of them have at most $6$ permutation (and thus is a permutations of $6$ numbers), giving us a maximum of $486$ valid numbers.
However, if $abc$ can be divided by $11$ , so can $cba$ , which is distinct if $c \neq a$ . And if $c = a$ then $abc$ and $cba$ have the same permutations. Either way, we have doubled counted.
This reduces the number of permutations to $486/2 =  243$
Furthermore, if $a=b$ or $c=0$ (which turn out to be equivalent conditions! for example $220$ ), not all (inverse) permutations are distinct ( $\mathbf{2}20 = 2\mathbf{2}0$ ) or valid ( $022$ ). (There are 9 of these.)
Similarly, for $a0b$ , not all (inverse) permutations are valid. (There are 8 of these.)
As long as you notice at least one example of one of these 3 cases, you may infer that the answer must be smaller than $243$ . This leaves us with only one possible answer: $\boxed{226}$


Answer: 4536.0

How many integers between $1000$ and $9999$ have four distinct digits?
$\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561$

Solution: There are $9$ choices for the first number, since it cannot be $0$ , there are only $9$ choices left for the second number since it must differ from the first, $8$ choices for the third number, since it must differ from the first two, and $7$ choices for the fourth number, since it must differ from all three.  This means there are $9 \times 9 \times 8 \times 7=\boxed{4536}$ integers between $1000$ and $9999$ with four distinct digits.


Answer: 15.0

How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.)
$\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}$

Solution: Firstly, observe that the second digit of such a number cannot be $1$ or $2$ , because the digits must be distinct and increasing. The second digit also cannot be $4$ as the number must be less than $2400$ , so it must be $3$ . It remains to choose the latter two digits, which must be $2$ distinct digits from $\left\{4,5,6,7,8,9\right\}$ . That can be done in $\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15$ ways; there is then only $1$ way to order the digits, namely in increasing order. This means the answer is $\boxed{15}$

Solution: As in Solution 1, we find that the first two digits must be $23$ , and the third digit must be at least $4$ . If it is $4$ , then there are $5$ choices for the last digit, namely $5$ $6$ $7$ $8$ , or $9$ . Similarly, if the third digit is $5$ , there are $4$ choices for the last digit, namely $6$ $7$ $8$ , and $9$ ; if $6$ , there are $3$ choices; if $7$ , there are $2$ choices; and if $8$ , there is $1$ choice. It follows that the total number of such integers is $5+4+3+2+1=\boxed{15}$


Answer: 3.0

How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution: Find the least common multiple of $15, 20, 25$ by turning the numbers into their prime factorization. \[15 = 3 * 5, 20 = 2^2 * 5, 25 = 5^2\] Gather all necessary multiples $3, 2^2, 5^2$ when multiplied gets $300$ . The multiples of $300 - 300, 600, 900, 1200, 1500, 1800, 2100$ . The number of multiples between 1000 and 2000 is $\boxed{3}$


Answer: 28.0

How many line segments have both their endpoints located at the vertices of a given cube
$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$

Solution: There are $8$ choices for the first endpoint of the line segment, and $7$ choices for the second endpoint, giving a total of $8\cdot 7 = 56$ segments.  However, both $\overline{AB}$ and $\overline{BA}$ were counted, while they really are the same line segment.  Every segment got double counted in a similar manner, so there are really $\frac{56}{2} = 28$ line segments, and the answer is $\boxed{28}$

Solution:
Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal.
A cube has $12$ edges:  Four on the top face, four on the bottom face, and four that connect the top face to the bottom face.
A cube has $6$ square faces, each of which has $2$ facial diagonals, for a total of $6\cdot 2 = 12$
A cube has $4$ spacial diagonals:  each diagonal goes from one of the bottom vertices to the "opposite" top vertex.
Thus, there are $12 + 12 + 4 = 28$ segments, and the answer is $\boxed{28}$


Answer: 2.0

How many non- congruent triangles with perimeter $7$ have integer side lengths?
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 5$

Solution:
By the triangle inequality , no side may have a length greater than the semiperimeter, which is $\frac{1}{2}\cdot7=3.5$
Since all sides must be integers, the largest possible length of a side is $3$ . Therefore, all such triangles must have all sides of length $1$ $2$ , or $3$ . Since $2+2+2=6<7$ , at least one side must have a length of $3$ . Thus, the remaining two sides have a combined length of $7-3=4$ . So, the remaining sides must be either $3$ and $1$ or $2$ and $2$ . Therefore, the number of triangles is $\boxed{2}$


Answer: 2.0

Solution:
By the triangle inequality , no side may have a length greater than the semiperimeter, which is $\frac{1}{2}\cdot7=3.5$
Since all sides must be integers, the largest possible length of a side is $3$ . Therefore, all such triangles must have all sides of length $1$ $2$ , or $3$ . Since $2+2+2=6<7$ , at least one side must have a length of $3$ . Thus, the remaining two sides have a combined length of $7-3=4$ . So, the remaining sides must be either $3$ and $1$ or $2$ and $2$ . Therefore, the number of triangles is $\boxed{2}$


Answer: 405.0

How many non- empty subsets $S$ of $\{1,2,3,\ldots ,15\}$ have the following two properties?
$(1)$ No two consecutive integers belong to $S$
$(2)$ If $S$ contains $k$ elements , then $S$ contains no number less than $k$
$\mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\qquad \mathrm{(E) \ }  405$

Solution:
This question can be solved fairly directly by casework and pattern-finding.  We give a somewhat more general attack, based on the solution to the following problem:
How many ways are there to choose $k$ elements from an ordered $n$ element set without choosing two consecutive members?
You want to choose $k$ numbers out of $n$ with no consecutive numbers. For each configuration, we can subtract $i-1$ from the $i$ -th element in your subset. This converts your configuration into a configuration with $k$ elements where the largest possible element is $n-k+1$ , with no restriction on consecutive numbers. Since this process is easily reversible, we have a bijection .
Without consideration of the second condition, we have: ${15 \choose 1} + {14 \choose 2} + {13 \choose 3} + ... + {9 \choose 7} + {8 \choose 8}$
Now we examine the second condition. It simply states that no element in our original configuration (and hence also the modified configuration, since we don't move the smallest element) can be less than $k$ , which translates to subtracting $k - 1$ from the "top" of each binomial coefficient .
Now we have, after we cancel all the terms ${n \choose k}$ where $n < k$ ${15 \choose 1} + {13 \choose 2} + {11 \choose 3} + {9 \choose 4} + {7 \choose 5}=  15 + 78 + 165 + 126 + 21 = \boxed{405}$

Solution:
Let $s$ be the numbers of elements in a subset. First we examine the second condition. No elements less than $s$ can be put in a subset of size $s$ , therefore the "lowest" element that can go into the subset is $s$ , whereas the "highest" element that can go into the subset is $15$ . This is a total of $15-s+1$ or $16-s$ possible elements.
Now we consider the first condition. No consecutive elements are allowed. This means that if an element $e$ is put into the subset, both $e-1$ and $e+1$ are no longer possibilities. Assume that all subsets are ordered from least to greatest (we are looking for the number of combinations, so we can order these combinations however we want). Then the first element will be $s$ (as a reminder, the lowest possible element in a subset is $s$ ), the second element will be at least $s+2$ , and so on. After $s$ elements are chosen, we will have skipped $s-1$ elements (these are the elements that were "eliminated" as they were consecutive). Therefore, we ignore exactly $s-1$ elements (if we ignore more or less elements, then $s$ changes) Since we must ignore $s-1$ elements, we can simply remove those beforehand. ( $16-s)-(s-1) = 17-2s$ possible elements
Now we look for the bounds of $s$ . We are looking for non-empty subsets, so $s\ge1$ . If $s$ is too large, there will not be enough non-consecutive terms between $s$ and $15$ . Specifically, the highest element in a subset using "optimal" selection will be $s+2(s-1)$ or $3s-2$ . If $3s-1>15$ , that means s is too large. Therefore $3s-2\le15$ ; solving for $s$ yields $s\le5$ . Now we know that $1 \le s \le 5$
We want to know the number of ways to arrange $17-2s$ "balls" into $s$ identical "boxes" with at most $1$ ball per box, for $s=1$ to $s=5$ . This is equivalent to $\sum_{s=1}^{5}{17-2s \choose s} = 405$ , or $\boxed{405}$

Solution:
We will split the problem into cases, and maybe one could then generalize this to arbitrary $n$
$\bold{Case 1}:$ $n=15, k=1$ ). Then this is easy. We have $\binom{15}{1}=15$
$\bold{Case 2}:$ $n=14, k=2$ ). Now we have something tricky. To get a good grasp on this case, let us consider the smallest element; $2$ , in the first spot. We then have $15-1=14$ numbers left. However, we cannot have the digit $3$ . Hence we have to choose $2$ numbers from $14-1=13$ integers left. This can be done in $\binom{13}{2}=78$ ways.
$\bold{Case 3}:$ $n=13, k=3$ ). As in the last case, we can use the smallest element to get a good grasp on this case. Put the digit $3$ in the first spot. Then there are $11$ integers left for the second spot. But in total, we have to avoid $2$ elements for each subset of cardinality $3$ ${a,b,c}$ . So we have to choose $3$ elements from $13-2=11$ elements, which can be done in $\binom{11}{3}=165$ ways.
$\bold{Case 4}:$ $n=12, k=4$ ). As in the previous case, we have to avoid $3$ numbers, and choose $4$ from them. Which is choosing $4$ elements from $12-3=9$ elements, which can be done in $\binom{9}{4}=126$ ways.
$\bold{Case 5}:$ $n=11, k=4$ ). Now you are accustomed to the strategy. We remove $4$ integers leaving $\binom{7}{5}=21$ ways.
Adding up all of the cases yields $405$ , as in $\boxed{405}$


Answer: 59.0

How many non- similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 59\qquad\textbf{(D) } 89\qquad\textbf{(E) } 178\qquad$

Solution: The sum of the angles of a triangle is $180$ degrees. For an arithmetic progression with an odd number of terms, the middle term is equal to the average of the sum of all of the terms, making it $\frac{180}{3} = 60$ degrees. The minimum possible value for the smallest angle is $1$ and the highest possible is $59$ (since the numbers are distinct), so there are $\boxed{59}$ possibilities.

Solution:
Let the first angle be $x$ , and the common difference be $d$ . The arithmetic progression can now be expressed as $x + (x + d) + (x + 2d) = 180$ . Simplifiying, $x + d = 60$ . Now, using stars and bars, we have $\binom{61}{1} = 61$ .
However, we must subtract the two cases in which either $x$ or $d$ equal $0$ , so we have $61 - 2$ $\boxed{59}$

Solution: Consider that we have $(a+n)+(a+n+1)+(a+n+2)=180 \Longleftrightarrow 3a+3(n+1)=180 \Longleftrightarrow a=59-n$ , where $n \geq 0$ and $n$ is an integer. Since $a \neq 0$ $n=0,1,2,3,\cdots, 58$ which is $\boxed{59}$ solutions.


Answer: 6.0

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?
$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$

Solution:
Let $a$ and $b$ be the two legs of the triangle, and $c$ be the hypotenuse.
By using $Area = \frac{r}{2} (a+b+c)$ , where $r$ is the in-radius, we get:
\[3(a+b+c) = \frac{r}{2} (a+b+c)\] \[r=6\]
In right triangle, $r = \frac{a+b-c}{2}$ \[a+b-c = 12\] \[c = a + b - 12\]
By the triangle's area we get:
\[\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}\] \[ab = 6(a+b+c)\]
By substituting $c$ in:
\[ab = 6(a+b+a + b - 12)\] \[ab - 12a - 12b + 72 = 0\] \[(a - 12)(b - 12) = 72\]
As $72 = 2^3 \cdot 3^2$ , there are $\frac{(3+1)(2+1)}{2} = 6$ solutions, $\boxed{6}$

Solution: Well, obviously MAA would try to make the answer choices trap some people. One way they could do that is by thinking "non-congruent" would be ignored, so the answer would be multiplied by 2. The only answer choice that can be divided by 2 to create an existing answer is 12, so the answer is $\boxed{6}$


Answer: 8.0

How many non-congruent triangles have vertices at three of the eight points in the array shown below?
[asy]dot((0,0)); dot((.5,.5)); dot((.5,0)); dot((.0,.5)); dot((1,0)); dot((1,.5)); dot((1.5,0)); dot((1.5,.5));[/asy]
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

Solution: Assume the base of the triangle is on the bottom four points because a congruent triangle can be made by reflecting the base on the top four points. For a triangle with a base of length $1$ , there are $3$ triangles. For a triangle with a base of length $2$ , there are $3$ triangles. For length $3$ , there are $2$ . In total, the number of non-congruent triangles is $3+3+2=\boxed{8}$


Answer: 5.0

How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?
$\textbf{(A)}\; 3 \qquad\textbf{(B)}\; 4 \qquad\textbf{(C)}\; 5 \qquad\textbf{(D)}\; 6 \qquad\textbf{(E)}\; 7$

Solution:
Since we want non-congruent triangles that are neither isosceles nor equilateral, we can just list side lengths $(a,b,c)$ with $a<b<c$ . Furthermore, "positive area" tells us that $c < a + b$ and the perimeter constraints means $a+b+c < 15$
There are no triangles when $a = 1$ because then $c$ must be less than $b+1$ , implying that $b \geq c$ , contrary to $b < c$
When $a=2$ , similar to above, $c$ must be less than $b+2$ , so this leaves the only possibility $c = b+1$ . This gives 3 triangles $(2,3,4), (2,4,5), (2,5,6)$ within our perimeter constraint.
When $a=3$ $c$ can be $b+1$ or $b+2$ , which gives triangles $(3,4,5), (3,4,6), (3,5,6)$ . Note that $(3,4,5)$ is a right triangle, so we get rid of it and we get only 2 triangles.
All in all, this gives us $3+2 = \boxed{5}$ triangles.


Answer: 144.0

How many nonempty subsets $B$ of ${0, 1, 2, 3, \cdots, 12}$ have the property that the number of elements in $B$ is equal to the least element of $B$ ? For example, $B = {4, 6, 8, 11}$ satisfies the condition.
$\textbf{(A) } 256 \qquad\textbf{(B) } 136 \qquad\textbf{(C) } 108 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 156$

Solution: There is no way to have a set with 0. If a set is to have its lowest element as 1, it must have only 1 element: 1. If a set is to have its lowest element as 2, it must have 2, and the other element will be chosen from the natural numbers between 3 and 12, inclusive. To calculate this, we do $\binom{10}{1}$ . If the set is the have its lowest element as 3, the other 2 elements will be chosen from the natural numbers between 4 and 12, inclusive. To calculate this, we do $\binom{9}{2}$ . We can see a pattern emerge, where the top is decreasing by 1 and the bottom is increasing by 1. In other words, we have to add $1 + \binom{10}{1} + \binom{9}{2} + \binom{8}{3} + \binom{7}{4} + \binom{6}{5}$ . This is $1+10+36+56+35+6 = \boxed{144}$


Answer: 3281.0

How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
$\textbf{(A) } 512 \qquad  \textbf{(B) } 729 \qquad  \textbf{(C) } 1094 \qquad  \textbf{(D) } 3281 \qquad  \textbf{(E) } 59,048$

Solution: This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of $|x|=3280.5$ , which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are $3280+1=\boxed{3281}$

Solution: Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all $a_i=0$ . The total number of ways to pick $a_i$ from $i=0, 1, 2, 3, ... 7$ is $3^8=6561$ $\frac{6561-1}{2}=3280$ gives the number of possible negative integers. The question asks for the number of non-negative integers, so subtracting from the total gives $6561-3280=\boxed{3281}$ . (RegularHexagon, KLBBC minor changes)

Solution:
Note that the number of total possibilities (ignoring the conditions set by the problem) is $3^8=6561$ . So, E is clearly unrealistic.
Note that if $a_7$ is 1, then it's impossible for \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] to be negative. Therefore, if $a_7$ is 1, there are $3^7=2187$ possibilities. (We also must convince ourselves that these $2187$ different sets of coefficients must necessarily yield $2187$ different integer results.)
As A, B, and C are all less than 2187, the answer must be $\boxed{3281}$

Solution:
Note that we can do some simple casework:
If $a_7=1$ , then we can choose anything for the other 7 variables, so this give us $3^7$ .
If $a_7=0$ and $a_6=1$ , then we can choose anything for the other 6 variables, giving us $3^6$ .
If $a_7=0$ $a_6=0$ , and $a_5=1$ , then we have $3^5$ .
Continuing in this vein, we have $3^7+3^6+\cdots+3^1+3^0$ ways to choose the variables' values, except we have to add 1
because we haven't counted the case where all variables are 0. So our total sum is $\boxed{3281}$ .
Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative.

Solution:
This solution has a similar idea to that of above.
We will start off with casework on $a_{7}$ :
If $a_7 = -1$ , we can show that there is no way to assign the values of $-1, 0, 1$ to the other $a_{i}$ 's for $0 \leq i \leq 6$ . This is because even if we make all of the other $a_{i}$ equal to $1$ , we will have that our number is $(-1)(3^{7}) + (3^{6}+3^{5}+...+3^{0}) = -3^{7} + \frac{3^{7}-1}{3-1}$ . However, note that since $\frac{3^{7}-1}{3-1} < 3^{7}$ $-3^{7} + \frac{3^{7}-1}{3-1} < 0$ .
If $a_{7}=0$ , we can ignore $a_{7}$ and pretend like it never even existed. This then simplifies the problem: the question remains the same, except that instead of having $a_{0}...a_{7}$ , we only have $a_{0}...a_{6}$ . Now, we introduce some notation: let the number of nonnegative integers that can be written in the form of $a_{n} \cdot 3^{n} + ... + a_{0} \cdot 3^{0}$ be denoted as $S_{n}$ . Then, the problem is asking us for $S_{7}$ , and the total number of nonnegative integers we get from this case is $S_{6}$ . Let's keep this in mind and revisit this idea later.
If $a_{7}=1$ , we can show that no matter what values we assign to the rest of the $a_{i}$ , we will always get a nonnegative integer (we can make this even stricter and say positive integer). This is because of something we figured out in the first case: $\frac{3^{7}-1}{3-1} < 3^{7}$ . So, even if we let $a_{i}=-1$ for $0 \leq i \leq 6$ , we will still have $3^{7} - (3^{6}+...+3^{0}) > 0$ . So, for this case, we have $3^7$ possible nonnegative integers.
Totalling up our values from the three cases, we see that $S_{7} = 0 + S_{6} + 3^{7}$ . In fact, using the reasoning from the three cases above, we can deduce that, for all positive integers $n$ $S_{n} = 3^{n} + S_{n-1}$ . We can now start building up our values for $S$ $S_{0} = 2$ $a_{0} = 0, 1$ ), $S_{1} = 3^{1}+S_{0} = 5, S_{2} = 14, ..., S_{6}=1094$ . So, we have $S_{7} = 2187 + 1094 = \boxed{3281}$ .
~advanture

Solution: The key is to realize that this question is basically taking place in $a\in\{0,1,2\}$ if each value of $a$ was increased by $1$ , essentially making it into base $3$ . Then the range would be from $0\cdot3^7+$ $0\cdot3^6+$ $0\cdot3^5+$ $0\cdot3^4+$ $0\cdot3^3+$ $0\cdot3^2+$ $0\cdot3^1+$ $0\cdot3^0=$ $0$ to $2\cdot3^7+$ $2\cdot3^6+$ $2\cdot3^5+$ $2\cdot3^4+$ $2\cdot3^3+$ $2\cdot3^2+$ $2\cdot3^1+$ $2\cdot3^0=$ $3^8-1=$ $6561-1=$ $6560$ , yielding $6561$ different values. Since the distribution for all $a_i\in \{-1,0,1\}$ the question originally gave is symmetrical, we retain the $3280$ positive integers and one $0$ but discard the $3280$ negative integers. Thus, we are left with the answer, $\boxed{3281}\qquad$ .  --anna0kear

Solution: First, set $a_i=0$ for all $i\geq1$ . The range would be the integers for which $[-1,1]$ . If $a_i=0$ for all $i\geq2$ , our set expands to include all integers such that $-4\leq\mathbb{Z}\leq4$ . Similarly, when $i\geq3$ we get $-13\leq\mathbb{Z}\leq13$ , and when $i\geq4$ the range is $-40\leq\mathbb{Z}\leq40$ . The pattern continues until we reach $i=7$ , where $-3280\leq\mathbb{Z}\leq3280$ . Because we are only looking for positive integers, we filter out all $\mathbb{Z}<0$ , leaving us with all integers between $0\leq\mathbb{Z}\leq3280$ , inclusive. The answer becomes $\boxed{3281}$ .  --anna0kear

Solution:
To get the number of integers, we can get the highest positive integer that can be represented using \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
Note that the least nonnegative integer that can be represented is $0$ , when all $a_i=0$ . The highest number will be the number when all $a_i=1$ . That will be \[3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}\] \[=3280\]
Therefore, there are $3280$ positive integers and $(3280+1)$ nonnegative integers (while including $0$ ) that can be represented. Our answer is $\boxed{3281}$

Solution:
Notice that there are $3^8$ options for $a_7, a_6, \cdots a_0$ since each $a_i$ can take on the value $-1$ $0$ , or $1$ . Now we want to find how many of them are positive and then we can add one in the end to account for $0$ (they are asking for non-negative).
By symmetry (look out for these on the contest), we see that exactly half of them are positive. So $\lfloor{\tfrac{3^8}{2}}\rfloor = 3280.$ Now we will add $1$ because of the $0$ to account for the non-negative solutions.
So our final answer is $3280 + 1 = 3281$ which is $\boxed{3281}$

Solution: Obviously, there are $3^8 = 6561$ possible, and one of them is 0, so other $6560$ are either positive or negative. By the symmetry, we can get the answer is $6560/2 + 1$ which is $\boxed{3281}$

Solution: The solution has to obviously be smaller than $3^{7-0+1}=3^8$ but larger than $3^{(7-0+1)-1} = 3^7.$ The only remaining answer is $\boxed{3281}.$ ~mathboy282


Answer: 3281.0

Solution: This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of $|x|=3280.5$ , which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are $3280+1=\boxed{3281}$

Solution: Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all $a_i=0$ . The total number of ways to pick $a_i$ from $i=0, 1, 2, 3, ... 7$ is $3^8=6561$ $\frac{6561-1}{2}=3280$ gives the number of possible negative integers. The question asks for the number of non-negative integers, so subtracting from the total gives $6561-3280=\boxed{3281}$ . (RegularHexagon, KLBBC minor changes)

Solution:
Note that the number of total possibilities (ignoring the conditions set by the problem) is $3^8=6561$ . So, E is clearly unrealistic.
Note that if $a_7$ is 1, then it's impossible for \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] to be negative. Therefore, if $a_7$ is 1, there are $3^7=2187$ possibilities. (We also must convince ourselves that these $2187$ different sets of coefficients must necessarily yield $2187$ different integer results.)
As A, B, and C are all less than 2187, the answer must be $\boxed{3281}$

Solution:
Note that we can do some simple casework:
If $a_7=1$ , then we can choose anything for the other 7 variables, so this give us $3^7$ .
If $a_7=0$ and $a_6=1$ , then we can choose anything for the other 6 variables, giving us $3^6$ .
If $a_7=0$ $a_6=0$ , and $a_5=1$ , then we have $3^5$ .
Continuing in this vein, we have $3^7+3^6+\cdots+3^1+3^0$ ways to choose the variables' values, except we have to add 1
because we haven't counted the case where all variables are 0. So our total sum is $\boxed{3281}$ .
Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative.

Solution:
This solution has a similar idea to that of above.
We will start off with casework on $a_{7}$ :
If $a_7 = -1$ , we can show that there is no way to assign the values of $-1, 0, 1$ to the other $a_{i}$ 's for $0 \leq i \leq 6$ . This is because even if we make all of the other $a_{i}$ equal to $1$ , we will have that our number is $(-1)(3^{7}) + (3^{6}+3^{5}+...+3^{0}) = -3^{7} + \frac{3^{7}-1}{3-1}$ . However, note that since $\frac{3^{7}-1}{3-1} < 3^{7}$ $-3^{7} + \frac{3^{7}-1}{3-1} < 0$ .
If $a_{7}=0$ , we can ignore $a_{7}$ and pretend like it never even existed. This then simplifies the problem: the question remains the same, except that instead of having $a_{0}...a_{7}$ , we only have $a_{0}...a_{6}$ . Now, we introduce some notation: let the number of nonnegative integers that can be written in the form of $a_{n} \cdot 3^{n} + ... + a_{0} \cdot 3^{0}$ be denoted as $S_{n}$ . Then, the problem is asking us for $S_{7}$ , and the total number of nonnegative integers we get from this case is $S_{6}$ . Let's keep this in mind and revisit this idea later.
If $a_{7}=1$ , we can show that no matter what values we assign to the rest of the $a_{i}$ , we will always get a nonnegative integer (we can make this even stricter and say positive integer). This is because of something we figured out in the first case: $\frac{3^{7}-1}{3-1} < 3^{7}$ . So, even if we let $a_{i}=-1$ for $0 \leq i \leq 6$ , we will still have $3^{7} - (3^{6}+...+3^{0}) > 0$ . So, for this case, we have $3^7$ possible nonnegative integers.
Totalling up our values from the three cases, we see that $S_{7} = 0 + S_{6} + 3^{7}$ . In fact, using the reasoning from the three cases above, we can deduce that, for all positive integers $n$ $S_{n} = 3^{n} + S_{n-1}$ . We can now start building up our values for $S$ $S_{0} = 2$ $a_{0} = 0, 1$ ), $S_{1} = 3^{1}+S_{0} = 5, S_{2} = 14, ..., S_{6}=1094$ . So, we have $S_{7} = 2187 + 1094 = \boxed{3281}$ .
~advanture

Solution: The key is to realize that this question is basically taking place in $a\in\{0,1,2\}$ if each value of $a$ was increased by $1$ , essentially making it into base $3$ . Then the range would be from $0\cdot3^7+$ $0\cdot3^6+$ $0\cdot3^5+$ $0\cdot3^4+$ $0\cdot3^3+$ $0\cdot3^2+$ $0\cdot3^1+$ $0\cdot3^0=$ $0$ to $2\cdot3^7+$ $2\cdot3^6+$ $2\cdot3^5+$ $2\cdot3^4+$ $2\cdot3^3+$ $2\cdot3^2+$ $2\cdot3^1+$ $2\cdot3^0=$ $3^8-1=$ $6561-1=$ $6560$ , yielding $6561$ different values. Since the distribution for all $a_i\in \{-1,0,1\}$ the question originally gave is symmetrical, we retain the $3280$ positive integers and one $0$ but discard the $3280$ negative integers. Thus, we are left with the answer, $\boxed{3281}\qquad$ .  --anna0kear

Solution: First, set $a_i=0$ for all $i\geq1$ . The range would be the integers for which $[-1,1]$ . If $a_i=0$ for all $i\geq2$ , our set expands to include all integers such that $-4\leq\mathbb{Z}\leq4$ . Similarly, when $i\geq3$ we get $-13\leq\mathbb{Z}\leq13$ , and when $i\geq4$ the range is $-40\leq\mathbb{Z}\leq40$ . The pattern continues until we reach $i=7$ , where $-3280\leq\mathbb{Z}\leq3280$ . Because we are only looking for positive integers, we filter out all $\mathbb{Z}<0$ , leaving us with all integers between $0\leq\mathbb{Z}\leq3280$ , inclusive. The answer becomes $\boxed{3281}$ .  --anna0kear

Solution:
To get the number of integers, we can get the highest positive integer that can be represented using \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$
Note that the least nonnegative integer that can be represented is $0$ , when all $a_i=0$ . The highest number will be the number when all $a_i=1$ . That will be \[3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}\] \[=3280\]
Therefore, there are $3280$ positive integers and $(3280+1)$ nonnegative integers (while including $0$ ) that can be represented. Our answer is $\boxed{3281}$

Solution:
Notice that there are $3^8$ options for $a_7, a_6, \cdots a_0$ since each $a_i$ can take on the value $-1$ $0$ , or $1$ . Now we want to find how many of them are positive and then we can add one in the end to account for $0$ (they are asking for non-negative).
By symmetry (look out for these on the contest), we see that exactly half of them are positive. So $\lfloor{\tfrac{3^8}{2}}\rfloor = 3280.$ Now we will add $1$ because of the $0$ to account for the non-negative solutions.
So our final answer is $3280 + 1 = 3281$ which is $\boxed{3281}$

Solution: Obviously, there are $3^8 = 6561$ possible, and one of them is 0, so other $6560$ are either positive or negative. By the symmetry, we can get the answer is $6560/2 + 1$ which is $\boxed{3281}$

Solution: The solution has to obviously be smaller than $3^{7-0+1}=3^8$ but larger than $3^{(7-0+1)-1} = 3^7.$ The only remaining answer is $\boxed{3281}.$ ~mathboy282


Answer: 4.0

How many nonzero complex numbers $z$ have the property that $0, z,$ and $z^3,$ when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}$

Solution:
Convert $z$ and $z^3$ into modulus-argument (polar) form, giving $z=r\text{cis}(\theta)$ for some $r$ and $\theta$ . Thus, by De Moivre's Theorem, $z^3=r^3\text{cis}(3\theta)$ . Since the distance from $0$ to $z$ is $r$ , and the triangle is equilateral, the distance from $0$ to $z^3$ must also be $r$ , so $r^3=r$ , giving $r=1$ . (We know $r \neq 0$ since the problem statement specifies that $z$ must be nonzero.)
Now, to get from $z$ to $z^3$ , which should be a rotation of $60^{\circ}$ if the triangle is equilateral, we multiply by $z^2 = r^2\text{cis}(2\theta)$ , again using De Moivre's Theorem. Thus we require $2\theta=\pm\frac{\pi}{3} + 2\pi k$ (where $k$ can be any integer). If $0 < \theta < \frac{\pi}{2}$ , we must have $\theta=\frac{\pi}{6}$ , while if $\frac{\pi}{2} \leq \theta < \pi$ , we must have $\theta = \frac{5\pi}{6}$ . Hence there are $2$ values that work for $0 < \theta < \pi$ . By symmetry, the interval $\pi \leq \theta < 2\pi$ will also give $2$ solutions. The answer is thus $2 + 2 = \boxed{4}$

Solution: As before, $r=1$ . Represent $z$ in polar form. By De Moivre's Theorem, $z^3=\text{cis}(3\theta)$ . To form an equilateral triangle, their difference in angle must be $\frac{\pi}{3}$ , so \[\frac{\text{cis}(3\theta)}{\text{cis}(\theta)}=\text{cis}(2\theta)=\text{cis}(\pm\frac{\pi}{3})\] From the polar form of $z$ , we know that $0\geq\theta\leq2\pi$ , so $\text{cis}(2\theta)$ cycles in a circle twice. By contrast, $\pm\frac{\pi}{3}$ represent $2$ fixed, distinct points. Thus, $\text{cis}(2\theta)$ intersects these points twice each $\implies\boxed{4}$

Solution:
For the triangle to be equilateral, the vector from $z$ to $z^3$ , i.e $z^3 - z$ , must be a $60^{\circ}$ rotation of the vector from $0$ to $z$ , i.e. just $z$ . Thus we must have
\[\frac{(z^3-z)}{(z-0)}=\text{cis}{(\pi/3)} \text{ or } \text{cis}(5\pi/3)\]
Simplifying gives \[z^2-1= \text{cis}(\pi/3) \text{ or } z^2-1= \text{cis}(5\pi/3)\] so \[z^2=1+\text{cis}(\pi/3) \text{ or } z^2=1+\text{cis}(5\pi/3)\]
Since any nonzero complex number will have two square roots, each equation gives two solutions. Thus, as before, the total number of possible values of $z$ is $\boxed{4}$

Solution:
Since the complex numbers $0,z,$ and $z^3$ form an equilateral triangle in the complex plane, we note that either $z^3$ is a 60 degrees counterclockwise rotation about the origin from $z$ or $z$ is a 60 degrees counterclockwise rotation about the origin from $z^3$
Therefore, we note that either $z^3 = z \text{cis} 60^\circ{}$ or $z \text{cis}(-60^\circ{}) = z^3$
The first equation in $z$ (meaning $z^3 = z \text{cis} 60^\circ{}$ ) gives us: $z^2 = cis 60^\circ{}$ , which gives 2 solutions in $z$
The second equation in $z$ (which is $z \text{cis} (-60^\circ{}) = z^3$ ) gives us $z^2 = \text{cis} (-60^\circ{})$ , which must give another 2 solutions in $z$
Therefore, there are $\boxed{4}$ solutions for $z$ . (Professor-Mom)

Solution:
Let the $z = re^{i\theta}$ , so $z^3 = r^3e^{3i\theta}$ . To have an equilateral triangle, we must have $|z^3| = |z|$ , so $r^3 = r$ , so $r= 1$
Note that the angle between $z^3$ and $z$ is $3\theta - \theta = 2\theta$ . Then, by the Law of Cosines,
\[|z^3 - z|^2 = |z|^2 + |z^3|^2 - 2|z||z^3|\cos 2\theta.\]
Since we have an equilateral triangle, it must be that $|z^3 - z| =|z| = |z^3| = r = 1$ . Hence, \[1 = 2 - 2\cos 2\theta = 2 - 2(1-2\sin^2\theta) = 4\sin^2\theta,\] \[\sin^2\theta = \frac{1}{4}\] \[\sin\theta = \pm\frac{1}{2}\] \[\theta = \pm\frac{\pi}{6}, \pm\frac{5\pi}{6}.\]
These $4$ values of $\theta$ correspond to $\boxed{4}$ distinct values of $z$ .
~ brainfertilzer


Answer: 835.0

How many numbers between $1$ and $2005$ are integer multiples of $3$ or $4$ but not $12$
$\textbf{(A) } 501 \qquad \textbf{(B) } 668 \qquad \textbf{(C) } 835 \qquad \textbf{(D) } 1002 \qquad \textbf{(E) } 1169$

Solution: To find the multiples of $3$ or $4$ but not $12$ , you need to find the number of multiples of $3$ and $4$ , and then subtract twice the number of multiples of $12$ , because you overcount and do not want to include them. The multiples of $3$ are $\frac{2005}{3} = 668\text{ }R1.$ The multiples of $4$ are $\frac{2005}{4} = 501 \text{ }R1$ . The multiples of $12$ are $\frac{2005}{12} = 167\text{ }R1.$ So, the answer is $668+501-167-167 = \boxed{835}$

Solution: From $1$ $12$ , the multiples of $3$ or $4$ but not $12$ are $3, 4, 6, 8,$ and $9$ , a total of five numbers. Since $\frac{5}{12}$ of positive integers are multiples of $3$ or $4$ but not $12$ from $1$ $12$ , the answer is approximately $\frac{5}{12} \cdot 2005$ $\boxed{835}$


Answer: 96.0

How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$

Solution:
Let $\underline{ABC}$ be one such odd positive $3$ -digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$
As $A\in\{1,2,4,5,6,7,8,9\}$ and $C\in\{1,5,7,9\},$ there are $8$ possibilities for $A$ and $4$ possibilities for $C.$ Note that each ordered pair $(A,C)$ determines the value of $B$ modulo $3,$ so $B$ can be any element in one of the sets $\{0,6,9\},\{1,4,7\},$ or $\{2,5,8\}.$ We conclude that there are always $3$ possibilities for $B.$
By the Multiplication Principle, the answer is $8\cdot4\cdot3=\boxed{96}.$

Solution:
Let $\underline{ABC}$ be one such odd positive $3$ -digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$
As $A\in\{1,2,4,5,6,7,8,9\},B\in\{0,1,2,4,5,6,7,8,9\},$ and $C\in\{1,5,7,9\},$ note that:
We apply casework to $A+B+C\equiv0\pmod3:$ \[\begin{array}{c|c|c||l} & & & \\ [-2.5ex] \boldsymbol{A\operatorname{mod}3} & \boldsymbol{B\operatorname{mod}3} & \boldsymbol{C\operatorname{mod}3} & \multicolumn{1}{c}{\textbf{Count}} \\ [0.5ex] \hline & & &  \\ [-2ex] 0 & 0 & 0 & 2\cdot3\cdot1=6 \\ 0 & 1 & 2 & 2\cdot3\cdot1=6 \\ 0 & 2 & 1 & 2\cdot3\cdot2=12 \\ 1 & 0 & 2 & 3\cdot3\cdot1=9 \\ 1 & 1 & 1 & 3\cdot3\cdot2=18 \\ 1 & 2 & 0 & 3\cdot3\cdot1=9 \\ 2 & 0 & 1 & 3\cdot3\cdot2=18 \\ 2 & 1 & 0 & 3\cdot3\cdot1=9 \\ 2 & 2 & 2 & 3\cdot3\cdot1=9 \end{array}\] Together, the answer is $6+6+12+9+18+9+18+9+9=\boxed{96}.$

Solution:
Analyze that the three-digit integers divisible by $3$ start from $102.$ In the $200$ 's, it starts from $201.$ In the $300$ 's, it starts from $300.$ We see that the units digits is $0, 1,$ and $2.$
Write out the $1$ - and $2$ -digit multiples of $3$ starting from $0, 1,$ and $2.$ Count up the ones that meet the conditions. Then, add up and multiply by $3,$ since there are three sets of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$ 's ll contain the digit $3.$
Together, the answer is $3(12+12+12)-12=\boxed{96}.$

Solution: Consider the number of $2$ -digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$ -digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$ -digit number. However, we have $7 \equiv 1\pmod{3},$ and thus we need to count any $2$ -digit number $\equiv 2\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \boxed{96}.$

Solution: We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$ -digit numbers. Exactly $\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\cdot\frac{1}{3}=150.$ Of these $150$ numbers, $\frac{4}{5}$ $\textbf{do not}$ have $3$ in their ones (units) digit, $\frac{9}{10}$ $\textbf{do not}$ have $3$ in their tens digit, and $\frac{8}{9}$ $\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$ -digit integers is \[900\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{4}{5}\cdot\frac{9}{10}\cdot\frac{8}{9}=\boxed{96}.\]

Solution:
We will start with the numbers that could work. This numbers include _ _ $1$ , _ _ $5$ , _ _ $7$ , _ _ $9$ . Let's work case by case.
Case $1$ : _ _ $1$ : The two blanks could be any number that is $2$ mod $3$ that does not include $3$ . We have $24$ cases for this case (we could count every case).
Case $2$ : _ _ $5$ : The $2$ blanks could be any number that is $1$ mod $3$ that does not include $3$ . But we could see that this case has exactly the same solutions to case $1$ because we have a $1-1$ correspondence. We can do the exact same for case $3$
Cases $4$ : _ _ $9$ : We need the blanks to be a multiple of $3$ , but does not contain 3. We have $(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87, 90, 96, 99)$ which also contains $24$ numbers. Therefore, we have $24 \cdot 4$ which is equal to $\boxed{96}.$

Solution:
This problem is solvable by inclusion exclusion principle. There are $\frac{999-105}{6} + 1 = 150$ odd $3$ -digit numbers divisible by $3$ . We consider the number of $3$ -digit numbers divisible by $3$ that contain either $1, 2$ or $3$ digits of $3$
For $\underline{AB3}$ $AB$ is any $2$ -digit number divisible by $3$ , which gives us $\frac{99-12}{3} + 1 = 30$ . For $\underline{A3B}$ , for each odd $B$ , we have $3$ values of $A$ that give a valid case, thus we have $5(3) = 15$ cases. For $\underline{3AB}$ , we also have $15$ cases, but when $B=3, 9$ $A$ can equal $0$ , so we have $17$ cases.
For $\underline{A33}$ , we have $3$ cases. For $\underline{3A3}$ , we have $4$ cases. For $\underline{33A}$ , we have $2$ cases. Finally, there is just one case for $\underline{333}$
By inclusion exclusion principle, we get $150 - 30 - 15 - 17 + 3 + 4 + 2 - 1 = \boxed{96}$ numbers.


Answer: 469.0

How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0$
$\textbf{(A)}\ 469\qquad\textbf{(B)}\ 471\qquad\textbf{(C)}\ 475\qquad\textbf{(D)}\ 478\qquad\textbf{(E)}\ 481$

Solution: We can use complementary counting. There are $2017$ positive integers in total to consider, and there are $9$ one-digit integers, $9 \cdot 9 = 81$ two digit integers without a zero, $9 \cdot 9 \cdot 9 = 729$ three digit integers without a zero, and $9 \cdot 9 \cdot 9 = 729$ four-digit integers starting with a 1 without a zero. Therefore, the answer is $2017 - 9 - 81 - 729 - 729 = \boxed{469}$

Solution: First, we notice there are no one-digit numbers that contain a zero. There are $9$ two-digit integers and $9 \cdot 9 + 9 \cdot 9 + 9 = 171$ three-digit integers containing at least one zero. Next, we consider the four-digit integers beginning with one. There are $3 \cdot 9 \cdot 9 = 243$ of these four-digit integers with one zero, ${3 \choose 2} \cdot 9 = 27$ with two zeros, and $1$ with three zeros $(1000)$ . Finally, we consider the numbers $2000$ to $2017$ which all contain at least one zero. Adding all of these together we get $9 + 171 + 243 + 27 + 1 + 18 = \boxed{469}$


Answer: 505.0

How many of the first $2018$ numbers in the sequence $101, 1001, 10001, 100001, \dots$ are divisible by $101$
$\textbf{(A) }253 \qquad \textbf{(B) }504 \qquad \textbf{(C) }505 \qquad \textbf{(D) }506  \qquad \textbf{(E) }1009 \qquad$

Solution:
The number $10^n+1$ is divisible by 101 if and only if $10^n\equiv -1\pmod{101}$ . We note that $(10,10^2,10^3,10^4)\equiv (10,-1,-10,1)\pmod{101}$ , so the powers of 10 are 4-periodic mod 101.
It follows that $10^n\equiv -1\pmod{101}$ if and only if $n\equiv 2\pmod 4$
In the given list, $10^2+1,10^3+1,10^4+1,\dots,10^{2019}+1$ , the desired exponents are $2,6,10,\dots,2018$ , and there are $\dfrac{2020}{4}=\boxed{505}$ numbers in that list.

Solution: Note that $10^{2k}+1$ for some odd $k$ will suffice $\mod {101}$ . Each $2k \in \{2,6,10,\dots,2018\}$ , so the answer is $\boxed{505}$

Solution: If we divide each number by $101$ , we see a pattern occuring in every 4 numbers. $101, 1000001, 10000000001, \dots$ . We divide $2018$ by $4$ to get $504$ with $2$ left over. Looking at our pattern of four numbers from above, the first number is divisible by $101$ . This means that the first of the $2$ left over will be divisible by $101$ , so our answer is $\boxed{505}$

Solution: Note that $909$ is divisible by $101$ , and thus $9999$ is too. We know that $101$ is divisible and $1001$ isn't so let us start from $10001$ . We subtract $9999$ to get 2. Likewise from $100001$ we subtract, but we instead subtract $9999$ times $10$ or $99990$ to get $11$ . We do it again and multiply the 9's by $10$ to get $101$ . Following the same knowledge, we can use mod $101$ to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is ${0,-9,-99 ( 2),11, 0, ...}$ . Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide $2017$ by four to get $504$ remainder $1$ . Thus the answer is $504$ plus the 1st term or $\boxed{505}$

Solution: Note that $101=x^2+1$ and $100...0001=x^n+1$ , where $x=10$ . We have that $\frac{x^n+1}{x^2+1}$ must have a remainder of $0$ . By the remainder theorem, the roots of $x^2+1$ must also be roots of $x^n+1$ . Plugging in $i,-i$ to $x^n+1$ yields that $n\equiv2\mod{4}$ . Because the sequence starts with $10^2+1$ , the answer is $\lceil 2018/4 \rceil=\boxed{505}$


Answer: 600.0

How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$
where $x$ is a real number , and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$

Solution:
Noting that all of the numbers are even, we can reduce this to any real number $x$ between $0$ to $\frac 12$ , as this will be equivalent to $\frac n2$ to $\frac {n+1}2$ for any integer $n$ (same reasoning as above). So now we only need to test every 10 numbers; and our answer will be 100 times the number of integers we can reach between 1 and 10.
We can now approach this by directly searching for the integers (this solution) or brute forcing all of the cases (next solution):
We can match up the greatest integer functions with one of the partitions of the integer. If we let $x = \frac 12$ then we get the solution $10$ ; now consider when $x < \frac 12$ $\lfloor 2x \rfloor = 0$ $\lfloor 4x \rfloor \le 1$ $\lfloor 6x \rfloor \le 2$ $\lfloor 8x \rfloor \le 3$ . But according to this the maximum we can get is $1+2+3 = 6$ , so we only need to try the first 6 numbers.
Out of these 6 cases, only 3 fails. So between 1 and 10 we can reach only the integers $1,2,4,5,6,10$ ; hence our solution is $6 \cdot 100 = \boxed{600}$

Solution:
As we change the value of $x$ , the value of our expression changes only when $x$ crosses rational number of the form $\frac{m}{n}$ , where $n$ is divisible by 2, 4, 6 or 8.  Thus, we need only see what happens at the numbers of the form $\frac{m}{\textrm{lcm}(2, 4, 6, 8)} = \frac{m}{24}$ .  This gives us 24 calculations to make; we summarize the results here:
$\frac{1}{24}, \frac{2}{24} \to 0$
$\frac{3}{24} \to 1$
$\frac{4}{24}, \frac{5}{24} \to 2$
$\frac{6}{24}, \frac{7}{24} \to 4$
$\frac{8}{24} \to 5$
$\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to 6$
$\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to 10$
$\frac{15}{24} \to 11$
$\frac{16}{24},\frac{17}{24} \to 12$
$\frac{18}{24}, \frac{19}{24} \to 14$
$\frac{20}{24}\to 15$
$\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to16$
$\frac{24}{24} \to 20$
Thus, we hit 12 of the first 20 integers and so we hit $50 \cdot 12 = \boxed{600}$ of the first $1000$

Solution: Because $2,4,6,8$ are all multiples of $2$ , we can speed things up. We only need to check up to $\frac{12}{24}$ , and the rest should repeat. As shown before, we hit 6 integers ( $1,2,4,5,6,10$ ) from $\frac{1}{24}$ to $\frac{12}{24}$ . Similarly, this should repeat 100 times, for $\boxed{600}$

Solution: We only need to check the numbers where it increments, namely $\frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}$ . As shown before, we hit 6 integers ( $1,2,4,5,6,10$ ) from $\frac{1}{24}$ to $\frac{1}{2}$ . Similarly, this should repeat 100 times, for $\boxed{600}$

Solution: Recall from Hermite's Identity that $\sum_{k = 0}^{n - 1}\left\lfloor x + \frac kn\right\rfloor = \lfloor nx\rfloor$ . Then we can rewrite $\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor = 4\lfloor x\rfloor + \left\lfloor x + \frac18\right\rfloor + \left\lfloor x + \frac16\right\rfloor + 2\left\lfloor x + \frac14\right\rfloor + \left\lfloor x + \frac13\right\rfloor$ $+ \left\lfloor x + \frac38\right\rfloor + 4\left\lfloor x + \frac12\right\rfloor + \left\lfloor x + \frac58\right\rfloor + \left\lfloor x + \frac23\right\rfloor + 2\left\lfloor x + \frac34\right\rfloor + \left\lfloor x + \frac56\right\rfloor + \left\lfloor x + \frac78\right\rfloor$ . There are $12$ terms here (we don't actually have to write all of it out; we can just see where there will be duplicates and subtract accordingly from $20$ ). Starting from every integer $x$ , we can keep adding to achieve one higher value for each of these terms, but after raising the last term, we will have raised the whole sum by $20$ while only achieving $12$ of those $20$ values. We can conveniently shift the $1000$ (since it can be achieved) to the position of the $0$ so that there are only complete cycles of $20$ , and the answer is $\frac {12}{20}\cdot1000 = \boxed{600}$

Solution:
Let $x=\lfloor x\rfloor+\{x\}$ then \begin{align*} \lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor&=\lfloor 2(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 4(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 6(\lfloor x\rfloor+\{x\})\rfloor+\lfloor 8(\lfloor x\rfloor+\{x\})\rfloor\\ &=2\lfloor x\rfloor+4\lfloor x\rfloor+6\lfloor x\rfloor+8\lfloor x\rfloor+\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor\\ &=20\lfloor x\rfloor+(\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor) \end{align*} Similar to the previous solutions, the value of $\lfloor 2\{x\}\rfloor+\lfloor 4\{x\}\rfloor+\lfloor 6\{x\}\rfloor+\lfloor 8\{x\}\rfloor$ changes when $\{x\}=\frac{m}{n}$ , where $m\in\{1,2,3,...,n-1\}$ $n\in\{2,4,6,8\}$ . Using Euler's Totient Function \[\sum\limits_{k=0}^4 \phi(2k)\] to obtain $12$ different values for $\{x\}=\frac{m}{n}$ . (note that here Euler's Totient Function counts the number of $\{x\}=\frac{m}{n}$ where $m$ $n$ are relatively prime so that the values of $\{x\}$ won't overlap.).
Thus if $k$ can be expressed as $\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 6x\rfloor+\lfloor 8x\rfloor$ , then $k=20a+b$ for some non-negative integers $a$ $b$ , where there are $12$ values for $b$
Exclusively, there are $49$ values for $a$ in the range $0<k<1000$ , or $49\cdot12=588$ ordered pairs $(a,b)$
If $a=0$ $b\neq0$ , which includes $11$ ordered pairs.
If $a=50$ $b=0$ , which includes $1$ ordered pair.
In total, there are $588+11+1=\boxed{600}$ values for $k$

Solution:
To simplify the question, let $y = 2x$ . Then, the expression in the question becomes $\lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor$
Let $\{x\}$ represent the non-integer part of $x$ (For example, $\{2.8\} = 0.8$ ). Then,
\begin{align*} \lfloor y \rfloor + \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor &= y - \{y\} + 2y - \{2y\} + 3y - \{3y\} + 4y - \{4y\} \\ &= 10y - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10(\lfloor y \rfloor + \{y\}) - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10\lfloor y \rfloor + 10\{y\} - (\{y\} + \{2y\} + \{3y\} + \{4y\}) \\ &= 10\lfloor y \rfloor + 9\{y\} - (\{2y\} + \{3y\} + \{4y\}) \\ \end{align*}
Since $\lfloor y \rfloor$ is always an integer, $10\lfloor y \rfloor$ will be a multiple of 10. Thus, we look for the range of the other part of the expression. We will be able to reach the same numbers when $y$ ranges from $0$ to $1$ , because the curly brackets ( $\{\}$ ) gets rid of any integer part. Let the combined integer part of $2y$ $3y$ , and $4y$ be $k$ (In other words, $k = \lfloor 2y \rfloor + \lfloor 3y \rfloor + \lfloor 4y \rfloor$ ). Then,
\begin{align*} 9\{y\} - (\{2y\} + \{3y\} + \{4y\}) &= 9\{y\} - (2\{y\} + 3\{y\} + 4\{y\} - k) \\ &= 9\{y\} - (9\{y\} - k) \\ &= k \end{align*}
The maximum value of $k$ will be when $y$ is slightly less than $1$ , which means $k = 1 + 2 + 3 = 6$ . As $y$ increases from $0$ to $1$ $k$ will increase whenever $2y$ $3y$ , or $4y$ is an integer, which happens when $y$ hits one of the numbers in the set $\left\{\dfrac14, \dfrac13, \dfrac12, \dfrac23, \dfrac34 \right\}$ . When $y$ reaches $\dfrac12$ , both $2y$ and $4y$ will be an integer, so $k$ will increase by $2$ . For all the other numbers in the set, $k$ increases by $1$ since only $1$ number in the set will be an integer. Thus, the possible values of $k$ are $\{0, 1, 2, 4, 5, 6\}$
Finally, looking back at the original expression, we plug in $k$ to get a multiple of $10$ plus any number in $\{0, 1, 2, 4, 5, 6\}$ . Thus, we hit all numbers ending in $\{0, 1, 2, 4, 5, 6\}$ , of which there are $\boxed{600}$


Answer: 0.0

How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$

Solution: The $n$ th term of this sequence is \[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\] It follows that the terms are \begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*} Therefore, there are $\boxed{0}$ prime numbers in this sequence.

Solution: Denote this sequence as $a_{n}$ , then we can find that \begin{align*} a_{1} &= 121 = 10^2 + 2\cdot10 + 1 = (10^2 + 10) + (10 + 1), \\ a_{2} &= 11211 = (10^4 + 10^3 + 10^2) + (10^2 + 10 + 1), \\ a_{3} &= 1112111 = (10^6 + 10^5 + 10^4 + 10^3) + (10^3 + 10^2 + 10 + 1), \\ & \ \vdots \end{align*} So, we can induct that the general term is \begin{align*} a_n &= (10^{2n} + 10^{2n-1} + \ldots + 10^{n+1} + 10^n) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= 10^n\cdot(10^n + 10^{n-1} + \ldots +10 + 1) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= \left(10^n+1\right)\sum_{k=0}^{n}10^k. \end{align*} Therefore, there are $\boxed{0}$ prime numbers in this sequence.

Solution: Observe how \begin{align*} 121 &= 110 + 11, \\ 11211 &= 11100 + 111, \\ 1112111 &= 1111000 + 1111, \\ & \ \vdots \end{align*} all take the form of \[\underbrace{111\ldots}_{n+1}\underbrace{00\ldots}_{n} + \underbrace{111\ldots}_{n+1} = \underbrace{111\ldots}_{n+1}(10^{n} + 1).\] Factoring each of the sums, we have \[11(10+1), 111(100+1), 1111(1000+1), \ldots\] respectively. With each number factored, there are $\boxed{0}$ primes in the set.

Solution: Note that $121$ is divisible by $11$ and $11211$ is divisible by $3$ . Because this is Problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is $\boxed{0}.$


Answer: 0.0

Solution: The $n$ th term of this sequence is \[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\] It follows that the terms are \begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*} Therefore, there are $\boxed{0}$ prime numbers in this sequence.

Solution: Denote this sequence as $a_{n}$ , then we can find that \begin{align*} a_{1} &= 121 = 10^2 + 2\cdot10 + 1 = (10^2 + 10) + (10 + 1), \\ a_{2} &= 11211 = (10^4 + 10^3 + 10^2) + (10^2 + 10 + 1), \\ a_{3} &= 1112111 = (10^6 + 10^5 + 10^4 + 10^3) + (10^3 + 10^2 + 10 + 1), \\ & \ \vdots \end{align*} So, we can induct that the general term is \begin{align*} a_n &= (10^{2n} + 10^{2n-1} + \ldots + 10^{n+1} + 10^n) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= 10^n\cdot(10^n + 10^{n-1} + \ldots +10 + 1) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= \left(10^n+1\right)\sum_{k=0}^{n}10^k. \end{align*} Therefore, there are $\boxed{0}$ prime numbers in this sequence.

Solution: Observe how \begin{align*} 121 &= 110 + 11, \\ 11211 &= 11100 + 111, \\ 1112111 &= 1111000 + 1111, \\ & \ \vdots \end{align*} all take the form of \[\underbrace{111\ldots}_{n+1}\underbrace{00\ldots}_{n} + \underbrace{111\ldots}_{n+1} = \underbrace{111\ldots}_{n+1}(10^{n} + 1).\] Factoring each of the sums, we have \[11(10+1), 111(100+1), 1111(1000+1), \ldots\] respectively. With each number factored, there are $\boxed{0}$ primes in the set.

Solution: Note that $121$ is divisible by $11$ and $11211$ is divisible by $3$ . Because this is Problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is $\boxed{0}.$


Answer: 1.0

How many of the following are equal to $x^x+x^x$ for all $x>0$
$\textbf{I:}\ 2x^x \qquad\textbf{II:}\ x^{2x} \qquad\textbf{III:}\ (2x)^x \qquad\textbf{IV:}\ (2x)^{2x}$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution:
We look at each statement individually.
$\textbf{I:}\ 2x^x$ . We note that $x^x+x^x=x^x(1+1)=2x^x$ . So statement $\textbf{I}$ is true.
$\textbf{II:}\ x^{2x}$ . We find a counter example which is $x=1$ $2\neq 1$ . So statement $\textbf{II}$ is false.
$\textbf{III:}\ (2x)^x$ . We see that this statement is equal to $2^xx^x$ $x=2$ is a counter example. $8\neq 16$ . So statement $\textbf{III}$ is false.
$\textbf{IV:}\ (2x)^{2x}$ . We see that $x=1$ is again a counter example. $2\neq 4$ . So statement $\textbf{IV}$ is false.
Therefore, our answer is $\boxed{1}$


Answer: 750.0

How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?

Solution: Notice that all odd numbers can be obtained by using $(a+1)^2-a^2=2a+1,$ where $a$ is a nonnegative integer. All multiples of $4$ can be obtained by using $(b+1)^2-(b-1)^2 = 4b$ , where $b$ is a positive integer. Numbers congruent to $2 \pmod 4$ cannot be obtained because squares are $0, 1 \pmod 4.$ Thus, the answer is $500+250 = \boxed{750}.$


Answer: 6.0

How many of the twelve pentominoes pictured below have at least one line of reflectional symmetry?
[asy] unitsize(5mm); defaultpen(linewidth(1pt)); draw(shift(2,0)*unitsquare); draw(shift(2,1)*unitsquare); draw(shift(2,2)*unitsquare); draw(shift(1,2)*unitsquare); draw(shift(0,2)*unitsquare); draw(shift(2,4)*unitsquare); draw(shift(2,5)*unitsquare); draw(shift(2,6)*unitsquare); draw(shift(1,5)*unitsquare); draw(shift(0,5)*unitsquare); draw(shift(4,8)*unitsquare); draw(shift(3,8)*unitsquare); draw(shift(2,8)*unitsquare); draw(shift(1,8)*unitsquare); draw(shift(0,8)*unitsquare); draw(shift(6,8)*unitsquare); draw(shift(7,8)*unitsquare); draw(shift(8,8)*unitsquare); draw(shift(9,8)*unitsquare); draw(shift(9,9)*unitsquare); draw(shift(6,5)*unitsquare); draw(shift(7,5)*unitsquare); draw(shift(8,5)*unitsquare); draw(shift(7,6)*unitsquare); draw(shift(7,4)*unitsquare); draw(shift(6,1)*unitsquare); draw(shift(7,1)*unitsquare); draw(shift(8,1)*unitsquare); draw(shift(6,0)*unitsquare); draw(shift(7,2)*unitsquare); draw(shift(11,8)*unitsquare); draw(shift(12,8)*unitsquare); draw(shift(13,8)*unitsquare); draw(shift(14,8)*unitsquare); draw(shift(13,9)*unitsquare); draw(shift(11,5)*unitsquare); draw(shift(12,5)*unitsquare); draw(shift(13,5)*unitsquare); draw(shift(11,6)*unitsquare); draw(shift(13,4)*unitsquare); draw(shift(11,1)*unitsquare); draw(shift(12,1)*unitsquare); draw(shift(13,1)*unitsquare); draw(shift(13,2)*unitsquare); draw(shift(14,2)*unitsquare); draw(shift(16,8)*unitsquare); draw(shift(17,8)*unitsquare); draw(shift(18,8)*unitsquare); draw(shift(17,9)*unitsquare); draw(shift(18,9)*unitsquare); draw(shift(16,5)*unitsquare); draw(shift(17,6)*unitsquare); draw(shift(18,5)*unitsquare); draw(shift(16,6)*unitsquare); draw(shift(18,6)*unitsquare); draw(shift(16,0)*unitsquare); draw(shift(17,0)*unitsquare); draw(shift(17,1)*unitsquare); draw(shift(18,1)*unitsquare); draw(shift(18,2)*unitsquare);[/asy]
$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$

Solution:
Pentonimoes.gif
The ones with lines over the shapes have at least one line of symmetry. Counting the number of shapes that have line(s) on them,
we find $\boxed{6}$ pentominoes.


Answer: 870.0

How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$

Solution:
Let $k = a + d = b + c$ so $d = k-a, b=k-c$ . It follows that $(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93$ . Hence $(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)$
Solve them in terms of $c$ to get $(a,b,c,d) = (c - 93,c - 92,c,c + 1),$ $(c - 31,c - 28,c,c + 3),$ $(c - 1,c + 92,c,c + 93),$ $(c - 3,c + 28,c,c + 31)$ . The last two solutions don't follow $a < b < c < d$ , so we only need to consider the first two solutions.
The first solution gives us $c - 93\geq 1$ and $c + 1\leq 499$ $\implies 94\leq c\leq 498$ , and the second one gives us $32\leq c\leq 496$
So the total number of such quadruples is $405 + 465 = \boxed{870}$

Solution:
Let $b = a + m$ and $c = a + m + n$ . From $a + d = b + c$ $d = b + c - a = a + 2m + n$
Substituting $b = a + m$ $c = a + m + n$ , and $d = b + c - a = a + 2m + n$ into $bc - ad = 93$ \[bc - ad = (a + m)(a + m + n) - a(a + 2m + n) = m(m + n). = 93 = 3(31)\] Hence, $(m,n) = (1,92)$ or $(3,28)$
For $(m,n) = (1,92)$ , we know that $0 < a < a + 1 < a + 93 < a + 94 < 500$ , so there are $405$ four-tuples. For $(m,n) = (3,28)$ $0 < a < a + 3 < a + 31 < a + 34 < 500$ , and there are $465$ four-tuples. In total, we have $405 + 465 = \boxed{870}$ four-tuples.

Solution:
Square both sides of the first equation in order to get $bc$ and $ad$ terms, which we can plug $93$ in for. \begin{align*} (a+d)^2 = (b+c)^2 &\implies a^2 + 2ad + d^2 = b^2 + 2bc + c^2 \\ &\implies 2bc-2ad = a^2-b^2 + d^2-c^2 \\ &\implies 2(bc-ad) = (a-b)(a+b)+(d-c)(d+c) \end{align*} We can plug $93$ in for $bc - ad$ to get $186$ on the left side, and also observe that $a-b = c-d$ after rearranging the first equation. Plug in $c-d$ for $a-b$
$186 = (c-d)(a+b) + (d-c)(d+c) \implies 186 = -(d-c)(a+b) + (d-c)(d+c) \implies 186 = (d-c)(d+c-a-b)$
Now observe the possible factors of $186$ , which are $1 \cdot 186, 2\cdot 93, 3 \cdot 62, 6\cdot 31$ $(d-c)$ and $(d+c-a-b)$ must be factors of $186$ , and $(d+c-a-b)$ must be greater than $(d-c)$
$1 \cdot 186$ work, and yields $405$ possible solutions. $2 \cdot 93$ does not work, because if $c-d = 2$ , then $a+b$ must differ by 2 as well, but an odd number $93$ can only result from two numbers of different parity. $c-d$ will be even, and $a+b$ will be even, so $c+d - (a+b)$ must be even. $3 \cdot 62$ works, and yields $465$ possible solutions, while $6 \cdot 31$ fails for the same reasoning above.
Thus, the answer is $405 + 465 = \boxed{870}$

Solution:
Add the two conditions together to get $a+d+ad+93=b+c+bc$ . Rearranging and factorising with SFFT, $(a+1)(d+1)+93=(b+1)(c+1)$ . This implies that for every quadruple $(a,b,c,d)$ , we can replace $a\longrightarrow a+1$ $b\longrightarrow b+1$ , etc. and this will still produce a valid quadruple. This means, that we can fix $a=1$ , and then just repeatedly add $1$ to get the other quadruples.
Now, our conditions are $b+c=d+1$ and $bc=d+93$ . Replacing $d$ in the first equation, we get $bc-b-c=92$ . Factorising again with SFFT gives $(b-1)(c-1)=93$ . Since $b<c$ , we have two possible cases to consider.
Case 1: $b=2$ $c=94$ . This produces the quadruple $(1,2,94,95)$ , which indeed works.
Case 2: $b=4$ $c=32$ . This produces the quadruple $(1,4,32,35)$ , which indeed works.
Now, for case 1, we can add $1$ to each term exactly $404$ times (until we get the quadruple $(405,406,498,499)$ ), until we violate $d<500$ . This gives $405$ quadruples for case 1.
For case 2, we can add $1$ to each term exactly $464$ times (until we get the quadruple $(465,468,496,499)$ ). this gives $465$ quadruples for case 2.
In conclusion, having exhausted all cases, we can finish. There are hence $405+465=\boxed{870}$ possible quadruples.

Solution:
Let $r = d-c$ . From the equation $a+d = b+c$ , we have \[r = d-c = b-a ,\] so $b = a+r$ and $c = d-r$ . We then have \[93 = (a+r)(d-r) - ad = rd - ra - r^2 = r(d-a-r) .\] Since $c > b$ $d-r > a+r$ , or $d-a-r > r$ . Since the prime factorization of 93 is $3 \cdot 31$ , we must either have $r=1$ and $d-a-r = 93$ , or $r=3$ and $d-a-r = 31$ . We consider these cases separately.
If $r=1$ , then $d-a = 94$ $b= a+1$ , and $c = d-1$ . Thus $d$ can be any integer between 95 and 499, inclusive, and our choice of $d$ determines the four-tuple $(a,b,c,d)$ . We therefore have $499-95+1 = 405$ possibilities in this case.
If $r=3$ , then $d-a = 34$ $b = a+3$ , and $c=d-3$ . Thus $d$ can be any integer between 35 and 499, inclusive, and our choice of $d$ determines the four-tuple $(a,b,c,d)$ , as before. We therefore have $499-35+1 = 465$ possibilities in this case.
Since there are 405 possibilities in the first case and 465 possibilities in the second case, in total there are $405 + 465 = \boxed{870}$ four-tuples.

Solution:
Assume $d = x+m, a = x-m, c = x+n$ , and $b = x-n$ . This clearly satisfies the condition that $a+d = b+c$ since ( $2x = 2x$ ) . Now plug this into $bc-ad = 93$ . You get $(x+n)(x-n) - (x+m)(x-m) = 93 \Rightarrow m^2 - n^2 = 93 \Rightarrow (m-n)(m+n) = 93$
Since $m>n$ (as given by the condition that $a<b<c<d$ ), $m+n>m-n$ and $m$ and $n$ are integers, there are two cases we have to consider since $93 = 3\cdot 31$ . We first have to consider $m-n = 1, m+n = 93$ , and then consider $m-n=3, m+n = 31$
In the first case, we get $m=47, n=46$ and in the second case we get $m=17, n=14$ . Now plug these values (in separate cases) back into $a,b,c,d$ . Since the only restriction is that all numbers have to be greater than $0$ or less than $500$ , we can write two inequalities. Either $x+47 < 500, x-47 > 0$ , or $x+17 < 500, x-17 > 0$ (using the inequalities given by $d$ and $a$ , and since $b$ and $c$ are squeezed in between $d$ and $a$ , we only have to consider these two inequalities).
This gives us either $47 < x < 453$ or $17 < x < 483$ , and using simple counting, there are $405$ values for $x$ in the first case and $465$ values for $x$ in the second case, and hence our answer is $405+465 = \boxed{870}$


Answer: 2.0

How many ordered pairs $(a, b)$ of positive integers satisfy the equation \[a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),\] where $\text{gcd}(a,b)$ denotes the greatest common divisor of $a$ and $b$ , and $\text{lcm}(a,b)$ denotes their least common multiple?
$\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}$

Solution:
Let $x = \text{lcm}(a, b)$ , and $y = \text{gcd}(a, b)$ . Therefore, $a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y$ . Thus, the equation becomes
\[x\cdot y + 63 = 20x + 12y\] \[x\cdot y - 20x - 12y + 63 = 0\]
Using Simon's Favorite Factoring Trick , we rewrite this equation as
\[(x - 12)(y - 20) - 240 + 63 = 0\] \[(x - 12)(y - 20) = 177\]
Since $177 = 3\cdot 59$ and $x > y$ , we have $x  - 12 = 59$ and $y - 20 = 3$ , or $x - 12 = 177$ and $y - 20 = 1$ . This gives us the solutions $(71, 23)$ and $(189, 21)$ . Since the $\text{GCD}$ must be a divisor of the $\text{LCM}$ , the first pair does not work. Assume $a>b$ . We must have $a = 21 \cdot 9$ and $b = 21$ , and we could then have $a<b$ , so there are $\boxed{2}$ solutions.
(awesomeag)


Answer: 597.0

How many ordered pairs $(a,b)$ such that $a$ is a positive real number and $b$ is an integer between $2$ and $200$ , inclusive, satisfy the equation $(\log_b a)^{2017}=\log_b(a^{2017})?$
$\textbf{(A)}\ 198\qquad\textbf{(B)}\ 199\qquad\textbf{(C)}\ 398\qquad\textbf{(D)}\ 399\qquad\textbf{(E)}\ 597$

Solution:
By the properties of logarithms, we can rearrange the equation to read $x^{2017}=2017x$ with $x=\log_b a$ . If $x\neq 0$ , we may divide by it and get $x^{2016}=2017$ , which implies $x=\pm \root{2016}\of{2017}$ . Hence, we have $3$ possible values $x$ , namely \[x=0,\qquad x=2017^{\frac1{2016}},\, \text{and}\quad  x=-2017^{\frac1{2016}}.\]
Since $\log_b a=x$ is equivalent to $a=b^x$ , each possible value $x$ yields exactly $199$ solutions $(b,a)$ , as we can assign $a=b^x$ to each $b=2,3,\dots,200$ . In total, we have $3\cdot 199=\boxed{597}$ solutions.


Answer: 4.0

How many ordered pairs $(m,n)$ of positive integers , with $m \ge n$ , have the property that their squares differ by $96$
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$

Solution: Find all of the factor pairs of $96$ $(1,96),(2,48),(3,32),(4,24),(6,16),(8,12).$ You can eliminate $(1,96)$ and ( $3,32)$ because you cannot have two numbers add to be an even number and have an odd difference at the same time without them being a decimal. You only have $4$ pairs left, so the answer is $\boxed{4}$


Answer: 5.0

How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations? \begin{align*} x^2+3y&=9 \\ (|x|+|y|-4)^2 &= 1 \end{align*} $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7$

Solution: The second equation is $(|x|+|y| - 4)^2 = 1$ . We know that the graph of $|x| + |y|$ is a very simple diamond shape, so let's see if we can reduce this equation to that form: \[(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.\] We now have two separate graphs for this equation and one graph for the first equation, so let's put it on the coordinate plane:  We see from the graph that there are $5$ intersections, so the answer is $\boxed{5}$


Answer: 3.0

How many ordered pairs of integers $(m, n)$ satisfy the equation $m^2+mn+n^2 = m^2n^2$
$\textbf{(A) }7\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }5$

Solution:
Clearly, $m=0,n=0$ is one of the solutions. However, we can be quite sure that there are more, so we apply Simon's Favorite Factoring Trick to get the following:
\begin{align*} m^2+mn+n^2 &= m^2n^2\\ m^2+mn+n^2 +mn &= m^2n^2 +mn\\ (m+n)^2 &= m^2n^2 +mn\\ (m+n)^2 &= mn(mn+1).\\ \end{align*}
Essentially, this says that the product of two consecutive numbers $mn,mn+1$ must be a perfect square. This is practically impossible except $mn=0$ or $mn+1=0$ $mn=0$ gives $(0,0)$ $mn=-1$ gives $(1,-1), (-1,1)$ . Answer: $\boxed{3}.$

Solution:
Case 1: $mn = 0$
In this case, $m = n = 0$
Case 2: $mn \neq 0$
Denote $k = {\rm gcd} \left( m, n \right)$ .
Denote $m = k u$ and $n = k v$ .
Thus, ${\rm gcd} \left( u, v \right) = 1$
Thus, the equation given in this problem can be written as \[ u^2 + uv + v^2 = k^2 u^2 v^2 . \]
Modulo $u$ , we have $v^2 \equiv 0 \pmod{u}$ .
Because ${\rm gcd} \left( u, v \right) = 1$ ., we must have $|u| = |v| = 1$ .
Plugging this into the above equation, we get $2 + uv = k^2$ .
Thus, we must have $uv = -1$ and $k = 1$
Thus, there are two solutions in this case: $\left( m , n \right) = \left( 1, -1 \right)$ and $\left( m , n \right) = \left( -1, 1 \right)$
Putting all cases together, the total number of solutions is $\boxed{3}$

Solution: We can move all terms to one side and write the equation as a quadratic in terms of $n$ to get \[(1-m^2)n^2+(m)n+(m^2)=0.\] The discriminant of this quadratic is \[\Delta = m^2-4(1-m^2)(m^2)=m^2(4m^2-3).\] For $n$ to be an integer, we must have $m^2(4m^2-3)$ be a perfect square. Thus, either $4m^2-3$ is a perfect square or $m^2 = 0$ and $m = 0$ . The first case gives $m=-1,1$ , which result in the equations $-n+1=0$ and $n-1=0$ , for a total of two pairs: $(-1,1)$ and $(1,-1)$ . The second case gives the equation $n^2=0$ , so it's only pair is $(0,0)$ . In total, the total number of solutions is $\boxed{3}$

Solution: Let $x=m+n, y=mn$ then \[x^2-y=y^2\] Completing the square then gives \[4x^2+1=(2y+1)^2\] Since the RHS is a square, clearly the only solutions are $x=0,y=0$ and $x=0,y=-1$ . The first gives $(0,0)$ while the second gives $(-1,1)$ and $(1,-1)$ by solving it as a quadratic with roots $m$ and $n$ . Thus there are $\boxed{3}$ solutions.


Answer: 4.0

How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Solution: Rearranging the terms and and completing the square for $y$ yields the result $x^{2020}+(y-1)^2=1$ . Then, notice that $x$ can only be $0$ $1$ and $-1$ because any value of $x^{2020}$ that is greater than 1 will cause the term $(y-1)^2$ to be less than $0$ , which is impossible as $y$ must be real. Therefore, plugging in the above values for $x$ gives the ordered pairs $(0,0)$ $(1,1)$ $(-1,1)$ , and $(0,2)$ gives  a  total  of $\boxed{4}$ ordered pairs.

Solution:
Bringing all of the terms to the LHS, we see a quadratic equation \[y^2 - 2y + x^{2020} = 0\] in terms of $y$ . Applying the quadratic formula, we get \[y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.\] In order for $y$ to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, $4(1-x^{2020})$ must be nonnegative. Therefore, \[4(1-x^{2020}) \geq 0 \implies x^{2020} \leq 1.\] Here, we see that we must split the inequality into a compound, resulting in $-1 \leq x \leq 1$
The only integers that satisfy this are $x \in \{-1,0,1\}$ . Plugging these values back into the quadratic equation, we see that $x = \{-1,1\}$ both produce a discriminant of $0$ , meaning that there is only 1 solution for $y$ . If $x = \{0\}$ , then the discriminant is nonzero, therefore resulting in two solutions for $y$
Thus, the answer is $2 \cdot 1 + 1 \cdot 2 = \boxed{4}$

Solution:
Set it up as a quadratic in terms of y: \[y^2-2y+x^{2020}=0\] Then the discriminant is \[\Delta = 4-4x^{2020}\] This will clearly only yield real solutions when $|x^{2020}| \leq 1$ , because the discriminant must be positive.
Then $x=-1,0,1$ . Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: \[y^2-2y+1=(y-1)^2=0\] This has only has solutions $1$ , so $(\pm 1,1)$ are solutions.
Next, if $x=0$ \[y^2-2y=0 \Rightarrow y(y-2)=0\] Which has 2 solutions, so $(0,2)$ and $(0,0)$
These are the only 4 solutions, so our answer is $\boxed{4}$

Solution:
Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$
Because $x^{2020} \geq 0$ for all $x$ , then $y(2-y) \geq 0 \Rightarrow y = 0,1,2$
If $y=0$ or $y=2$ , the right side is $0$ and therefore $x=0$
When $y=1$ , the right side become $1$ , therefore $x=1,-1$
Our solutions are $(0,2)$ $(0,0)$ $(1,1)$ $(-1,1)$ . These are the only $4$ solutions, so the answer is $\boxed{4}$

Solution:
Since $x^{2020}$ and $y^2$ are perfect squares, they are both nonnegative. That means $y^2$ plus a nonnegative number equals $2y$ , which means $y^2 \leq 2y.$ The only possible integer values for $y$ are $0, 1, 2$
For $y=0$ $x$ can only be $0$
For $y=1$ $x^2=1$ so $x=1, -1$
For $y=2$ $x$ can only be $0$ as well.
This gives us the solutions $(0, 0)$ $(1, 1)$ $(-1, 1)$ , and $(0, 2)$ . These are the only solutions, so there is a total of $\boxed{4}$ ordered pairs.


Answer: 4.0

Solution: Rearranging the terms and and completing the square for $y$ yields the result $x^{2020}+(y-1)^2=1$ . Then, notice that $x$ can only be $0$ $1$ and $-1$ because any value of $x^{2020}$ that is greater than 1 will cause the term $(y-1)^2$ to be less than $0$ , which is impossible as $y$ must be real. Therefore, plugging in the above values for $x$ gives the ordered pairs $(0,0)$ $(1,1)$ $(-1,1)$ , and $(0,2)$ gives  a  total  of $\boxed{4}$ ordered pairs.

Solution:
Bringing all of the terms to the LHS, we see a quadratic equation \[y^2 - 2y + x^{2020} = 0\] in terms of $y$ . Applying the quadratic formula, we get \[y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.\] In order for $y$ to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, $4(1-x^{2020})$ must be nonnegative. Therefore, \[4(1-x^{2020}) \geq 0 \implies x^{2020} \leq 1.\] Here, we see that we must split the inequality into a compound, resulting in $-1 \leq x \leq 1$
The only integers that satisfy this are $x \in \{-1,0,1\}$ . Plugging these values back into the quadratic equation, we see that $x = \{-1,1\}$ both produce a discriminant of $0$ , meaning that there is only 1 solution for $y$ . If $x = \{0\}$ , then the discriminant is nonzero, therefore resulting in two solutions for $y$
Thus, the answer is $2 \cdot 1 + 1 \cdot 2 = \boxed{4}$

Solution:
Set it up as a quadratic in terms of y: \[y^2-2y+x^{2020}=0\] Then the discriminant is \[\Delta = 4-4x^{2020}\] This will clearly only yield real solutions when $|x^{2020}| \leq 1$ , because the discriminant must be positive.
Then $x=-1,0,1$ . Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: \[y^2-2y+1=(y-1)^2=0\] This has only has solutions $1$ , so $(\pm 1,1)$ are solutions.
Next, if $x=0$ \[y^2-2y=0 \Rightarrow y(y-2)=0\] Which has 2 solutions, so $(0,2)$ and $(0,0)$
These are the only 4 solutions, so our answer is $\boxed{4}$

Solution:
Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$
Because $x^{2020} \geq 0$ for all $x$ , then $y(2-y) \geq 0 \Rightarrow y = 0,1,2$
If $y=0$ or $y=2$ , the right side is $0$ and therefore $x=0$
When $y=1$ , the right side become $1$ , therefore $x=1,-1$
Our solutions are $(0,2)$ $(0,0)$ $(1,1)$ $(-1,1)$ . These are the only $4$ solutions, so the answer is $\boxed{4}$

Solution:
Since $x^{2020}$ and $y^2$ are perfect squares, they are both nonnegative. That means $y^2$ plus a nonnegative number equals $2y$ , which means $y^2 \leq 2y.$ The only possible integer values for $y$ are $0, 1, 2$
For $y=0$ $x$ can only be $0$
For $y=1$ $x^2=1$ so $x=1, -1$
For $y=2$ $x$ can only be $0$ as well.
This gives us the solutions $(0, 0)$ $(1, 1)$ $(-1, 1)$ , and $(0, 2)$ . These are the only solutions, so there is a total of $\boxed{4}$ ordered pairs.


Answer: 9.0

How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: Cross-multiplying gives $MN=36.$ We write $36$ as a product of two positive integers: \begin{align*} 36 &= 1\cdot36 \\ &= 2\cdot18 \\ &= 3\cdot12 \\ &= 4\cdot9 \\ &= 6\cdot6. \end{align*} The products $1\cdot36, 2\cdot18, 3\cdot12,$ and $4\cdot9$ each produce $2$ ordered pairs $(M,N),$ as we can switch the order of the factors. The product $6\cdot6$ produces $1$ ordered pair $(M,N).$ Together, we have $4\cdot2+1=\boxed{9}$ ordered pairs $(M,N).$

Solution: Cross-multiplying gives $MN=36.$ From the prime factorization \[36=2^2\cdot3^2,\] we conclude that $36$ has $(2+1)(2+1)=9$ positive divisors. There are $9$ values of $M,$ and each value generates $1$ ordered pair $(M,N).$ So, there are $\boxed{9}$ ordered pairs $(M,N)$ in total.


Answer: 1.0

How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation \[(1+2a)(2+2b)(2a+b) = 32ab?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}$

Solution: Equation $(1+2a)(2+2b)(2a+b)=32ab$ is equivalent to \[b(2a-1)^2+2a(b-1)^2+(2a-b)^2=0,\] where $a$ $b>0$ . Therefore $2a-1=b-1=2a-b=0$ , so $(a,b)=\left(\tfrac12,1\right)$ . Hence the answer is $\boxed{1}$


Answer: 1.0

Solution: Using AM-GM on the two terms in each factor on the left, we get \[(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,\] meaning the equality condition must be satisfied. This means $1 = 2a = b$ , so we only have $\boxed{1}$ solution.


Answer: 3.0

How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*} $\textbf{(A) } 1 \qquad  \textbf{(B) } 2 \qquad  \textbf{(C) } 3 \qquad  \textbf{(D) } 4 \qquad  \textbf{(E) } 8$

Solution:
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this:  Now, it becomes clear that there are $\boxed{3}$ intersection points.

Solution:
$x+3y=3$ can be rewritten to $x=3-3y$ . Substituting $3-3y$ for $x$ in the second equation will give $||3-3y|-y|=1$ . Splitting this question into casework for the ranges of $y$ will give us the total number of solutions.
$\textbf{Case 1:}$ $y>1$ $3-3y$ will be negative so $|3-3y| = 3y-3.$ $|3y-3-y| = |2y-3| = 1$
$2y-3$ is positive so $2y-3 = 1$ and $y = 2$ and $x = 3-3(2) = -3$
$2y-3$ is negative so $|2y-3| = 3-2y = 1$ $2y = 2$ and so there are no solutions ( $y$ can't equal to $1$
$\textbf{Case 2:}$ $y = 1$ :
It is fairly clear that $x = 0.$
$\textbf{Case 3:}$ $y<1$ $3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$
$3-4y$ will be negative so $4y-3 = 1$ $\rightarrow$ $4y = 4$ . We already have this solution from Case 2 as $y = 1$
$3-4y$ will be positive so $3-4y = 1$ $\rightarrow$ $4y = 2$ $y = \frac{1}{2}$ and $x = \frac{3}{2}$ .
Thus, the solutions are: $(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)$ , and the answer is $\boxed{3}$

Solution:
Note that $||x| - |y||$ can take on one of four values: $x + y$ $x - y$ $-x + y$ $-x -y$ . So we have 4 cases:
Case 1: ||x| - |y|| = x+y \[x+3y=3\] \[x+y=1\]
Subtracting:
$2y=2 \Rightarrow y=1$ and $x=0$
$\text{Result: } (0,1)$
Case 2: ||x| - |y|| = x-y \[x+3y=3\] \[x-y=1\]
Subtacting:
$4y=2 \Rightarrow y=\dfrac{1}{2}$ an $x=\dfrac{3}{2}$
$\text{Result: } \left(\dfrac{3}{2},\dfrac{1}{2}\right)$
Case 3: ||x| - |y|| = -x+y \[x+3y=3\] \[-x+y=1\]
Adding:
$4y=4 \Rightarrow y=1$ and $x=0$
$\text{Result: } (0,1)$ . Since this is the same solution as we got in Case 1, we can not count this (otherwise, we would have overcounted).
Case 4: ||x| - |y|| = -x-y \[x+3y=3\] \[-x-y=1\]
Adding:
$2y=4 \Rightarrow y=2$ and $x=-3$
$\text{Result: } (-3,2)$
Answer
We solved each case by elimination (either adding the two equations or subtracting), to obtain three solutions: $(0, 1)$ $(-3,2)$ $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$
Our answer is $\boxed{3}$

Solution: Just as in solution $2$ , we derive the equation $||3-3y|-|y||=1$ . If we remove the absolute values, the equation collapses into four different possible values. $3-2y$ $3-4y$ $2y-3$ , and $4y-3$ , each equal to either $1$ or $-1$ . Remember that if $P-Q=a$ , then $Q-P=-a$ . Because we have already taken $1$ and $-1$ into account, we can eliminate one of the conjugates of each pair, namely $3-2y$ and $2y-3$ , and $3-4y$ and $4y-3$ . Find the values of $y$ when $3-2y=1$ $3-2y=-1$ $3-4y=1$ and $3-4y=-1$ . We see that $3-2y=1$ and $3-4y=-1$ give us the same value for $y$ , so the answer is $\boxed{3}$

Solution:
Just as in solution $2$ , we derive the equation $x=3-3y$ . Squaring both sides in the second equation gives $x^2+y^2-2|xy|=1$ . Putting $x=3-3y$ and doing a little calculation gives $10y^2-18y+9-2|3y-3y^2|=1$ . From here we know that $3y-3y^2$ is either positive or negative.
When positive, we get $2y^2-3y+1=0$ and then, $y=1/2$ or $y=1$ .
When negative, we get $y^2-3y+2=0$ and then, $y=2$ or $y=1$ . Clearly, there are $3$ different pairs of values and that gives us $\boxed{3}$

Solution: Since the absolute value is the square root of the square, we get that the first equation is quartic(degree $4$ ) and the other is linear. Subtract to get $\boxed{3}$