Answer: 550.0

Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .)

Solution: Recall that the arithmetic mean of all the $n$ digit palindromes is just the average of the largest and smallest $n$ digit palindromes, and in this case the $2$ palindromes are $101$ and $999$ and $\frac{101+999}{2}=\boxed{550},$ which is the final answer.

Solution: For any palindrome $\underline{ABA},$ note that $\underline{ABA}$ is $100A + 10B + A = 101A + 10B.$ The average for $A$ is $5$ since $A$ can be any of $1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ The average for $B$ is $4.5$ since $B$ is either $0, 1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ Therefore, the answer is $505 + 45 = \boxed{550}.$

Solution:
For every three-digit palindrome $\underline{ABA}$ with $A\in\{1,2,3,4,5,6,7,8,9\}$ and $B\in\{0,1,2,3,4,5,6,7,8,9\},$ note that $\underline{(10-A)(9-B)(10-A)}$ must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to \begin{align*} \underline{ABA}+\underline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\ &=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\ &=1000+90+10 \\ &=1100. \end{align*} For instances: \begin{align*} 171+929&=1100, \\ 262+838&=1100, \\ 303+797&=1100, \\ 414+686&=1100, \\ 545+555&=1100, \end{align*} and so on.
From this symmetry, the arithmetic mean of all the three-digit palindromes is $\frac{1100}{2}=\boxed{550}.$

Solution:
We notice that a three-digit palindrome looks like this: $\underline{aba}.$
And we know $a$ can be any digit from $1$ through $9,$ and $b$ can be any digit from $0$ through $9,$ so there are $9\times{10}=90$ three-digit palindromes.
We want to find the sum of these $90$ palindromes and divide it by $90$ to find the arithmetic mean.
How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: $101a+10b.$
Thus, all of these $90$ palindromes can be broken into this form.
Thus, the sum of these $90$ palindromes will be $101\times{(1+2+...+9)}\times{10}+10\times{(0+1+2+...+9)}\times{9},$ because each $a$ will be in $10$ different palindromes (since for each $a,$ there are $10$ choices for $b$ ). The same logic explains why we multiply by $9$ when computing the total sum of $b.$
We get a sum of $45\times{1100},$ but don't compute this! Divide this by $90$ and you will get $\boxed{550}.$

Solution: The possible values of the first and last digits each are $1, 2, ..., 8, 9$ with a sum of $45$ so the average value is $5.$ The middle digit can be any digit from $0$ to $9$ with a sum of $45,$ so the average value is $4.5.$ The average of all three-digit palindromes is $5\cdot 10^2+4.5\cdot 10+5=\boxed{550}.$


Answer: 348.0

Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

Solution:
If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic $ax^2+bx+c$ such that $f(1)=1440$ $f(2)=1716$ , and $f(3)=1848$ . Plugging in the values for x gives us a system of three equations:
$a+b+c=1440$
$4a+2b+c=1716$
$9a+3b+c=1848$
Solving gives $a=-72, b=492,$ and $c=1020$ . Thus, the answer is $-72(8)^2+492\cdot8+1020= \boxed{348}.$

Solution:
Use the same rationale as in solution 1; instead of using terms $1,2,3$ , we use $-1,0,1$ and solve the $6$ th term.
$a-b+c=1440$
$c=1716$
$a+b+c=1848$
Accordingly we will solve
$a=-72, b=204, c=1716$
$36a+6b+c= \boxed{348}.$

Solution:
Setting one of the sequences as $a+nr_1$ and the other as $b+nr_2$ , we can set up the following equalities
$ab = 1440$
$(a+r_1)(b+r_2)=1716$
$(a+2r_1)(b+2r_2)=1848$
We want to find $(a+7r_1)(b+7r_2)$
Foiling out the two above, we have
$ab + ar_2 + br_1 + r_1r_2 = 1716$ and $ab + 2ar_2 + 2br_1 + 4r_1r_2 = 1848$
Plugging in $ab=1440$ and bringing the constant over yields
$ar_2 + br_1 + r_1r_2 = 276$
$ar_2 + br_1 + 2r_1r_2 = 204$
Subtracting the two yields $r_1r_2 = -72$ and plugging that back in yields $ar_2 + br_1 = 348$
Now we find
$(a+7r_1)(b+7r_2) = ab + 7(ar_2 + br_1) + 49r_1r_2 = 1440 + 7(348) + 49(-72) = \boxed{348}$


Answer: 181.0

Find the largest integer $n$ satisfying the following conditions:

Solution:
Write $n^2 = (m + 1)^3 - m^3 = 3m^2 + 3m + 1$ , or equivalently, $(2n + 1)(2n - 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2$
Since $2n + 1$ and $2n - 1$ are both odd and their difference is $2$ , they are relatively prime . But since their product is three times a square, one of them must be a square and the other three times a square. We cannot have $2n - 1$ be three times a square, for then $2n + 1$ would be a square congruent to $2$ modulo $3$ , which is impossible.
Thus $2n - 1$ is a square, say $b^2$ . But $2n + 79$ is also a square, say $a^2$ . Then $(a + b)(a - b) = a^2 - b^2 = 80$ . Since $a + b$ and $a - b$ have the same parity and their product is even, they are both even. To maximize $n$ , it suffices to maximize $2b = (a + b) - (a - b)$ and check that this yields an integral value for $m$ . This occurs when $a + b = 40$ and $a - b = 2$ , that is, when $a = 21$ and $b = 19$ . This yields $n = 181$ and $m = 104$ , so the answer is $\boxed{181}$

Solution: Suppose that the consecutive cubes are $m$ and $m + 1$ . We can use completing the square and the first condition to get: \[(2n)^2 - 3(2m + 1)^2 = 1\equiv a^2 - 3b^2\] where $a$ and $b$ are non-negative integers. Now this is a Pell equation , with solutions in the form $(2 + \sqrt {3})^k = a_k + \sqrt {3}b_k,$ $k = 0,1,2,3,...$ . However, $a$ is even and $b$ is odd. It is easy to see that the parity of $a$ and $b$ switch each time (by induction). Hence all solutions to the first condition are in the form: \[(2 + \sqrt {3})^{2k + 1} = a_k + \sqrt {3}b_k\] where $k = 0,1,2,..$ . So we can (with very little effort) obtain the following: $(k,a_k = 2n) = (0,2),(1,26),(2,362),(3,5042)$ . It is an AIME problem so it is implicit that $n < 1000$ , so $2n < 2000$ . It is easy to see that $a_n$ is strictly increasing by induction. Checking $2n = 362\implies n =\boxed{181}$ in the second condition works (we know $b_k$ is odd so we don't need to find $m$ ). So we're done.

Solution:
Let us generate numbers $1$ to $1000$ for the second condition, for squares. We know for $N$ to be integer, the squares must be odd. So we generate $N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801$ $N$ cannot exceed $1000$ since it is AIME problem. Now take the first criterion, let $a$ be the smaller consecutive cube. We then get:
$N^2 = (A + 1)^3 - A^3$
$N^2 - 1 = 3A^2 + 3A$
$(N + 1)(N - 1) = 3A(A + 1)$
Now we know either $N + 1$ or $N - 1$ must be factor of $3$ , hence $N = 1 \pmod 3$ or $N = 2 \pmod 3$ . Only $1, 73, 181, 721$ satisfy this criterion. Testing each of the numbers in the condition yields $181$ as the largest that fits both, thus answer $= \boxed{181}$


Answer: 486.0

Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers

Solution:
Let us write down one such sum, with $m$ terms and first term $n + 1$
$3^{11} = (n + 1) + (n + 2) + \ldots + (n + m) = \frac{1}{2} m(2n + m + 1)$
Thus $m(2n + m + 1) = 2 \cdot 3^{11}$ so $m$ is a divisor of $2\cdot 3^{11}$ .  However, because $n \geq 0$ we have $m^2 < m(m + 1) \leq 2\cdot 3^{11}$ so $m < \sqrt{2\cdot 3^{11}} < 3^6$ .  Thus, we are looking for large factors of $2\cdot 3^{11}$ which are less than $3^6$ .  The largest such factor is clearly $2\cdot 3^5 = 486$ ; for this value of $m$ we do indeed have the valid expression $3^{11} = 122 + 123 + \ldots + 607$ , for which $k=\boxed{486}$

Solution: Proceed as in Solution 1 until it is noted that $m$ is a divisor of $2\cdot 3^{11}$ . The divisors of $2\cdot 3^{11}$ are $3^{1} , 2\cdot 3^{1} , 3^{2} , 2\cdot 3^{2} , \ldots , 2\cdot 3^{10} , 3^{11}$ . Note that the factors of $m(2n + m + 1)$ are of opposite parity (if $m$ is odd, then $(2n + m + 1)$ is even and vice versa). Thus, one of the two factors will be a power of three, and the other will be twice a power of three. $(2n + m + 1)$ will represent the greater factor while $m$ will represent the lesser factor. Given this information, we need to find the factor pair that maximizes the lesser of the two factors, as this will maximize the value of $m$ . The factor pair which maximizes the lesser factor is $2\cdot 3^{5}$ and $3^{6}$ . It follows that $m$ $2\cdot 3^{5}$ $\boxed{486}$


Answer: 349.0

Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying

Solution:
Assume that $z=a+bi$ . Then, \[z^3=(a^3-3ab^2)+(3a^2b-b^3)i\] Note that by the Triangle Inequality, \[|(a^3-3ab^2)-(3a^2b-b^3)|<p\implies |a^3+b^3-3ab^2-3a^2b|<a^2+b^2\] Thus, we know \[|a+b||a^2+b^2-4ab|<a^2+b^2\] Without loss of generality, assume $a>b$ (as otherwise, consider $i^3\overline z=b+ai$ ). If $|a/b|\geq 4$ , then \[17b^2\geq a^2+b^2>|a+b||a^2+b^2-4ab|\geq |b-4b||16b^2-16b^2+b^2|=3b^3\] `Thus, this means $b\leq\frac{17}3$ or $b\leq 5$ . Also note that the roots of $x^2-4x+1$ are $2\pm\sqrt 3$ , so thus if $b\geq 6$ \[2\sqrt 3b=(2(2-\sqrt 3)-4)b<a<4b\] Note that \[1000>p=a^2+b^2\geq 12b^2+b^2=13b^2\] so $b^2<81$ , and $b<9$ . If $b=8$ , then $16\sqrt 3\leq a\leq 32$ . Note that $\gcd(a,b)=1$ , and $a\not\equiv b\pmod 2$ , so $a=29$ or $31$ . However, then $5\mid a^2+b^2$ , absurd.
If $b=7$ , by similar logic, we have that $14\sqrt 3 <a< 28$ , so $b=26$ . However, once again, $5\mid a^2+b^2$ . If $b=6$ , by the same logic, $12\sqrt3<a<24$ , so $a=23$ , where we run into the same problem. Thus $b\leq 5$ indeed.
If $b=5$ , note that \[(a+5)(a^2+25-20a)<a^2+25\implies a<20\] We note that $p=5^2+18^2=349$ works. Thus, we just need to make sure that if $b\leq 4$ $a\leq 18$ . But this is easy, as \[p>(a+b)(a^2+b^2-4ab)\geq (4+18)(4^2+18^2-4\cdot 4\cdot 18)>1000\] absurd. Thus, the answer is $\boxed{349}$

Solution:
Denote $z = a + i b$ . Thus, $a^2 + b^2 = p$
Thus, \[z^3 = a \left( a^2 - 3 b^2 \right) + i b \left( - b^2 + 3 a^2 \right) .\]
Because $p$ ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have ${\rm Re} \left( z^3 \right) > 0$ and ${\rm Im} \left( z^3 \right) > 0$ .
Thus, \begin{align*} a \left( a^2 - 3 b^2 \right) & > 0 , \hspace{1cm} (1) \\ b \left( - b^2 + 3 a^2 \right) & > 0. \hspace{1cm} (2) \end{align*}
Because $p$ ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have the following triangle inequalities: \begin{align*} {\rm Re} \left( z^3 \right) + {\rm Im} \left( z^3 \right) & > p \hspace{1cm} (3) \\ p + {\rm Re} \left( z^3 \right) & > {\rm Im} \left( z^3 \right) \hspace{1cm} (4) \\ p + {\rm Im} \left( z^3 \right) & > {\rm Re} \left( z^3 \right) \hspace{1cm} (5) \end{align*}
We notice that $| z^3 | = p^{3/2}$ , and ${\rm Re} \left( z^3 \right)$ ${\rm Im} \left( z^3 \right)$ , and $| z^3 |$ form a right triangle. Thus, ${\rm Re} z^3 + {\rm Im} z^3 > p^{3/2}$ .
Because $p > 1$ $p^{3/2} > p$ .
Therefore, (3) holds.
Conditions (4) and (5) can be written in the joint form as \[\left| {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) \right| < p . \hspace{1cm} (4)\]
We have \begin{align*} {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) & = \left( a^3 - 3 a b^2 \right) - \left( - b^3 + 3 a^2 b \right) \\ & = \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \end{align*} and $p = a^2 + b^2$
Thus, (5) can be written as \[\left| \left( a + b \right) \left( a^2 - 4 ab + b^2 \right)  \right| < a^2 + b^2 . \hspace{1cm} (6)\]
Therefore, we need to jointly solve (1), (2), (6).
From (1) and (2), we have either $a, b >0$ , or $a, b < 0$ .
In (6), by symmetry, without loss of generality, we assume $a, b > 0$
Thus, (1) and (2) are reduced to \[a > \sqrt{3} b . \hspace{1cm} (7)\]
Let $a = \lambda b$ . Plugging this into (6), we get \begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right)  \right| < \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} .  \hspace{1cm} (8) \end{align*}
Because $p= a^2 + b^2$ is a prime, $a$ and $b$ are relatively prime.
Therefore, we can use (7), (8), $a^2 + b^2 <1000$ , and $a$ and $b$ are relatively prime to solve the problem.
To facilitate efficient search, we apply the following criteria:
\begin{enumerate*}
\item To satisfy (7) and $a^2 + b^2 < 1000$ , we have $1 \leq b \leq 15$ .
In the outer layer, we search for $b$ in a decreasing order.
In the inner layer, for each given $b$ , we search for $a$ .
\item Given $b$ , we search for $a$ in the range $\sqrt{3} b < a < \sqrt{1000 - b^2}$ .
\item We can prove that for $b \geq 9$ , there is no feasible $a$ .
The proof is as follows.
For $b \geq 9$ , to satisfy $a^2 + b^2 < 1000$ , we have $a \leq 30$ .
Thus, $\sqrt{3} < \lambda \leq \frac{30}{9}$ .
Thus, the R.H.S. of (8) has the following upper bound \begin{align*} \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} & < \frac{1}{b} \frac{\lambda^2 + \lambda}{\lambda + 1} \\ & = \frac{\lambda}{b} \\ & \leq \frac{\frac{30}{9}}{9} \\ & < \frac{10}{27} . \end{align*}
Hence, to satisfy (8), a necessary condition is \begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right)  \right| < \frac{10}{27} . \end{align*}
However, this cannot be satisfied for $\sqrt{3} < \lambda \leq \frac{30}{9}$ .
Therefore, there is no feasible solution for $b \geq 9$ .
Therefore, we only need to consider $b \leq 8$
\item We eliminate $a$ that are not relatively prime to $b$
\item We use the following criteria to quickly eliminate $a$ that make $a^2 + b^2$ a composite number.
\item For the remaining $\left( b, a \right)$ , check whether (8) and the condition that $a^2 + b^2$ is prime are both satisfied.
The first feasible solution is $b = 5$ and $a = 18$ .
Thus, $a^2 + b^2 = 349$
\item For the remaining search, given $b$ , we only search for $a \geq \sqrt{349 - b^2}$
Following the above search criteria, we find the final answer as $b = 5$ and $a = 18$ .
Thus, the largest prime $p$ is $p = a^2 + b^2 = \boxed{349}$


Answer: 25.0

Find the last three digits of the product of the positive roots of $\sqrt{1995}x^{\log_{1995}x}=x^2$

Solution:
Instead of taking $\log_{1995}$ , we take $\log_x$ of both sides and simplify:
$\log_x(\sqrt{1995}x^{\log_{1995}x})=\log_x(x^{2})$
$\log_x\sqrt{1995}+\log_x x^{\log_{1995}x}=2$
$\dfrac{1}{2} \log_x 1995 + \log_{1995} x = 2$
We know that $\log_x 1995$ and $\log_{1995} x$ are reciprocals, so let $a=\log_{1995} x$ .  Then we have $\dfrac{1}{2}\left(\dfrac{1}{a}\right) + a = 2$ .  Multiplying by $2a$ and simplifying gives us $2a^2-4a+1=0$ , as shown above.
Because $a=\log_{1995} x$ $x=1995^a$ . By the quadratic formula, the two roots of our equation are $a=\frac{2\pm\sqrt2}{2}$ . This means our two roots in terms of $x$ are $1995^\frac{2+\sqrt2}{2}$ and $1995^\frac{2-\sqrt2}{2}.$ Multiplying these gives $1995^2$
$1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}$ , so our answer is $\boxed{025}$

Solution: Let $y=\log_{1995}x$ . Rewriting the equation in terms of $y$ , we have \[\sqrt{1995}\left(1995^y\right)^y=1995^{2y}\] \[1995^{y^2+\frac{1}{2}}=1995^{2y}\] \[y^2+\frac{1}{2}=2y\] \[2y^2-4y+1=0\] \[y=\frac{4\pm\sqrt{16-\left(4\right)\left(2\right)\left(1\right)}}{4}=\frac{4\pm\sqrt{8}}{4}=\frac{2\pm\sqrt{8}}{2}={1\pm\sqrt{2}}\] Thus, the product of the positive roots is $\left(1995^{\frac{2+\sqrt{8}}{2}}\right)\left(1995^{\frac{2-\sqrt{8}}{2}}\right)=1995^2=\left(2000-5\right)^2$ , so the last three digits are $\boxed{025}$


Answer: 282.0

Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer.

Solution:
The difference between consecutive integral squares must be greater than 1000. $(x+1)^2-x^2\geq1000$ , so $x\geq\frac{999}{2}\implies x\geq500$ $x=500$ does not work, so $x>500$ . Let $n=x-500$ . By inspection, $n^2$ should end in a number close to but less than 1000 such that there exists $1000N$ within the difference of the two squares. Examine when $n^2=1000$ . Then, $n=10\sqrt{10}$ . One example way to estimate $\sqrt{10}$ follows.
Then, $n\approx 31.6$ $n^2<1000$ , so $n$ could be $31$ . Add 500 to get the first square and 501 to get the second. Then, the two integral squares are $531^2$ and $532^2$ . Checking, $531^2=281961$ and $532^2=283024$ $282,000$ straddles the two squares, which have a difference of 1063. The difference has been minimized, so $N$ is minimized $N=282000\implies\boxed{282}$

Solution:
Let us first observe the difference between $x^2$ and $(x+1)^2$ , for any arbitrary $x\ge 0$ $(x+1)^2-x^2=2x+1$ . So that means for every $x\ge 0$ , the difference between that square and the next square have a difference of $2x+1$ . Now, we need to find an $x$ such that $2x+1\ge 1000$ . Solving gives $x\ge \frac{999}{2}$ , so $x\ge 500$ . Now we need to find what range of numbers has to be square-free: $\overline{N000}\rightarrow \overline{N999}$ have to all be square-free.
Let us first plug in a few values of $x$ to see if we can figure anything out. $x=500$ $x^2=250000$ , and $(x+1)^2=251001$ . Notice that this does not fit the criteria, because $250000$ is a square, whereas $\overline{N000}$ cannot be a square. This means, we must find a square, such that the last $3$ digits are close to $1000$ , but not there, such as $961$ or $974$ . Now, the best we can do is to keep on listing squares until we hit one that fits. We do not need to solve for each square: remember that the difference between consecutive squares are $2x+1$ , so all we need to do is addition. After making a list, we find that $531^2=281961$ , while $532^2=283024$ . It skipped $282000$ , so our answer is $\boxed{282}$

Solution: Let $x$ be the number being squared. Based on the reasoning above, we know that $N$ must be at least $250$ , so $x$ has to be at least $500$ . Let $k$ be $x-500$ . We can write $x^2$ as $(500+k)^2$ , or $250000+1000k+k^2$ . We can disregard $250000$ and $1000k$ , since they won't affect the last three digits, which determines if there are any squares between $\overline{N000}\rightarrow \overline{N999}$ . So we must find a square, $k^2$ , such that it is under $1000$ , but the next square is over $1000$ . We find that $k=31$ gives $k^2=961$ , and so $(k+1)^2=32^2=1024$ . We can be sure that this skips a thousand because the $1000k$ increments it up $1000$ each time. Now we can solve for $x$ $(500+31)^2=281961$ , while $(500+32)^2=283024$ . We skipped $282000$ , so the answer is $\boxed{282}$

Solution:
The goal is to find the least $N \in \mathbb{N}$ such that $\exists m \in \mathbb{N}$ where $m^2 + 1 \leq 1000N, 1000N + 1000 \leq (m+1)^2$
Combining the two inequalities leads to $(m+1)^2 \geq m^2 + 1001, m \geq 500$
Let $m = k + 500$ , where $k \in \mathbb{W}$ , then the inequalities become,
$N \geq \frac{(k+500)^2 + 1}{1000} = \frac{k^2 + 1}{1000} + k + 250$ , and
$N \leq \frac{(k+501)^2}{1000} - 1 = \frac{(k+1)^2}{1000} + k + 250.$
For $k=31$ , one can verify that $N = 282$ is the unique integer satisfying the inequalities.
For $k \leq 30$ $k + 250 < \frac{k^2 + 1}{1000} + k + 250 \leq N$ $\leq \frac{(k+1)^2}{1000} + k + 250 \leq \frac{(30+1)^2}{1000} + k + 250 < k + 251$
i.e., $k + 250 < N < k + 251$ , a contradiction.
Note $k \geq 32$ leads to larger $N$ (s).
Hence, the answer is $\boxed{282}$


Answer: 49.0

Find the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$ . (The notation $\lfloor x\rfloor$ means the greatest integer less than or equal to $x$ .)

Solution:
Note that if $\frac{2002}n - \frac{2002}{n+1}\leq 1$ , then either $\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor$ ,
or $\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor+1$ . Either way, we won't skip any natural numbers.
The greatest $n$ such that $\frac{2002}n - \frac{2002}{n+1} > 1$ is $n=44$ . (The inequality simplifies to $n(n+1)<2002$ , which is easy to solve by trial, as the solution is obviously $\simeq \sqrt{2002}$ .)
We can now compute: \[\left\lfloor\frac{2002}{45}\right\rfloor=44\] \[\left\lfloor\frac{2002}{44}\right\rfloor=45\] \[\left\lfloor\frac{2002}{43}\right\rfloor=46\] \[\left\lfloor\frac{2002}{42}\right\rfloor=47\] \[\left\lfloor\frac{2002}{41}\right\rfloor=48\] \[\left\lfloor\frac{2002}{40}\right\rfloor=50\]
From the observation above (and the fact that $\left\lfloor\frac{2002}{2002}\right\rfloor=1$ ) we know that all integers between $1$ and $44$ will be achieved for some values of $n$ . Similarly, for $n<40$ we obviously have $\left\lfloor\frac{2002}{n}\right\rfloor > 50$
Hence the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$ is $\boxed{049}$

Solution:
Rewriting the given information and simplifying it a bit, we have \begin{align*}  k \le \frac{2002}{n} < k+1 &\implies \frac{1}{k} \ge \frac{n}{2002} > \frac{1}{k+1}. \\ &\implies \frac{2002}{k} \ge n > \frac{2002}{k+1}.   \end{align*}
Now note that in order for there to be no integer solutions to $n,$ we must have $\left\lfloor \frac{2002}{k} \right\rfloor = \left\lfloor \frac{2002}{k+1} \right\rfloor.$ We seek the smallest such $k.$ A bit of experimentation yields that $k=49$ is the smallest solution, as for $k=49,$ it is true that $\left\lfloor \frac{2002}{49} \right\rfloor = \left\lfloor \frac{2002}{50} \right\rfloor = 40.$ Furthermore, $k=49$ is the smallest such case. (If unsure, we could check if the result holds for $k=48,$ and as it turns out, it doesn't.) Therefore, the answer is $\boxed{049}.$

Solution:
In this solution we use inductive reasoning and a lot of trial and error. Depending on how accurately you can estimate, the solution will come quicker or slower.
Using values of $k$ as $1, 2, 3, 4,$ and $5,$ we can find the corresponding values of $n$ relatively easily. For $k = 1$ $n$ is in the range $[2002-1002]$ ; for $k = 2$ $n$ is the the range $[1001-668]$ , etc: $3, [667,501]; 4, [500-401]; 5, [400-334]$ . For any positive integer $k, n$ is in a range of $\left\lfloor \frac{2002}{k} \right\rfloor -\left\lceil \frac{2002}{k+1} \right\rceil$
Now we try testing $k = 1002$ to get a better understanding of what our solution will look like. Obviously, there will be no solution for $n$ , but we are more interested in how the range will compute to. Using the formula we got above, the range will be $1-2$ . Testing any integer $k$ from $1002-2000$ will result in the same range. Also, notice that each and every one of them have no solution for $n$ . Testing $1001$ gives a range of $2-2$ , and $2002$ gives $1-1$ . They each have a solution for $n$ , and their range is only one value. Therefore, we can assume with relative safety that the integer $k$ we want is the lowest integer that follows this equation
\[\left\lfloor\frac{2002}{k}\right\rfloor + 1 = \left\lceil \frac{2002}{k+1}\right\rceil\]
Now we can easily guess and check starting from $k = 1$ . After a few tests it's not difficult to estimate a few jumps, and it took me only a few minutes to realize the answer was somewhere in the forties (You could also use the fact that $45^2=2025$ ). Then it's just a matter of checking them until we get $\boxed{049}$ .
Alternatively, you could use the equation above and proceed with one of the other two solutions listed.


Answer: 132.0

Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes.

Solution:
Suppose $p=11$ ; then $m^2-m+11=11qrs$ . Reducing modulo 11, we get $m\equiv  1,0 \pmod{11}$ so $k(11k\pm 1)+1 = qrs$
Suppose $q=11$ . Then we must have $11k^2\pm k + 1 = 11rs$ , which leads to $k\equiv \mp 1 \pmod{11}$ , i.e., $k\in \{1,10,12,21,23,\ldots\}$
$k=1$ leads to $rs=1$ (impossible)! Then $k=10$ leads to $rs=103$ , a prime (impossible). Finally, for $k=12$ we get $rs=143=11\cdot 13$
Thus our answer is $m=11k= \boxed{132}$

Solution: Let $m^2 - m + 11 = pqrs$ for primes $p, q, r, s\ge11$ . If $p, q, r, s = 11$ , then $m^2-m+11=11^4$ . We can multiply this by $4$ and complete the square to find $(2m-1)^2=4\cdot 11^4-43$ . But \[(2\cdot 11^2-1)^2=4\cdot 11^4-4\cdot 11^2+1 <4\cdot 11^4-43<(2\cdot 11^2)^2,\] hence we have pinned a perfect square $(2m-1)^2=4\cdot 11^4-43$ strictly between two consecutive perfect squares, a contradiction. Hence $pqrs \ge 11^3 \cdot 13$ . Thus $m^2-m+11\ge 11^3\cdot 13$ , or $(m-132)(m+131)\ge0$ . From the inequality, we see that $m \ge 132$ $132^2 - 132 + 11 = 11^3 \cdot 13$ , so $m = \boxed{132}$ and we are done.


Answer: 797.0

Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$

Solution:
Recall that $1000$ divides this expression if $8$ and $125$ both divide it. It should be fairly obvious that $n \geq 3$ ; so we may break up the initial condition into two sub-conditions.
(1) $5^n \equiv n \pmod{8}$ . Notice that the square of any odd integer is $1$ modulo $8$ (proof by plugging in $1^2,3^2,5^2,7^2$ into modulo $8$ ), so the LHS of this expression goes $5,1,5,1,\ldots$ , while the RHS goes $1,2,3,4,5,6,7,8,1,\ldots$ . The cycle length of the LHS is $2$ , RHS is $8$ , so the cycle length of the solution is $\operatorname{lcm}(2,8)=8$ . Indeed, the $n$ that solve this congruence are exactly those such that $n \equiv 5 \pmod{8}$
(2) $2^n \equiv n \pmod{125}$ . This is extremely computationally intensive if we try to calculate all $2^1,2^2,\ldots,2^{100} \pmod{125}$ , so we take a divide-and-conquer approach instead. In order for this expression to be true, $2^n \equiv n \pmod{5}$ is necessary; it shouldn't take too long for us to go through the $20$ possible LHS-RHS combinations, considering that $n$ must be odd. We only need to test $10$ values of $n$ and obtain $n \equiv 3 \pmod{20}$ or $n \equiv 17 \pmod{20}$
With this in mind we consider $2^n \equiv n \pmod{25}$ . By the Generalized Fermat's Little Theorem, $2^{20} \equiv 1 \pmod{25}$ , but we already have $n$ modulo $20$ . Our calculation is greatly simplified. The LHS's cycle length is $20$ and the RHS's cycle length is $25$ , from which their least common multiple is $100$ . In this step we need to test all the numbers between $1$ to $100$ that $n \equiv 3 \pmod{20}$ or $n \equiv 17 \pmod{20}$ . In the case that $n \equiv 3 \pmod{20}$ , the RHS goes $3,23,43,63,83$ , and we need $2^n \equiv n \equiv 2^3 \pmod{25}$ ; clearly $n \equiv 83 \pmod{100}$ . In the case that $n \equiv 17 \pmod{20}$ , by a similar argument, $n \equiv 97 \pmod{100}$
In the final step, we need to calculate $2^{97}$ and $2^{83}$ modulo $125$
Note that $2^{97}\equiv2^{-3}$ ; because $8\cdot47=376\equiv1\pmod{125},$ we get $2^{97} \equiv 47\pmod{125}$
Note that $2^{83}$ is $2^{-17}=2^{-16}\cdot2^{-1}$ . We have \begin{align*} 2^{-1}&\equiv63, \\ 2^{-2}&\equiv63^2=3969\equiv-31, \\ 2^{-4}&\equiv(-31)^2=961\equiv-39, \\ 2^{-8}&\equiv1521\equiv21, \\ 2^{-16}&\equiv441, \\ 2^{-17}&\equiv63\cdot441\equiv7\cdot(-31)=-217\equiv33. \end{align*} This time, LHS cycle is $100$ , RHS cycle is $125$ , so we need to figure out $n$ modulo $500$ . It should be $n \equiv 283,297 \pmod{500}$
Put everything together. By the second subcondition, the only candidates less than $1000$ are $283,297,783,797$ . Apply the first subcondition, $n=\boxed{797}$ is the desired answer.

Solution:
We have that $2^n + 5^n \equiv n\pmod{1000}$ , or $2^n + 5^n \equiv n \pmod{8}$ and $2^n + 5^n \equiv n \pmod{125}$ by CRT. It is easy to check $n < 3$ don't work, so we have that $n \geq 3$ . Then, $2^n \equiv 0 \pmod{8}$ and $5^n \equiv 0 \pmod{125}$ , so we just have $5^n \equiv n \pmod{8}$ and $2^n \equiv n \pmod{125}$ . Let us consider both of these congruences separately.
First, we look at $5^n \equiv n \pmod{8}$ . By Euler's Totient Theorem (ETT), we have $5^4 \equiv 1 \pmod{8}$ , so $5^5 \equiv 5 \pmod{8}$ . On the RHS of the congruence, the possible values of $n$ are all nonnegative integers less than $8$ and on the RHS the only possible values are $5$ and $1$ . However, for $5^n$ to be $1 \pmod{8}$ we must have $n \equiv 0 \pmod{4}$ , a contradiction. So, the only possible values of $n$ are when $n \equiv 5 \pmod{8} \implies n = 8k+5$
Now we look at $2^n \equiv n \pmod{125}$ . Plugging in $n = 8k+5$ , we get $2^{8k+5} \equiv 8k+5 \pmod{125} \implies 2^{8k} \cdot 32 \equiv 8k+5 \pmod{125}$ . Note, for $2^n \equiv n\pmod{125}$ to be satisfied, we must have $2^n \equiv n \pmod{5}$ and $2^n \equiv n\pmod{25}$ . Since $2^{8k} \equiv 1\pmod{5}$ as $8k \equiv 0\pmod{4}$ , we have $2 \equiv -2k \pmod{5} \implies k = 5m-1$ . Then, $n = 8(5m-1) + 5 = 40m-3$ . Now, we get $2^{40m-3} \equiv 40m-3 \pmod{125}$ . Using the fact that $2^n \equiv n\pmod{25}$ , we get $2^{-3} \equiv 15m-3 \pmod{25}$ . The inverse of $2$ modulo $25$ is obviously $13$ , so $2^{-3} \equiv 13^3 \equiv 22 \pmod{25}$ , so $15m \equiv 0 \pmod{25} \implies m = 5s$ . Plugging in $m = 5s$ , we get $n = 200s - 3$
Now, we are finally ready to plug $n$ into the congruence modulo $125$ . Plugging in, we get $2^{200s-3} \equiv 200s - 3 \pmod{125}$ . By ETT, we get $2^{100} \equiv 1 \pmod{125}$ , so $2^{200s- 3} \equiv 2^{-3} \equiv 47 \pmod{125}$ . Then, $200s \equiv 50 \pmod{125} \implies s \equiv 4 \pmod{5} \implies s = 5y+4$ . Plugging this in, we get $n = 200(5y+4) - 3 = 1000y+797$ , implying the smallest value of $n$ is simply $\boxed{797}$

Solution:
We wish to find the least positive integer $n$ for which $2^n+5^n-n\equiv0\pmod{1000}.$ Rearranging gives \[2^n+5^n\equiv n\pmod{1000}.\] Applying the Chinese Remainder Theorem, we get the following system of congruences: \begin{align*} 2^n+5^n &\equiv n \pmod{8}, \\ 2^n+5^n &\equiv n \pmod{125}. \end{align*} It is clear that $n\geq3,$ from which we simplify to \begin{align*} 5^n &\equiv n \pmod{8}, \hspace{15mm} &(1) \\ 2^n &\equiv n \pmod{125}. &(2) \end{align*} We solve each congruence separately:
Finally, combining the two results from above (bolded) generates the least such positive integer $n$ at $s=19:$ \begin{align*} n&=8r+5 \\ &=8(5s+4)+5 \\ &=40s+37 \\ &=\boxed{797} ~MRENTHUSIASM (inspired by Math Jams's 2021 AIME II Discussion


Answer: 1.0

Find the least positive integer $n$ such that

Solution:
We apply the identity
\begin{align*} \frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \sin (n+1)} \\ &= \frac{1}{\sin 1} \cdot \left(\frac{\cos n}{\sin n} - \frac{\cos (n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{align*}
The motivation for this identity arises from the need to decompose those fractions, possibly into telescoping
Thus our summation becomes
\[\sum_{k=23}^{67} \frac{1}{\sin (2k-1) \sin 2k} = \frac{1}{\sin 1} \left(\cot 45 - \cot 46 + \cot 47 - \cdots + \cot 133 - \cot 134 \right).\]
Since $\cot (180 - x) = - \cot x$ , the summation simply reduces to $\frac{1}{\sin 1} \cdot \left( \cot 45 - \cot 90 \right) = \frac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}$ . Therefore, the answer is $\boxed{001}$

Solution:
Let S be the sum of the sequence. We begin the same as in Solution 1 to get $S\sin(1)=\cot(45)-\cot(46)+\cot(47)-\cot(48)+...+\cot(133)-\cot(134)$ . Observe that this "almost telescopes," if only we had some extra terms. Consider adding the sequence $\frac{1}{\sin(46)\sin(47)}+\frac{1}{\sin(48)\sin(49)}+...+\frac{1}{\sin(134)\sin(135)}$ . By the identity $\sin(x)=\sin(180-x)$ , this sequence is equal to the original one, simply written backwards. By the same logic as before, we may rewrite this sequence as $S\sin(1)=\cot(46)-\cot(47)+\cot(48)-\cot(49)+...+\cot(134)-\cot(135)$ ,
and when we add the two sequences, they telescope to give $2S\sin(1)=\cot(45)-\cot(135)=2$ .
Hence, $S=\sin(1)$ , and our angle is $\boxed{001}$


Answer: 8.0

Find the least positive integer $n$ such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit $0$

Solution:
If a factor of $10^{n}$ has a $2$ and a $5$ in its prime factorization , then that factor will end in a $0$ . Therefore, we have left to consider the case when the two factors have the $2$ s and the $5$ s separated, so we need to find the first power of 2 or 5 that contains a 0.
For $n = 1:$ \[2^1 = 2 , 5^1 = 5\] $n = 2:$ \[2^2 = 4 , 5 ^ 2 =25\] $n = 3:$ \[2^3 = 8 , 5 ^3 = 125\]
and so on, until,
$n = 8:$ $2^8 = 256$ $5^8 = 390625$
We see that $5^8$ contains the first zero, so $n = \boxed{8}$


Answer: 195.0

Find the least positive integer $n$ such that when $3^n$ is written in base $143$ , its two right-most digits in base $143$ are $01$

Solution:
Note that the given condition is equivalent to $3^n \equiv 1 \pmod{143^2}$ and $143=11\cdot 13$ . Because $\gcd(11^2, 13^2) = 1$ , the desired condition is equivalent to $3^n \equiv 1 \pmod{121}$ and $3^n \equiv 1 \pmod{169}$
If $3^n \equiv 1 \pmod{121}$ , one can see the sequence $1, 3, 9, 27, 81, 1, 3, 9...$ so $5|n$
Now if $3^n \equiv 1 \pmod{169}$ , it is harder. But we do observe that $3^3 \equiv 1 \pmod{13}$ , therefore $3^3 = 13a + 1$ for some integer $a$ . So our goal is to find the first number $p_1$ such that $(13a+1)^ {p_1} \equiv 1 \pmod{169}$ . Then, $p_1 \equiv 0 \pmod{13},$ which follows from the binomial theorem. It is not difficult to see that the smallest $p_1=13$ , so ultimately $3^{39} \equiv 1 \pmod{169}$ . Therefore, $39|n$
The first $n$ satisfying both criteria is thus $5\cdot 39=\boxed{195}$

Solution:
Note that Euler's Totient Theorem would not necessarily lead to the smallest $n$ and that in this case that $n$ is greater than $1000$
We wish to find the least $n$ such that $3^n \equiv 1 \pmod{143^2}$ . This factors as $143^2=11^{2}*13^{2}$ . Because $\gcd(121, 169) = 1$ , we can simply find the least $n$ such that $3^n \equiv 1 \pmod{121}$ and $3^n \equiv 1 \pmod{169}$
Quick inspection yields $3^5 \equiv 1 \pmod{121}$ and $3^3 \equiv 1 \pmod{13}$ . Now we must find the smallest $k$ such that $3^{3k} \equiv 1 \pmod{169}$ . Euler's gives $3^{156} \equiv 1 \pmod{169}$ . So $3k$ is a factor of $156$ . This gives $k=1,2, 4, 13, 26, 52$ . Some more inspection yields $k=13$ is the smallest valid $k$ . So $3^5 \equiv 1 \pmod{121}$ and $3^{39} \equiv 1 \pmod{169}$ . The least $n$ satisfying both is $lcm(5, 39)=\boxed{195}$ . (RegularHexagon)

Solution:
Listing out the powers of $3$ , modulo $169$ and modulo $121$ , we have: \[\begin{array}{c|c|c} n & 3^n\mod{169} & 3^n\mod{121}\\ \hline 0 & 1 & 1\\ 1 & 3 & 3\\ 2 & 9 & 9\\ 3 & 27 & 27\\ 4 & 81 & 81\\ 5 & 74 & 1\\ 6 & 53\\ 7 & 159\\ 8 & 139\\ 9 & 79\\ 10 & 68\\ 11 & 35\\ 12 & 105\\ 13 & 146\\ 14 & 100\\ 15 & 131\\ 16 & 55\\ 17 & 165\\ 18 & 157\\ 19 & 133\\ 20 & 61\\ 21 & 14\\ 22 & 42\\ 23 & 126\\ 24 & 40\\ 25 & 120\\ 26 & 22\\ 27 & 66\\ 28 & 29\\ 29 & 87\\ 30 & 92\\ 31 & 107\\ 32 & 152\\ 33 & 118\\ 34 & 16\\ 35 & 48\\ 36 & 144\\ 37 & 94\\ 38 & 113\\ 39 & 1\\ \end{array}\]
The powers of $3$ repeat in cycles of $5$ and $39$ in modulo $121$ and modulo $169$ , respectively. The answer is $\text{lcm}(5, 39) = \boxed{195}$

Solution: We have that \[3^n \equiv 1 \pmod{143^2}.\] Now, $3^{110} \equiv 1 \pmod{11^2}$ so by the Fundamental Theorem of Orders, $\text{ord}_{11^2}(3)|110$ and with some bashing, we get that it is $5$ . Similarly, we get that $\text{ord}_{13^2}(3)=39$ . Now, $\text{lcm}(39,5)=\boxed{195}$ which is our desired solution.

Solution:
We wish to find the smallest $n$ such that $3^n\equiv 1\pmod{143^2}$ , so we want $n\equiv 1\pmod{121}$ and $n\equiv 1\pmod{169}$ . Note that $3^5\equiv 1\pmod{121}$ ,  so $3^n$ repeats $121$ with a period of $5$ , so $5|n$ . Now, in order for $n\equiv 1\pmod{169}$ , then $n\equiv 1\pmod{13}$ . Because $3^3\equiv 1\pmod{13}$ $3^n$ repeats with a period of $3$ , so $3|n$ .
Hence, we have that for some positive integer $p$ $3^n\equiv (3^3)^p\equiv (26+1)^p\equiv \binom{p}{0}26^p+\binom{p}{1}26^{p-1}....+\binom{p}{p-2}26^2+\binom{p}{p-1}26+\binom{p}{p}\equiv 26p+1\equiv 1\pmod{169}$ , so $26p\equiv 0\pmod{169}$ and $p\equiv 0\pmod{13}$ . Thus, we have that $5|n$ $3|n$ , and $13|n$ , so the smallest possible value of $n$ is $3\times5\times13=\boxed{195}$ .
-Stormersyle

Solution: We can see that $3^n-1 = 143^2*x$ , which means that $v_{11}(3^n-1) \geq 2$ $v_{13}(3^n-1) \geq 2$ $v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})$ $v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})$ by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as $v_{13}(3^n-1) \geq 2$ . Therefore the minimum possible value of n is $3\times5\times13=\boxed{195}$

Solution:
Note that the problem is basically asking for the least positive integer $n$ such that $11^2 \cdot 13^2 | 3^n - 1.$ It is easy to see that $n = \text{lcm } (a, b),$ where $a$ is the least positive integer satisfying $11^2 | 3^a - 1$ and $b$ the least positive integer satisfying $13^2 | 3^b - 1$ . Luckily, finding $a$ is a relatively trivial task, as one can simply notice that $3^5 = 243 \equiv 1 \mod 121$ . However, finding $b$ is slightly more nontrivial. The order of $3^k$ modulo $13$ (which is $3$ ) is trivial to find, as one can either bash out a pattern of remainders upon dividing powers of $3$ by $13$ , or one can notice that $3^3 = 27 \equiv 1 \mod 13$ (the latter which is the definition of period/orders by FLT). We can thus rewrite $3^3$ as $(2 \cdot 13 + 1) \mod 13^2$ . Now suppose that \[3^{3k} \equiv (13n + 1) \mod 13^2.\] I claim that $3^{3(k+1)} \equiv (13(n+2) + 1) \mod 13^2.$
Proof:
To find $3^{3(k+1)},$ we can simply multiply $3^{3k}$ by $3^3,$ which is congruent to $2 \cdot 13 + 1$ modulo $13^2$ .
By expanding the product out, we obtain \[3^{3(k+1)} \equiv (13n + 1)(2 \cdot 13 + 1) = 13^2 \cdot 2n + 13n + 2 \cdot 13 + 1 \mod 13^2,\] and since the $13^2$ on the LHS cancels out, we're left with \[13n + 2 \cdot 13 + 1 \mod 13^2 \implies 13(n+2) + 1 \mod 13^2\] . Thus, our claim is proven.
Let $f(n)$ be the second to last digit when $3^{3k}$ is written in base $13^2$ .
Using our proof, it is easy to see that $f(n)$ satisfies the recurrence $f(1) = 2$ and $f(n+1) = f(n) + 2$ . Since this implies $f(n) = 2n,$ we just have to find the least positive integer $n$ such that $2n$ is a multiple of $13$ , which is trivially obtained as $13$ .
The least integer $n$ such that $3^n - 1$ is divisible by $13^2$ is $3 \cdot 13 = 39,$ so our final answer is $\text{lcm } (5, 39) = \boxed{195}.$

Solution:
The requested positive integer is the least value of $n>0$ such that $3^n\equiv 1\pmod{143^2}.$ Note that $143=11\cdot 13.$ The least power of $3$ that is congruent to $1$ modulo $11^2$ is $3^5=243=2\cdot 11^2+1.$ It follows that. $3^n\equiv 1\pmod {11^2}$ if and only if $n=5j$ for some positive integer $j$
The least power of $3$ that is congruent to $1$ modulo $13$ is $3^3=27=2\cdot 13+1.$ It follows that $3^n\equiv 1\pmod{13}$ if and only if $n=3k$ for some positive integer $k$ . Additionally, for some positive integer $k$ , the Binomial Theorem shows that $3^{3k}=(26+1)^k=26\cdot k+1 \pmod{13^2}$ . In particular, $3^n=3^{3k}\equiv 1\pmod {13^2}$ if and only if $k=13m$ for some positive integer $m$ , that is, if and only if $n=39m.$
Because $11^2$ and $13^2$ are relatively prime, $3^n\equiv 1\pmod {143^2}$ if and only if $3^n\equiv 1\pmod{11^2}$ and $3^n \equiv 1\pmod {13^2}$ . This occurs if and only if $n$ is a multiple of both of the relatively prime integers $5$ and $39$ , so the least possible value of $n$ is $5\cdot 39=\boxed{195}.$

Solution: We first note that we wish to find $3^n \equiv 1 \pmod{11^2}$ and $3^n \equiv 1 \pmod{13^2}.$ Not thinking of anything else, we try a few numbers for the first condition to get that $5 \mid n.$ For the second condition, upon testing up to 729, we find that it doesn't yield a solution readily, so we use Lifting the Exponent from our toolkit to get that \[v_{13}(27^m-1^m)=v_{13}(26)+v(m)=1+v(m), 3m = n\] which clearly implies $m=13$ and $39 | n.$ Our answer is then obviously $39 \cdot 5 = \boxed{195}.$


Answer: 725.0

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.

Solution: Suppose the original number is $N = \overline{a_na_{n-1}\ldots a_1a_0},$ where the $a_i$ are digits and the first digit, $a_n,$ is nonzero. Then the number we create is $N_0 = \overline{a_{n-1}\ldots a_1a_0},$ so \[N = 29N_0.\] But $N$ is $N_0$ with the digit $a_n$ added to the left, so $N = N_0 + a_n \cdot 10^n.$ Thus, \[N_0 + a_n\cdot 10^n = 29N_0\] \[a_n \cdot 10^n = 28N_0.\] The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number $10^n$ is never divisible by $7,$ so $a_n$ must be divisible by $7.$ But $a_n$ is a nonzero digit, so the only possibility is $a_n = 7.$ This gives \[7 \cdot 10^n = 28N_0\] or \[10^n = 4N_0.\] Now, we want to minimize both $n$ and $N_0,$ so we take $N_0 = 25$ and $n = 2.$ Then \[N = 7 \cdot 10^2 + 25 = \boxed{725},\] and indeed, $725 = 29 \cdot 25.$ $\square$

Solution:
Let $N$ be the required number, and $N'$ be $N$ with the first digit deleted. Now, we know that $N<1000$ (because this is an AIME problem). Thus, $N$ has $1,$ $2$ or $3$ digits. Checking the other cases, we see that it must have $3$ digits.
Let $N=\overline{abc}$ , so $N=100a+10b+c$ . Thus, $N'=\overline{bc}=10b+c$ . By the constraints of the problem, we see that $N=29N'$ , so \[100a+10b+c=29(10b+c).\] Now, we subtract and divide to get \[100a=28(10b+c)\] \[25a=70b+7c.\] Clearly, $c$ must be a multiple of $5$ because both $25a$ and $70b$ are multiples of $5$ . Thus, $c=5$ . Now, we plug that into the equation: \[25a=70b+7(5)\] \[25a=70b+35\] \[5a=14b+7.\] By the same line of reasoning as earlier, $a=7$ . We again plug that into the equation to get \[35=14b+7\] \[b=2.\] Now, since $a=7$ $b=2$ , and $c=5$ , our number $N=100a+10b+c=\boxed{725}$


Answer: 12.0

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$

Solution:
We can rewrite the numerator to be a perfect square by adding $-\dfrac{12x \sin x}{x \sin x}$ . Thus, we must also add back $12$
This results in $\dfrac{(3x \sin x-2)^2}{x \sin x}+12$
Thus, if $3x \sin x-2=0$ , then the minimum is obviously $12$ . We show this possible with the same methods in Solution 1; thus the answer is $\boxed{012}$

Solution:
Let $y = x\sin{x}$ and rewrite the expression as $f(y) = 9y + \frac{4}{y}$ , similar to the previous solution. To minimize $f(y)$ , take the derivative of $f(y)$ and set it equal to zero.
The derivative of $f(y)$ , using the Power Rule, is
$f'(y)$ $9 - 4y^{-2}$
$f'(y)$ is zero only when $y = \frac{2}{3}$ or $y = -\frac{2}{3}$ . It can further be verified that $\frac{2}{3}$ and $-\frac{2}{3}$ are relative minima by finding the derivatives at other points near the critical points, or by checking that the second derivative $f''(y)=8y^{-3}$ is positive. However, since $x \sin{x}$ is always positive in the given domain, $y = \frac{2}{3}$ . Therefore, $x\sin{x}$ $\frac{2}{3}$ , and the answer is $\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}$

Solution: As above, let $y = x\sin{x}$ . Add $\frac{12y}{y}$ to the expression and subtract $12$ , giving $f(x) = \frac{(3y+2)^2}{y} - 12$ . Taking the derivative of $f(x)$ using the Chain Rule and Quotient Rule , we have $\frac{\text{d}f(x)}{\text{d}x} = \frac{6y(3y+2)-(3y+2)^2}{y^2}$ . We find the minimum value by setting this to $0$ . Simplifying, we have $6y(3y+2) = (3y+2)^2$ and $y = \pm{\frac{2}{3}} = x\sin{x}$ . Since both $x$ and $\sin{x}$ are positive on the given interval, we can ignore the negative root. Plugging $y = \frac{2}{3}$ into our expression for $f(x)$ , we have $\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\left(\frac{2}{3}\right)}-12 = \boxed{012}$

Solution: Set $\frac{9x^2\sin^2 x + 4}{x\sin x}$ equal to $y$ . Then multiply by $x\sin x$ on both sides to get $9x^2\sin^2 x + 4 = y\cdot x\sin x$ .  We then subtract $yx\sin x$ from both sides to get $9x^2\sin^2 x + 4 - yx\sin x = 0$ .  This looks like a quadratic so set $z= x\sin x$ and use quadratic equation on $9z^2 - yz + 4 = 0$ to see that $z = \frac{y\pm\sqrt{y^2-144}}{18}$ . We know that $y$ must be an integer and as small as it can be, so $y$ = 12. We plug this back in to see that $x\sin x = \frac{2}{3}$ which we can prove works using methods from solution 1. This makes the answer $\boxed{012}$


Answer: 96.0

Find the number of $7$ -tuples of positive integers $(a,b,c,d,e,f,g)$ that satisfy the following systems of equations: \begin{align*} abc&=70,\\ cde&=71,\\ efg&=72. \end{align*}

Solution:
As 71 is prime, $c$ $d$ , and $e$ must be 1, 1, and 71 (in some order). However, since $c$ and $e$ are divisors of 70 and 72 respectively, the only possibility is $(c,d,e) = (1,71,1)$ . Now we are left with finding the number of solutions $(a,b,f,g)$ satisfying $ab = 70$ and $fg = 72$ , which separates easily into two subproblems. The number of positive integer solutions to $ab = 70$ simply equals the number of divisors of 70 (as we can choose a divisor for $a$ , which uniquely determines $b$ ). As $70 = 2^1 \cdot 5^1 \cdot 7^1$ , we have $d(70) = (1+1)(1+1)(1+1) = 8$ solutions. Similarly, $72 = 2^3 \cdot 3^2$ , so $d(72) = 4 \times 3 = 12$
Then the answer is simply $8 \times 12 = \boxed{096}$

Solution: We know that any two consecutive numbers are coprime. Using this, we can figure out that $c=1$ and $e=1$ $d$ then has to be $71$ . Now we have two equations left. $ab=70$ and $fg=72$ . To solve these we just need to figure out all of the factors. Doing the prime factorization of $70$ and $72$ , we find that they have $8$ and $12$ factors, respectively. The answer is $8 \times 12=\boxed{096}$


Answer: 81.0

Find the number of collections of $16$ distinct subsets of $\{1,2,3,4,5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X \cap Y \not= \emptyset.$

Solution:
Denote by $\mathcal C$ a collection of 16 distinct subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ .
Denote $N = \min \left\{ |S|: S \in \mathcal C \right\}$
Case 1: $N = 0$
This entails $\emptyset \in \mathcal C$ .
Hence, for any other set $A \in \mathcal C$ , we have $\emptyset \cap A = \emptyset$ . This is infeasible.
Case 2: $N = 1$
Let $\{a_1\} \in \mathcal C$ .
To get $\{a_1\} \cap A \neq \emptyset$ for all $A \in \mathcal C$ .
We must have $a_1 \in \mathcal A$
The total number of subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain $a_1$ is $2^4 = 16$ .
Because $\mathcal C$ contains 16 subsets.
We must have $\mathcal C = \left\{ \{a_1\} \cup A : \forall \ A \subseteq \left\{ 1, 2, 3, 4, 5 \right\} \backslash \left\{a_1 \right\} \right\}$ .
Therefore, for any $X, Y \in \mathcal C$ , we must have $X \cap Y \supseteq \{a_1\}$ .
So this is feasible.
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $a_1$ .
Therefore, the number of solutions is 5.
Case 3: $N = 2$
Case 3.1: There is exactly one subset in $\mathcal C$ that contains 2 elements.
Denote this subset as $\left\{ a_1, a_2 \right\}$ .
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ .
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ .
Therefore, the number of solutions is $\binom{5}{2} = 10$
Case 3.2: There are exactly two subsets in $\mathcal C$ that contain 2 elements.
They must take the form $\left\{ a_1, a_2 \right\}$ and $\left\{ a_1, a_3 \right\}$
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ and $\left\{ a_2, a_4, a_5 \right\}$ .
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ and $\left\{ a_1, a_3 \right\}$ .
Therefore, the number of solutions is $5 \cdot \binom{4}{2} = 30$
Case 3.3: There are exactly three subsets in $\mathcal C$ that contain 2 elements.
They take the form $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_1, a_4 \right\}$
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ $\left\{ a_2, a_4, a_5 \right\}$ $\left\{ a_2, a_3, a_5 \right\}$ .
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_1, a_4 \right\}$ .
Therefore, the number of solutions is $5 \cdot \binom{4}{3} = 20$
Case 3.4: There are exactly three subsets in $\mathcal C$ that contain 2 elements.
They take the form $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_2, a_3 \right\}$
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ $\left\{ a_2, a_4, a_5 \right\}$ $\left\{ a_1, a_4, a_5 \right\}$ .
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_2, a_3 \right\}$ .
Therefore, the number of solutions is $\binom{5}{3} = 10$
Case 3.5: There are exactly four subsets in $\mathcal C$ that contain 2 elements.
They take the form $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_1, a_4 \right\}$ $\left\{ a_1, a_5 \right\}$
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$ , except $\left\{ a_3, a_4, a_5 \right\}$ $\left\{ a_2, a_4, a_5 \right\}$ $\left\{ a_1, a_4, a_5 \right\}$ $\left\{ a_2, a_3, a_4 \right\}$ .
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ $\left\{ a_1, a_3 \right\}$ $\left\{ a_1, a_4 \right\}$ $\left\{ a_1, a_5 \right\}$ .
Therefore, the number of solutions is 5.
Putting all subcases together, the number of solutions is this case is $10 + 30 + 20 + 10 + 5 = 75$
Case 4: $N \geq 3$
The number of subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements is $\sum_{i=3}^5 \binom{5}{i} = 16$ .
Because $\mathcal C$ has 16 elements, we must select all such subsets into $\mathcal C$ .
Therefore, the number of solutions in this case is 1.
Putting all cases together, the total number of $\mathcal C$ is $5 + 75 + 1 = \boxed{081}$

Solution:
Denote the $A$ as $\{ 1,2,3,4,5 \}$ and the collection of subsets as $S$
Case 1: There are only sets of size $3$ or higher in $S$ Any two sets in $S$ must have at least one element common to both of them (since $3+3>5$ ). Since there are $16$ subsets of $A$ that have size $3$ or higher, there is only one possibility for this case.
Case 2: There are only sets of size $2$ or higher in $S$ Firstly, there cannot be a no size $2$ set $S$ , or it will be overcounting the first case.
If there is only one such size $2$ set, there are $10$ ways to choose it. That size $2$ set, say $X$ , cannot be in $S$ with $Y = A/X$ (a three element set). Thus, there are only $15$ possible size $3$ subsets that can be in $S$ , giving us $10$ for this case.
If there are two sets with size $2$ in $S$ , we can choose the common elements of these two subsets in $5$ ways, giving us a total of $5\cdot 6 = 30$
If there are three sets with size $2$ , they can either share one common element, which can be done in $5 \cdot 4 = 20$ ways, or they can share pairwise common elements (sort of like a cycle), which can be done $\binom{5}{2} = 10$ ways. In total, we have $30$ possibilities.
If there are four sets with size $2$ , they all have to share one common element, which can be done in $5\cdot 1$ ways.
Thus, summing everything up, this will give us $75$ possible sets $S$
Case 3: There is a set with size $1$ in $S$ Notice that be at most one size $1$ subset. There are $5$ ways to choose that single element set. Say it's $\{ 1\}$ . All other subsets in $S$ must have a $1$ in them, but there are only $2^4-1 = 15$ of them. Thus this case yields $5\cdot 1 = 5$ possibilities.
Thus, the total number of sets $S$ would be $1+75+5 = \boxed{081}$

Solution:
Firstly, there cannot be two subsets with cardinality 1, or they will not intersect.
If there is one subset $A$ with cardinality $1$ ; let the element in $A$ be $a$ , then there are $2^4=16$ subsets that do not include $a$ so they do not work. Every remaining subsets $S$ will have $a$ as an element so $S\cap{A}\geq1$ , since we just excluded all that do not. Since there are 15 subsets left, all are forced into the group of 16 subsets, so we just choose the number of $a$ to determine the set $A$ $A\in\{1,2,3,4,5\}$ so there are 5 ways.
For the rest of the cases, we assume there are no sets $A$ with cardinality 1.  Notice that the only way to "violate" the condition is to have subsets $X$ and $Y$ with cardinalities 2 and 3 in some order. Otherwise, by the Pigeonhole Principle, if two sets both have cardinalities more than 3, they are bound to have one element of intersection. Say a set $S$ has $|S|=2$ , then there is clearly only one set $s$ that will make $S\cap{s}=0$ . By our previous claim, all other subsets that have cardinality $c\geq{3}$ will work.
Now if we generalize a bit: If a subset has $N$ 2-element subsets which belong to set $M$ , then there are exactly $N$ subsets with cardinality 3 that don't work. Therefore, the number of "violating subsets" are all subsets with cardinality $c\leq{1}$ , all 2-element subsets that are not in $M$ , and all corresponding cardinality 3 subsets. Subtracting from the total 32 subsets, we get that $32-(1+5+(10-N)+N)=16$ subsets that do work. This includes all subsets in $M$ , so the remaining non-violating subsets are forced. This is equivalent now to choosing $N$ 2 element subsets.
Following casework on the number of 2-element subsets:
If $N=1$ : There are $\binom{5}{2}=10$ ways.
If $N=2$ : There are $5$ ways to choose the intersection between the 2 sets (remember they have to have at least one element of intersection) and $\binom{4}{2}=6$ ways to choose the distinct elements in the subsets, so there are $5*6=30$ ways.
If $N=3$ : It can be a cycle, where WLOG let the elements be $a,b,c$ so the sets are $\{a,b\}$ $\{b,c\}$ , and $\{c,a\}$ . This is just $\binom{5}{3}=10$ . Alternatively, it can also be the case where all sets share one element. There are 5 ways to choose this element and $\binom{4}{2}=6$ ways to choose the remaining elements to assign to each set. There are $30$ ways.
If $N=4$ : By the Pigeonhole Principle, the only way all pairwise sets have at one common intersection is if all share one element in common. There are 5 ways to choose this element and the remaining numbers are forced. There are 5 ways.
$N\geq{5}$ does not provide any valid cases since to have all pairwise elements to intersect one element, they must be the same element by the Pigeonhole Principle, but there are not enough subsets.
If $N=0$ , then there is only one way since $\binom{5}{3}+\binom{5}{4}+\binom{5}{5}=16$
Adding all cases yields $5+10+30+30+5+1=\boxed{81}$ ways!


Answer: 738.0

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$

Solution:
Plugging $2$ and $m$ into $P(x)$ and equating them, we get $8+4a+2b+c = m^3+am^2+bm+c$ . Rearranging, we have \[(m^3-8) + (m^2 - 4)a + (m-2)b = 0.\] Note that the value of $c$ won't matter as it can be anything in the provided range, giving a total of $41$ possible choices for $c.$ So what we just need to do is to just find the number of ordered pairs $(a, b)$ that work, and multiply it by $41.$ We can start by first dividing both sides by $m-2.$ (Note that this is valid since $m\neq2:$ \[m^2 + 2m + 4 + (m+2)a + b = 0.\] We can rearrange this so it is a quadratic in $m$ \[m^2 + (a+2)m + (4 + 2a + b) = 0.\] Remember that $m$ has to be unique and not equal to $2.$ We can split this into two cases: case $1$ being that $m$ has exactly one solution, and it isn't equal to $2$ ; case $2$ being that $m$ has two solutions, one being equal to $2,$ but the other is a unique solution not equal to $2.$
$\textbf{Case 1:}$
There is exactly one solution for $m,$ and that solution is not $2.$ This means that the discriminant of the quadratic equation is $0,$ using that, we have $(a+2)^2 = 4(4 + 2a + b),$ rearranging in a neat way, we have \[(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2\sqrt{4+b}.\] Using the fact that $4+b$ must be a perfect square, we can easily see that the values for $b$ can be $-4, -3, 0, 5,$ and $12.$ Also since it's a " $\pm$ " there will usually be $2$ solutions for $a$ for each value of $b.$ The two exceptions for this would be if $b = -4$ and $b = 12.$ For $b=-4$ because it would be a $\pm0,$ which only gives one solution, instead of two. And for $b=12$ because then $a = -6$ and the solution for $m$ would equal to $2,$ and we don't want this. (We can know this by putting the solutions back into the quadratic formula).
So we have $5$ solutions for $b,$ each of which give $2$ values for $a,$ except for $2,$ which only give one. So in total, there are $5*2 - 2 = 8$ ordered pairs of $(a,b)$ in this case.
$\textbf{Case 2:}$
$m$ has two solutions, but exactly one of them isn't equal to $2.$ This ensures that $1$ of the solutions is equal to $2.$
Let $r$ be the other value of $m$ that isn't $2.$ By Vieta: \begin{align*} r+2 &= -a-2\\ 2r &= 4+2a+b. \end{align*} From the first equation, we subtract both sides by $2$ and double both sides to get $2r = -2a - 8$ which also equals to $4+2a+b$ from the second equation. Equating both, we have $4a + b + 12 = 0.$ We can easily count that there would be $11$ ordered pairs $(a,b)$ that satisfy that.
However, there's an outlier case in which $r$ happens to also equal to $2,$ and we don't want that. We can reverse engineer and find out that $r=2$ when $(a,b) = (-6, 12),$ which we overcounted. So we subtract by one and we conclude that there are $10$ ordered pairs of $(a,b)$ that satisfy this case.
This all shows that there are a total of $8+10 = 18$ amount of ordered pairs $(a,b).$ Multiplying this by $41$ (the amount of values for $c$ ) we get $18\cdot41=\boxed{738}$ as our final answer.

Solution:
$p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$ , with $m\neq 2$
In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$ , with $|4+4m|\leq 20$ (which entails $|4+m|\leq 20$ ), so $m$ can be $-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$ , giving $410$ polynomials (since the coefficients are given by linear functions of $m$ and thus are distinct).
In the second case $p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)$ , and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$ , giving $328$ polynomials.
The total is $\boxed{738}$


Answer: 200.0

Find the number of five-digit positive integers, $n$ , that satisfy the following conditions:

Solution: The number takes a form of $5\text{x,y,z}5$ , in which $5|(x+y+z)$ . Let $x$ and $y$ be arbitrary digits. For each pair of $x,y$ , there are exactly two values of $z$ that satisfy the condition of $5|(x+y+z)$ . Therefore, the answer is $10\times10\times2=\boxed{200}$


Answer: 210.0

Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$ , and two distinct elements of a subset have a sum of $24$ . For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ are two such subsets.

Solution:
This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set $\{a, b, c, d\}$
Note that there are only two cases: 1 where $a + b = 16$ and $c + d = 24$ or 2 where $a + b = 16$ and $a + c = 24$ . Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you $a=d$ , which cannot be true.
Case 1.
This is probably the simplest: just make a list of possible combinations for $\{a, b\}$ and $\{c, d\}$ . We get $\{1, 15\}\dots\{7, 9\}$ for the first and $\{4, 20\}\dots\{11, 13\}$ for the second. That appears to give us $7*8=56$ solutions, right? NO. Because elements can't repeat, take out the supposed sets \[\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 12, 4, 20\}, \{5, 11, 5, 19\},\] \[\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}\] That's ten cases gone. So $46$ for Case 1.
Case 2.
We can look for solutions by listing possible $a$ values and filling in the blanks. Start with $a=4$ , as that is the minimum. We find $\{4, 12, 20, ?\}$ , and likewise up to $a=15$ . But we can't have $a=8$ or $a=12$ because $a=b$ or $a=c$ , respectively! Now, it would seem like there are $10$ values for $a$ and $17$ unique values for each $?$ , giving a total of $170$ , but that is once again not true because there are some repeated values!
There are two cases of overcounting:
case 1) (5,11,13,19) & (5.11.19.13)
The same is for (6,10,14,18) and (7,9,15,17)
case 2) those that have the same b and c values
this case includes:
(1,15,9,7) and (7,9,15,1)
(2,14,10,6) and (6,10,14,2)
(3,13,11,5) and (5,11,13,3)
So we need to subtract 6 overcounts.
So, that's $164$ for Case 2.
Total gives $\boxed{210}$

Solution:
Let's say our four elements in our subset are $a,b,c,d$ . We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.
$\textrm{Case } 1 \textrm{:}$ $a+b = 16$ and $c+d = 24$
List out possibilities for $a+b$ $(\text{i.e. } 1+15, 2+14, 3+13 \text{ etc.})$ but don't list $8+8$ because those are the same elements and that is restricted.
Then list out the possibilities for $c+d \text{ }(\text{i.e. } 4+20, 5+19, 6+18, \text{ etc.})$ but don't list $12+12$ because they are the same elements.
This will give you $7 \cdot 8$ elements, which is $56$ . However, as stated above, we have overlap. Just count starting from $a+b$ $15,14,13,11,10,9,7,6,5,4$ all overlap once, which is $10$ , thus $56 - 10 = 46$ cases in this case. Note that $12$ wasn't included because again, if $c+d = 24$ $c$ and $d$ cannot be $12$
$\textrm{Case } 2 \textrm{:}$ $a+b = 16$ and $b+c = 24$
Here, $b$ is included in both equations. We can easily see that $a, b, c$ will never equal each other.
Furthermore, there are 17 choices for $d$ $20 - 3$ included elements) for each $b$ . Listing out the possible $b$ s, we go from $15,14,13,11,10,9,7,6,5,4$ . Do not include $8$ or $12$ because if they are included, then $a/c$ will be the same as $b$ , which is restricted.
There are $10$ options there, and thus $10 \cdot 17 = 170$ . But, note that if $d = b+8$ $a+d = a+b+8 = 24$ , and so we have a double-counted set. Starting with $b=15$ , we have $15, 14, 13, 11, 10, 9$ (where $d$ is $7, 6, 5, 3, 2, 1)$ . That means there are $6$ double-counted cases. Thus $170 - 6 = 164$ cases in this case.
Adding these up, we get $46+164 = \boxed{210}.$


Answer: 185.0

Find the number of functions $f$ from $\{0, 1, 2, 3, 4, 5, 6\}$ to the integers such that $f(0) = 0$ $f(6) = 12$ , and \[|x - y| \leq |f(x) - f(y)| \leq 3|x - y|\] for all $x$ and $y$ in $\{0, 1, 2, 3, 4, 5, 6\}$

Solution:
Because $f(n)$ and $f(n+1)$ can differ by at most 3 for $n=0,1,2,3,4,5$ $f$ can decrease at most once. This is because if $f$ decreases $2$ times (or more) then we get a contradiction: $12 = f(6) \le 3\cdot 4 + (-1)\cdot 2 = 10$
If $f$ never decreases, then $f(n+1)-f(n)\in \{1,2,3\}$ for all $n$ . Let $a, b$ , and $c$ denote the number of times this difference is $1, 2$ , and $3$ , respectively. Then $a+b+c=6$ and $a+2b+3c = 12$ . Subtracting the first equation from the second yields $b+2c=6$ , so $(b,c)=(6,0), (4,1), (2,2)$ , or $(0,3)$ . These yield $a=0,1,2$ , or $3$ , respectively, so the number of possibilities in this case is \[\binom{6}{0,6,0}+\binom{6}{1,4,1}+\binom{6}{2,2,2}+\binom{6}{3,0,3}=141.\] If $f$ decreases from $f(0)$ to $f(1)$ or from $f(5)$ to $f(6)$ , then $f(2)$ or $f(4)$ , respectively, is determined. The only solutions to $a+b+c=4$ and $a+2b+3c=10$ are $(a,b,c)=(1,0,3)$ and $(a,b,c)=(0,2,2)$ , so the number of functions is \[2\left[\binom{4}{1,0,3}+\binom{4}{0,2,2}\right]=20.\] Finally, suppose that $f(n+1)<f(n)$ for some $n=1,2,3,4$ . Note that the condition $|(n+1)-(n-1)|\le |f(n+1)-f(n-1)|$ implies that $f(n+1)-f(n-1)\ge 2$ , so it must be that $f(n)-f(n+1)=1$ and \[f(n+2)-f(n+1)=f(n)-f(n-1)=3.\] This means that $f(n-1)$ and $f(n+2)$ are uniquely determined by the value of $f(n)$ , and, in particular, that $f(n+2)-f(n-1)=5$ . As a result, there are three more values of $f$ to determine, and they must provide a total increase of $7$ . The only ways to do this are either to have two differences of $3$ and one difference of $1$ , which can be arranged in $3$ ways, or to have one difference of $3$ and two differences of $2$ , which can be arranged in $3$ ways. Thus for each of the $4$ possibilities for $n$ , there are $6$ ways to arrange the increases, giving a total of $24$ ways.
The total number of functions is $141+20+24=\boxed{185}$


Answer: 756.0

Solution:
We perform casework on the number of fixed points (the number of points where $f(x) = x$ ).  To better visualize this, use the grid from Solution 1.
Case 1: 5 fixed points
Case 2: 4 fixed points
Case 3: 3 fixed points
Case 4: 2 fixed points
Case 5: 1 fixed point
Therefore, the answer is $1+20+150+380+205 = \boxed{756}$


Answer: 501.0

Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution.

Solution: Note the equation $\log(kx)=2\log(x+2)$ is valid for $kx>0$ and $x>-2$ $\log(kx)=2\log(x+2)=\log((x+2)^2)$ . The equation $kx=(x+2)^2$ is derived by taking away the outside logs from the previous equation. Because $(x+2)^2$ is always non-negative, $kx$ must also be non-negative; therefore this takes care of the $kx>0$ condition as long as $k\neq0$ , i.e. $k$ cannot be $0$ . Now, we graph both $(x+2)^2$ (the green graph) and $kx$ (the red graph for $k=-1,k=-2,k=-3,k=8$ ) for $x>-2$ . It is easy to see that all negative values of $k$ make the equation $\log(kx)=2\log(x+2)$ have only one solution. However, there is also one positive value of $k$ that makes the equation only have one solution, as shown by the steepest line in the diagram. We can show that the slope of this line is a positive integer by setting the discriminant of the equation $(x+2)^2=kx$ to be $0$ and solving for $k$ . Therefore, there are $500$ negative solutions and $1$ positive solution, for a total of $\boxed{501}$

Solution:
We use an algebraic approach. Since $\log(kx)=2\log(x+2)$ , then $kx = (x+2)^2$ (the converse isn't necessarily true!), or $x^2+(4-k)x+4=0$ . Our original equation has exactly one solution if and only if there is only one solution to the above equation, or one of the solutions is extraneous; it involves the computation of the log of a nonpositive number.
For the first case, we note that this can only occur when it is a perfect square trinomal, or $k = 0, 8$ . However, $k = 0$ results in $\log(0)$ on the LHS, which is invalid. $k = 8$ yields $x = 2$ , so that is one solution.
For the second case, we can use the quadratic formula. We have \[x = \frac{k-4 \pm \sqrt{k^2-8k}}2,\] so in order for there to be at least one real solution, the discriminant must be nonnegative, or $k < 0$ or $k > 8$
Note that if $k > 8$ , then both solutions to $x$ will be positive, and therefore both valid, which means all such $k$ are unsatisfactory.
We now wish to show that if $k < 0$ , then there is exactly one solution that works. Note that whenever $k < 0$ , both "solutions" in $x$ are negative. One of the solutions to the equation is $x = \frac{k-4 + \sqrt{k^2-8k}}2$ . We wish to prove that $x + 2 > 0$ , or $x > -2$ (therefore the RHS in the original equation will be defined). Substituting, we have $\frac{k-4 + \sqrt{k^2-8k}}2 > -2$ , or $\sqrt{k^2 - 8k} > -k$ . Since both sides are positive, we can square both sides (if $k < 0$ , then $-k > 0$ ) to get $k^2-8k > k^2$ , or $8k < 0 \implies k < 0$ , which was our original assumption, so this solution satisfies the original equation. The other case is when $x = \frac{k-4 - \sqrt{k^2-8k}}2$ , which we wish to show is less that $-2$ , or $\frac{k-4 - \sqrt{k^2-8k}}2 < -2 \implies k < \sqrt{k^2-8k}$ . However, since the square root is defined to be positive, then this is always true, which implies that whenever $k < 0$ , there is exactly one real solution that satisfies the original equation. Combining this with $k \in [-500, 500]$ , we find that the answer is $500 + 1 = \boxed{501}$

Solution: Immediately we notice $k$ is non-zero, in fact we must have $kx, (x+2) > 0$ for our sole solution $x = x_0$ . Simplifying the logarithmic equation we get $kx = (x+2)^2 \rightarrow 0 = x^2 + (4-k)x + 4 \rightarrow x = \frac{k-4 \pm \sqrt{k^2 - 8k}}{2}$ . Then $k \leq 0$ or $k \geq 8$ . When $k = 8$ we have exactly one real solution (easily verifiable). Notice when $k > 8$ , both solutions to $x$ are positive, and so all such $k$ are not satisfactory. When $k < 0$ it can be shown that the greater solution to $x$ is in the interval $(0,-2)$ and the lesser solution is in the interval $(-2,-\infty)$ which is satisfactory. Then $k = 8,-1,-2,...,-500 \rightarrow \boxed{501}$ satisfactory integer values of $k$


Answer: 57.0

Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions.

Solution:
We take cases for the outermost absolute value, then rearrange: \[\left|20|x|-x^2\right|=c\pm21.\] Let $f(x)=\left|20|x|-x^2\right|.$ We rewrite $f(x)$ as a piecewise function without using absolute values: \[f(x) = \begin{cases} \left|-20x-x^2\right| & \mathrm{if} \ x \le 0  \begin{cases} 20x+x^2 & \mathrm{if} \ x\le-20 \\ -20x-x^2 & \mathrm{if} \ -20<x\leq0 \end{cases} \\ \left|20x-x^2\right| & \mathrm{if} \ x > 0 \begin{cases} 20x-x^2 & \mathrm{if} \ 0<x\leq20 \\ -20x+x^2 & \mathrm{if} \ x>20 \end{cases} \end{cases}.\] We graph $y=f(x)$ with all extremum points labeled, as shown below. The fact that $f(x)$ is an even function ( $f(x)=f(-x)$ holds for all real numbers $x,$ so the graph of $y=f(x)$ is symmetric about the $y$ -axis) should facilitate the process of graphing.  Since $f(x)=c\pm21$ has $12$ distinct real solutions, it is clear that each case has $6$ distinct real solutions geometrically. We shift the graphs of $y=\pm21$ up $c$ units, where $c\geq0:$
Taking the intersection of these two cases gives $21<c<79,$ from which there are $79-21-1=\boxed{057}$ such integers $c.$

Solution:
Graph $y=|20|x|-x^2|$ (If you are having trouble, look at the description in the next two lines and/or the diagram in Solution 1). Notice that we want this to be equal to $c-21$ and $c+21$
We see that from left to right, the graph first dips from very positive to $0$ at $x=-20$ , then rebounds up to $100$ at $x=-10$ , then falls back down to $0$ at $x=0$
The positive $x$ are symmetric, so the graph re-ascends to $100$ at $x=10$ , falls back to $0$ at $x=10$ , and rises to arbitrarily large values afterwards.
Now we analyze the $y$ (varied by $c$ ) values. At $y=k<0$ , we will have no solutions, as the line $y=k$ will have no intersections with our graph.
At $y=0$ , we will have exactly $3$ solutions for the three zeroes.
At $y=n$ for any $n$ strictly between $0$ and $100$ , we will have exactly $6$ solutions.
At $y=100$ , we will have $4$ solutions, because local maxima are reached at $x= \pm 10$
At $y=m>100$ , we will have exactly $2$ solutions.
To get $12$ distinct solutions for $y=|20|x|-x^2|=c \pm 21$ , both $c +21$ and $c-21$ must produce $6$ solutions.
Thus $0<c-21$ and $c+21<100$ , so $c \in \{ 22, 23, \dots , 77, 78 \}$ is required.
It is easy to verify that all of these choices of $c$ produce $12$ distinct solutions (none overlap), so our answer is $\boxed{057}$

Solution:
Let $y = |x|.$ Then the equation becomes $\left|\left|20y-y^2\right|-c\right| = 21$ , or $\left|y^2-20y\right| = c \pm 21$ . Note that since $y = |x|$ $y$ is nonnegative, so we only care about nonnegative solutions in $y$ . Notice that each positive solution in $y$ gives two solutions in $x$ $x = \pm y$ ), whereas if $y = 0$ is a solution, this only gives one solution in $x$ $x = 0$ . Since the total number of solutions in $x$ is even, $y = 0$ must not be a solution. Hence, we require that $\left|y^2-20y\right| = c \pm 21$ has exactly $6$ positive solutions and is not solved by $y = 0.$
If $c < 21$ , then $c - 21$ is negative, and therefore cannot be the absolute value of $y^2 - 20y$ . This means the equation's only solutions are in $\left|y^2-20y\right| = c + 21$ . There is no way for this equation to have $6$ solutions, since the quadratic $y^2-20y$ can only take on each of the two values $\pm(c + 21)$ at most twice, yielding at most $4$ solutions. Hence, $c \ge 21$ $c$ also can't equal $21$ , since this would mean $y = 0$ would solve the equation. Hence, $c > 21.$
At this point, the equation $y^2-20y = c \pm 21$ will always have exactly $2$ positive solutions, since $y^2-20y$ takes on each positive value exactly once when $y$ is restricted to positive values (graph it to see this), and $c \pm 21$ are both positive. Therefore, we just need $y^2-20y = -(c \pm 21)$ to have the remaining $4$ solutions exactly. This means the horizontal lines at $-(c \pm 21)$ each intersect the parabola $y^2 - 20y$ in two places. This occurs when the two lines are above the parabola's vertex $(10,-100)$ . Hence we have \begin{align*} -(c + 21) &> -100 \\ c + 21 &< 100 \\ c &< 79. \end{align*} Hence, the integers $c$ satisfying the conditions are those satisfying $21 < c < 79.$ There are $\boxed{057}$ such integers.

Solution:
Removing the absolute value bars from the equation successively, we get \begin{align*} \left|\left|20|x|-x^2\right|-c\right|&=21 \\ \left|20|x|-x^2\right|&= c \pm21 \\ 20|x|-x^2 &= \pm c \pm 21 \\ x^2 \pm 20x \pm c \pm21 &= 0. \end{align*} The discriminant of this equation is \[\sqrt{400-4(\pm c \pm 21)}.\] Equating the discriminant to $0$ , we see that there will be two distinct solutions to each of the possible quadratics above only in the interval $-79 < c < 79$ . However, the number of zeros the equation $ax^2+b|x|+k$ has is determined by where $ax^2+bx+k$ and $ax^2-bx+k$ intersect, namely at $(0,k)$ . When $k<0$ $a>0$ $ax^2+b|x|+k$ will have only $2$ solutions, and when $k>0$ $a>0$ , then there will be $4$ real solutions, if they exist at all.
In order to have $12$ solutions here, we thus need to ensure $-c+21<0$ , so that exactly $2$ out of the $4$ possible equations of the form $ax^2+b|x|+k$ given above have y-intercepts below $0$ and only $2$ real solutions, while the remaining $2$ equations have $4$ solutions. This occurs when $c>21$ , so our final bounds are $21<c<79$ , giving us $\boxed{057}$ valid values of $c$


Answer: 295.0

Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$

Solution:
This solution refers to the Remarks section.
By the Euclidean Algorithm, we have \[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.\] We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1\right)$ gives \begin{align*} \gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ \gcd\left(\left(2^m+1\right)\left(2^m-1\right),2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \hspace{12mm} &&\text{by }\textbf{Claim 1} \\ \gcd\left(2^{2m}-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ 2^{\gcd(2m,n)}-1&>2^{\gcd(m,n)}-1 &&\text{by }\textbf{Claim 2} \\ \gcd(2m,n)&>\gcd(m,n), \end{align*} which implies that $n$ must have more factors of $2$ than $m$ does.
We construct the following table for the first $30$ positive integers: \[\begin{array}{c|c|c} && \\ [-2.5ex] \boldsymbol{\#}\textbf{ of Factors of }\boldsymbol{2} & \textbf{Numbers} & \textbf{Count} \\ \hline && \\ [-2.25ex]  0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\  && \\ [-2.25ex]  1 & 2,6,10,14,18,22,26,30 & 8 \\ && \\ [-2.25ex]  2 & 4,12,20,28 & 4 \\    && \\ [-2.25ex]  3 & 8,24 & 2 \\  && \\ [-2.25ex]  4 & 16 & 1 \\ \end{array}\] To count the ordered pairs $(m,n),$ we perform casework on the number of factors of $2$ that $m$ has:
Together, the answer is $225+56+12+2=\boxed{295}.$


Answer: 228.0

Find the number of ordered pairs of integers $(a, b)$ such that the sequence \[3, 4, 5, a, b, 30, 40, 50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.

Solution:
Since $3,4,5,a$ and $3,4,5,b$ cannot be an arithmetic progression, $a$ or $b$ can never be $6$ . Since $b, 30, 40, 50$ and $a, 30, 40, 50$ cannot be an arithmetic progression, $a$ and $b$ can never be $20$ . Since $a < b$ , there are ${24 - 2 \choose 2} = 231$ ways to choose $a$ and $b$ with these two restrictions in mind.
However, there are still specific invalid cases counted in these $231$ pairs $(a,b)$ . Since \[3,5,a,b\] cannot form an arithmetic progression, $\underline{(a,b) \neq (7,9)}$ \[a,b,30,50\] cannot be an arithmetic progression, so $(a,b) \neq (-10,10)$ ; however, since this pair was not counted in our $231$ , we do not need to subtract it off. \[3,a,b,30\] cannot form an arithmetic progression, so $\underline{(a,b) \neq (12,21)}$ \[4, a, b, 40\] cannot form an arithmetic progression, so $\underline{(a,b) \neq (16,28)}$ \[5, a,b, 50\] cannot form an arithmetic progression, $(a,b) \neq 20, 35$ ; however, since this pair was not counted in our $231$ (since we disallowed $a$ or $b$ to be $20$ ), we do not to subtract it off.
Also, the sequences $(3,a,b,40)$ $(3,a,b,50)$ $(4,a,b,30)$ $(4,a,b,50)$ $(5,a,b,30)$ and $(5,a,b,40)$ will never be arithmetic, since that would require $a$ and $b$ to be non-integers.
So, we need to subtract off $3$ progressions from the $231$ we counted, to get our final answer of $\boxed{228}$

Solution:
We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract $\textbf{1}$ from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50.
(Note if we take only 1 at a time, there will have to be 3 of $a, b$ , which is impossible.)
If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.
Finally, we count the sequences that are something like (one of 3,4,5,), $a, b$ , (one of 30, 40, 50). If this is to be the case, then let $a$ be the starting value in the sequence. The sequence will be $a, a+d, a+2d, a+3d$ ; We see that if we subtract the largest term by the smallest term we have $3d$ , so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are $3,a,b,30; 4,a,b,40; 5,a,b,50$ . Of these, only the last is invalid because it gives $b = 35$ , larger than our bounds $6<a<b<30$ . Therefore, we subtract $\textbf{2}$ from this case.
Our final answer is $231 - 1 - 2 = \boxed{228}$

Solution:
Denote $S = \left\{ (a, b) : 6 \leq a < b \leq 29 \right\}$
Denote by $A$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $a$ but not $b$
Denote by $B$ a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes $b$ but not $a$
Hence, $C$ is a subset of $S$ , such that there exists an arithmetic sequence that has 4 terms and includes both $a$ and $b$
Hence, this problem asks us to compute \[ | S | - \left( | A | + | B | + | C | \right) . \]
First, we compute $| S |$
We have $| S | = \binom{29 - 6 + 1}{2} = \binom{24}{2} = 276$
Second, we compute $| A |$
$\textbf{Case 1}$ $a = 6$
We have $b = 8 , \cdots , 19, 21, 22, \cdots, 29$ .
Thus, the number of solutions is 22.
$\textbf{Case 2}$ $a = 20$
We have $b = 21, 22, \cdots , 29$ .
Thus, the number of solutions is 9.
Thus, $| A | = 22 + 9 = 31$
Third, we compute $| B |$
In $B$ , we have $b = 6, 20$ . However, because $6 \leq a < b$ , we have $b \geq 7$ .
Thus, $b = 20$
This implies $a = 7, 8, 9, 11, 12, \cdots , 19$ . Note that $(a, b)=(10, 20)$ belongs in $C$
Thus, $| B | = 12$
Fourth, we compute $| C |$
$\textbf{Case 1}$ : In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the same side of $a$ and $b$
Hence, $(a, b) = (6 , 7), (7, 9) , (10, 20)$ .
Therefore, the number solutions in this case is 3.
$\textbf{Case 2}$ : In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the opposite sides of $a$ and $b$
$\textbf{Case 2.1}$ : The arithmetic sequence is $3, a, b, 30$
Hence, $(a, b) = (12, 21)$
$\textbf{Case 2.2}$ : The arithmetic sequence is $4, a, b, 40$
Hence, $(a, b) = (16, 28)$
$\textbf{Case 2.3}$ : The arithmetic sequence is $5, a, b, 50$
Hence, $(a, b) = (20, 35)$ . However, the sequence $... 20, 35, 30, 40, 50$ is not strictly increasing.
Putting two cases together, $| C | = 65.$
Therefore, \[| S | - \left( | A | + | B | + | C | \right) = 276 - \left( 31 + 12 + 5 \right) = \boxed{228}.\]

Solution:
divide cases into $7\leq a<20; 21\leq a\leq28$ .(Notice that $a$ can't be equal to $6,20$ , that's why I divide them into two parts.
There are three cases that arithmetic sequence forms: $3,12,21,30;4,16,28,40;3,5,7,9$ .(NOTICE that $5,20,35,50$ IS NOT A VALID SEQUENCE!)
So when $7\leq a<20$ , there are $10+11+12+...+22-3-13=192$ possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)
When $21\leq a \leq 28$ , there are $1+2+\cdots+8=36$ ways.
In all, there are $192+36=\boxed{228}$ possible sequences.


Answer: 34.0

Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$

Solution: Solving for $m$ gives us $m = \frac{503-3n}{5},$ so in order for $m$ to be an integer, we must have $3n \equiv 503 \mod 5 \longrightarrow n \equiv 1 \mod 5.$ The smallest possible value of $n$ is obviously $1,$ and the greatest is $\frac{503 - 5}{3} = 166,$ so the total number of solutions is $\frac{166-1}{5}+1 = \boxed{34}$

Solution:
Dividing by $4$ gives us $5m + 3n = 503$ . Thus, we have $503-5m\equiv 0 \pmod 3$ since $n$ is an integer. Rearranging then gives $m\equiv 1\pmod 3.$ Since $503-5m>0,$ we know that $m<503/5.$ Because $m$ is an integer, we can rewrite this as $m\le 100.$ Therefore, $m$ ranges from \[0\cdot 3+1\quad \text{to} \quad 33\cdot 3+1,\] giving $\boxed{034}$ values.
~vaporwave

Solution: Because the x-intercept of the equation is $\frac{2012}{20}$ , and the y-intercept is $\frac{2012}{12}$ , the slope is $\frac{\frac{-2012}{12}}{\frac{2012}{20}} = \frac{-5}{3}$ . Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: $(100,1), (97,6), (94,11)...$ Because the solutions are only positive, we can generate only 33 more solutions, so in total we have $33+1=\boxed{34}$ solutions.


Answer: 738.0

Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.

Solution:
There are $\left\lfloor\frac{999}{10}\right\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that the only time both $a$ and $b$ can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting ).
Excluding the numbers divisible by 100, which were counted already, there are $9$ numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling $9 \cdot 9 = 81$ such numbers; considering $b$ also and we have $81 \cdot 2 = 162$ . Therefore, there are $999 - (99 + 162) = \boxed{738}$ such ordered pairs.

Solution:
We proceed by casework on the number of digits of $a.$
Case 1: Both $a$ and $b$ have three digits
We now use constructive counting. For the hundreds digit of $a,$ we see that there are $8$ options - the numbers $1$ through $8.$ (If $a = 9,$ that means that $b$ will be a two digit number, and if $a = 0,$ $a$ will have two digits). Similarly, the tens digit can be $1-8$ as well because a tens digit of $0$ is obviously prohibited and a tens digit of $9$ will lead to a tens digit of $0$ in the other number. The units digit can be anything from $1-9.$ Hence, there are $8 \cdot 8 \cdot 9 = 576$ possible values in this case.
Case 2: $a$ (or $b$ ) has two digits
If $a$ has two digits, the only restrictions are that the units digit must not be $0$ and the tens digit must not be $9$ (because then that would lead to $b$ beginning with $90...$ ). There thus are $8 \cdot 9 = 72$ possibilities for $a,$ and we have to multiply by $2$ because there are the same number of possibilities for $b.$ Thus, there are $72 \cdot 2 = 144$ possible values in this case.
Case 3: $a$ (or $b$ ) has one digit
This is easy -- $a$ can be anything from $1$ to $9,$ for a total of $9$ possible values. We multiply this by $2$ to account for the single digit $b$ values, so we have $9 \cdot 2 = 18$ possible values for this case.
Adding them all up, we get $576 + 144 + 18 = \boxed{738},$ and we're done.

Solution:
For every $a \in [1, 999]$ $(a, 1000 - a)$ is a potential candidate for a solution, barring the cases where $a$ or $1000 -a$ has zero digits.
First, let's consider all viable $a$ that do not have a zero digit. As there are 9 non-zero digits, we have:
However, we have still overlooked the cases where $1000 - a$ contains zero digits:
Adding up the cases with $a \in [1, 999]$ with no zero digits and removing the cases with $1000 - a$ with zero digits gives us \[(9^1 + 9^2 + 9^3) - 9^2 = \boxed{738}\]


Answer: 231.0

Find the number of ordered pairs of positive integers $(m,n)$ such that ${m^2n = 20 ^{20}}$

Solution:
In this problem, we want to find the number of ordered pairs $(m, n)$ such that $m^2n = 20^{20}$ . Let $x = m^2$ . Therefore, we want two numbers, $x$ and $n$ , such that their product is $20^{20}$ and $x$ is a perfect square. Note that there is exactly one valid $n$ for a unique $x$ , which is $\tfrac{20^{20}}{x}$ . This reduces the problem to finding the number of unique perfect square factors of $20^{20}$
$20^{20} = 2^{40} \cdot 5^{20} = \left(2^2\right)^{20}\cdot\left(5^2\right)^{10}.$ Therefore, the answer is $21 \cdot 11 = \boxed{231}.$


Answer: 2004.0

Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$
$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

Solution:
As in the other solution, split the problem into when $s=0$ and when $s=1$ . When $s=1$ and $a+bi=\cos\theta+i\sin\theta$
$(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta)$ $=a-bi= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta)$
so we must have $2002\theta=-\theta+2\pi k$ and hence $\theta=\frac{2\pi k}{2003}$ . Since $\theta$ is restricted to $[0,2\pi)$ $k$ can range from $0$ to $2002$ inclusive, which is $2002-0+1=2003$ values. Thus the total is $1+2003 = \boxed{2004}$

Solution: Let $z = a + bi = re^{i\theta}$ where $r = |z|$ and $0\leq \theta < 2\pi.$ We want to solve $z^{2002} = \overline{z}.$ Since $z\overline{z} = |z|^2,$ we multiply by $z$ on both sides to get \[z^{2003} = |z|^2 = r^{2003}e^{2023i\theta} = r^2,\] from which we get $r\in \{0, 1\}$ (since $r\in \mathbb{R}$ ), immediately yielding $z = 0$ as an answer. If $r=1$ then we have $z^{2003} = 1$ , which each of the $2003$ rd roots of unity satisfies. Altogether, there are $1 + 2003 = \boxed{2004}$ values of $z$ , each of which correspond to a unique ordered pair $(a, b)$


Answer: 2004.0

Solution:
Let $s=\sqrt{a^2+b^2}$ be the magnitude of $a+bi$ . Then the magnitude of $(a+bi)^{2002}$ is $s^{2002}$ , while the magnitude of $a-bi$ is $s$ . We get that $s^{2002}=s$ , hence either $s=0$ or $s=1$
For $s=0$ we get a single solution $(a,b)=(0,0)$
Let's now assume that $s=1$ . Multiply both sides by $a+bi$ . The left hand side becomes $(a+bi)^{2003}$ , the right hand side becomes $(a-bi)(a+bi)=a^2 + b^2 = 1$ . Hence the solutions for this case are precisely all the $2003$ rd complex roots of unity, and there are $2003$ of those.
The total number of solutions is therefore $1+2003 = \boxed{2004}$

Solution:
Let $r$ be the magnitude of $a+bi$ . Notice that $r$ must be either $0$ or $1$ for this to be true, as shown in the above solutions. We know this because we are taking magnitude to the $2003$ rd power, and if the magnitude of $a+bi$ is larger than $1$ , it will increase and if it is smaller than $1$ it will decrease. However, the magnitude on the RHS is still $r$ , so this is not possible. Again, only $r=0$ and $r=1$ satisfy the equation.
Now if $r=0$ , then $(a,b)$ must be $(0,0)$
However if $r=1$ , we then have:
$\cos(2002 \theta) = \cos(-\theta)$ . This has solution of $\theta = 0$ . This would represent the number $1+0i$ , with conjugate $1-0i$ . This works because the magnitude is the same and the angle is nothing anyways. We multiply angle by $2002$ through De Moivre's Theorem and also we do $-\theta$ because it is a reflection, angles therefore is negative.
We then write:
$\cos(2002 \theta) = \cos(360-\theta)$ which has solution of $\theta = \frac{360}{2003}$
We can also write:
$\cos(2002 \theta) = \cos(720-\theta)$ which has solution $\theta = \frac{720}{2003}$
We notice that it is simply headed upwards and the answer is of the form $\frac{720}{2003} n$ , where n is some integer from $0$ to infinity, inclusive.
Well wait, it repeats itself $n=2003$ , that is $360$ which is also $0$ ! Hence we only have $n=0$ to $2002$ as original solutions, or $2003$ solutions.
$1+2003 = \boxed{2004}$


Answer: 200.0

Find the number of ordered triples $(a,b,c)$ where $a$ $b$ , and $c$ are positive integers $a$ is a factor of $b$ $a$ is a factor of $c$ , and $a+b+c=100$

Solution:
Denote $x = \frac{b}{a}$ and $y = \frac{c}{a}$ . The last condition reduces to $a(1 + x + y) = 100$ . Therefore, $1 + x + y$ is equal to one of the 9 factors of $100 = 2^25^2$
Subtracting the one, we see that $x + y = \{0,1,3,4,9,19,24,49,99\}$ . There are exactly $n - 1$ ways to find pairs of $(x,y)$ if $x + y = n$ . Thus, there are $0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = \boxed{200}$ solutions of $(a,b,c)$


Answer: 0.0

Find the number of pairs $(m, n)$ of integers which satisfy the equation $m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1$
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }3\qquad \textbf{(D) }9\qquad \textbf{(E) }\infty$

Solution: The equation is equivalent to $m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1$ . Taking mod 3, we get $m(m+1)(m+2)=1 (\bmod 3)$ . However, $m(m+1)(m+2)$ is always divisible by $3$ for any integer $m$ . Thus, the answer is $\boxed{0}$ Solution by mickyboy789


Answer: 63.0

Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\]

Solution:
It is trivial that the maximum value of $\sin \theta$ is $1$ , is achieved at $\theta = \frac{\pi}{2}+2k\pi$ for some integer $k$
This implies that $\sin(mx) = \sin(nx) = 1$ , and that $mx = \frac{\pi}{2}+2a\pi$ and $nx = \frac{\pi}{2}+2b\pi$ , for integers $a, b$
Taking their ratio, we have \[\frac{mx}{nx} = \frac{\frac{\pi}{2}+2a\pi}{\frac{\pi}{2}+2b\pi} \implies \frac{m}{n} = \frac{4a + 1}{4b + 1} \implies \frac{m}{4a + 1} = \frac{n}{4b + 1} = k.\] It remains to find all $m, n$ that satisfy this equation.
If $k = 1$ , then $m \equiv n \equiv 1 \pmod 4$ . This corresponds to choosing two elements from the set $\{1, 5, 9, 13, 17, 21, 25, 29\}$ . There are $\binom 82$ ways to do so.
If $k < 1$ , by multiplying $m$ and $n$ by the same constant $c = \frac{1}{k}$ , we have that $mc \equiv nc \equiv 1 \pmod 4$ . Then either $m \equiv n \equiv 1 \pmod 4$ , or $m \equiv n \equiv 3 \pmod 4$ . But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set $\{3, 7, 11, 15, 19, 23, 27\}$ . There are $\binom 72$ ways here. (This argument seems to have a logical flaw)
Finally, if $k > 1$ , note that $k$ must be an integer. This means that $m, n$ belong to the set $\{k, 5k, 9k, \dots\}$ , or $\{3k, 7k, 11k, \dots\}$ . Taking casework on $k$ , we get the sets $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}$ . Some sets have been omitted; this is because they were counted in the other cases already. This sums to $\binom 42 + \binom 42 + \binom 22 + \binom 22$
In total, there are $\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{063}$ pairs of $(m, n)$

Solution:
In order for $\sin(mx) + \sin(nx) = 2$ $\sin(mx) = \sin(nx) = 1$
This happens when $mx \equiv nx \equiv \frac{\pi}{2} ($ mod $2\pi).$
This means that $mx = \frac{\pi}{2} + 2\pi\alpha$ and $nx = \frac{\pi}{2} + 2\pi\beta$ for any integers $\alpha$ and $\beta$
As in Solution 1, take the ratio of the two equations: \[\frac{mx}{nx} = \frac{\frac{\pi}{2}+2\pi\alpha}{\frac{\pi}{2}+2\pi\beta} \implies \frac{m}{n} = \frac{\frac{1}{2}+2\alpha}{\frac{1}{2}+2\beta} \implies \frac{m}{n} = \frac{4\alpha+1}{4\beta+1}\]
Now notice that the numerator and denominator of $\frac{4\alpha+1}{4\beta+1}$ are both odd, which means that $m$ and $n$ have the same power of two (the powers of 2 cancel out).
Let the common power be $p$ : then $m = 2^p\cdot a$ , and $n = 2^p\cdot b$ where $a$ and $b$ are integers between 1 and 30.
We can now rewrite the equation: \[\frac{2^p\cdot a}{2^p\cdot b} = \frac{4\alpha+1}{4\beta+1} \implies \frac{a}{b} = \frac{4\alpha+1}{4\beta+1}\]
Now it is easy to tell that $a \equiv 1 ($ mod $4)$ and $b \equiv 1 ($ mod $4)$ . However, there is another case: that
$a \equiv 3 ($ mod $4)$ and $b \equiv 3 ($ mod $4)$ . This is because multiplying both $4\alpha+1$ and $4\beta+1$ by $-1$ will not change the fraction, but each congruence will be changed to $-1 ($ mod $4) \equiv 3 ($ mod $4)$
From the first set of congruences, we find that $a$ and $b$ can be two of $\{1, 5, 9, \ldots, 29\}$
From the second set of congruences, we find that $a$ and $b$ can be two of $\{3, 7, 11, \ldots, 27\}$
Now all we have to do is multiply by $2^p$ to get back to $m$ and $n$ .
Let’s organize the solutions in order of increasing values of $p$ , keeping in mind that $m$ and $n$ are bounded between 1 and 30.
For $p = 0$ we get $\{1, 5, 9, \ldots, 29\}, \{3, 7, 11, \ldots, 27\}$
For $p = 1$ we get $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}$
For $p = 2$ we get $\{4, 20\}, \{12, 28\}$
Note that $16\mid{a}$ since $m$ will cancel out a factor of 4 from $a$ , and $\frac{a}{m}$ must contain a factor of 4. Again, $1-4X$ will never contribute a factor of 2. Simply inspecting, we see two feasible values for $a$ and $m$ such that $a+m\leq30$
If we increase the value of $p$ more, there will be less than two integers in our sets, so we are done there.
There are 8 numbers in the first set, 7 in the second, 4 in the third, 4 in the fourth, 2 in the fifth, and 2 in the sixth.
In each of these sets we can choose 2 numbers to be $m$ and $n$ and then assign them in increasing order. Thus there are:
\[\dbinom{8}{2}+\dbinom{7}{2}+\dbinom{4}{2}+\dbinom{4}{2}+\dbinom{2}{2}+\dbinom{2}{2} = 28+21+6+6+1+1 = \boxed{063}\] possible pairs $(m,n)$ that satisfy the conditions.

Solution:
We know that the range of sine is between $-1$ and $1$ , inclusive.
Thus, the only way for the sum to be $2$ is for $\sin(mx)=\sin(nx)=1$
Note that $\sin(90+360k)=1$
Assuming $mx$ and $nx$ are both positive, $m$ and $n$ could be $1,5,9,13,17,21,25,29$ . There are $8$ ways, so $\dbinom{8}{2}$
If both are negative, $m$ and $n$ could be $3,7,11,15,19,23,27$ . There are $7$ ways, so $\dbinom{7}{2}$
However, the pair $(1,5)$ could also be $(2, 10)$ and so on. The same goes for some other pairs.
In total there are $14$ of these extra pairs.
The answer is $28+21+14 = \boxed{063}$

Solution:
The equation implies that $\sin(mx)=\sin(nx)=1$ . Therefore, we can write $mx$ as $2{\pi}k_1+\frac{\pi}{2}$ and $nx$ as $2{\pi}k_2+\frac{\pi}{2}$ for integers $k_1$ and $k_2$ . Then, $\frac{mx}{nx}=\frac{m}{n}=\frac{2k_1+\frac{1}{2}}{2k_2+\frac{1}{2}}$ . Cross multiplying, we get $m\cdot{(2k_2+\frac{1}{2})}=n\cdot{(2k_1+\frac{1}{2})} \Longrightarrow 4k_2m-4k_1n=n-m$ . Let $n-m=a$ so the equation becomes $4(m(k_2-k_1)+k_1a)=a$ . Let $k_2-k_1=X$ and $k_1=Y$ , then the equation becomes $a=4Ym+4Xa \Longrightarrow \frac{a(1-4X)}{m}=4Y$ . Note that $X$ and $Y$ can vary accordingly, and $4\mid{a}$ . Next, we do casework on $m\pmod{4}$
If $m\equiv 1\pmod{4}$
Once $a$ and $m$ are determined, $n$ is determined, so $a+m\leq30$ $a\in \{4,8,12,\dots,28\}$ and $m\in \{1,5,9,\dots,29\}$ . Therefore, there are $\sum_{i=1}^{7}{i}=28$ ways for this case such that $a+m\leq30$
If $m\equiv 3\pmod{4}$
$a\in \{4,8,12,\dots,28\}$ and $m\in \{3,7,11,\dots,27\}$ . Therefore, there are $\sum_{i=1}^{6}{i}=21$ ways such that $a+m\leq30$
If $m\equiv 2\pmod{4}$
Note that $8\mid{a}$ since $m$ in this case will have a factor of 2, which will cancel out a factor of 2 in $a$ , and we need the left hand side to divide 4. Also, $1-4X\equiv 1\pmod{4}$ so it is odd and will therefore never contribute a factor of 2. $a\in \{8,16,24\}$ and $m\in \{2,6,10,\dots,30\}$ . Following the condition $a+m\leq30$ , we conclude that there are $6+4+2=12$ ways for this case.
If $m\equiv 0\pmod{4}$
Adding all the cases up, we obtain $28+21+12+2=\boxed{063}$