Answer: 52.0

Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all positive real numbers $x$ and $y$ . If $f(500) =3$ , what is the value of $f(600)$
$(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5$

Solution: Letting $x = 500$ and $y = \dfrac65$ in the given equation, we get $f(500\cdot\frac65) = \frac3{\frac65} = \frac52$ , or $f(600) = \boxed{52}$

Solution: The only function that satisfies the given condition is $y = \frac{k}{x}$ , for some constant $k$ . Thus, the answer is $\frac{500 \cdot 3}{600} = \boxed{52}$

Solution: Note that the equation given above is symmetric, so we have $x \cdot f(x)=y \cdot f(y)$ . Plugging in $x=500$ and $y=600$ gives $f(y)=\boxed{52}$


Answer: 2000.0

Let $f$ be a real-valued function such that
\[f(x) + 2f(\frac{2002}{x}) = 3x\]
for all $x>0.$ Find $f(2).$
$\text{(A) }1000 \qquad \text{(B) }2000 \qquad \text{(C) }3000 \qquad \text{(D) }4000 \qquad \text{(E) }6000$

Solution:
Setting $x = 2$ gives $f(2) + 2f(1001) = 6$ .
Setting $x = 1001$ gives $2f(2) + f(1001) = 3003$
Adding these 2 equations and dividing by 3 gives $f(2) + f(1001) = \frac{6+3003}{3} = 1003$
Subtracting these 2 equations gives $f(2) - f(1001) = 2997$
Therefore, $f(2) = \frac{1003+2997}{2} = \boxed{2000}$


Answer: 96.0

Let $f$ be the unique function defined on the positive integers such that \[\sum_{d\mid n}d\cdot f\left(\frac{n}{d}\right)=1\] for all positive integers $n$ . What is $f(2023)$
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$

Solution:
First, we note that $f(1) = 1$ , since the only divisor of $1$ is itself.
Then, let's look at $f(p)$ for $p$ a prime. We see that \[\sum_{d \mid p} d \cdot f\left(\frac{p}{d}\right) = 1\] \[1 \cdot f(p) + p \cdot f(1) = 1\] \[f(p) = 1 - p \cdot f(1)\] \[f(p) = 1-p\] Nice.
Now consider $f(p^k)$ , for $k \in \mathbb{N}$ \[\sum_{d \mid p^k} d \cdot f\left(\frac{p^k}{d}\right) = 1\] \[1 \cdot f(p^k) + p \cdot f(p^{k-1}) + p^2 \cdot f(p^{k-2}) + \dotsc + p^k f(1) = 1\]
It can be (strongly) inductively shown that $f(p^k) = f(p) = 1-p$ . Here's how.
We already showed $k=1$ works. Suppose it holds for $k = n$ , then
\[1 \cdot f(p^n) + p \cdot f(p^{n-1}) + p^2 \cdot f(p^{n-2}) + \dotsc + p^n f(1) = 1 \implies f(p^m) = 1-p \; \forall \; m \leqslant n\]
For $k = n+1$ , we have
\[1 \cdot f(p^{n+1}) + p \cdot f(p^{n}) + p^2 \cdot f(p^{n-1}) + \dotsc + p^{n+1} f(1) = 1\] , then using $f(p^m) = 1-p \; \forall \; m \leqslant n$ , we simplify to
\[1 \cdot f(p^{n+1}) + p \cdot (1-p) + p^2 \cdot (1-p) + \dotsc + p^n \cdot (1-p) + p^{n+1} f(1) = 1\] \[f(p^{n+1}) + \sum_{i=1}^n p^i (1-p) + p^{n+1} = 1\] \[f(p^{n+1}) + p(1 - p^n) + p^{n+1} = 1\] \[f(p^{n+1}) + p = 1 \implies f(p^{n+1}) = 1-p\]
Very nice! Now, we need to show that this function is multiplicative, i.e. $f(pq) = f(p) \cdot f(q)$ for $p,q$ prime.
It's pretty standard, let's go through it quickly. \[\sum_{d \mid pq} d \cdot f\left(\frac{pq}{d}\right) = 1\] \[1 \cdot f(pq) + p \cdot f(q) + q \cdot f(p) + pq \cdot f(1) = 1\] Using our formulas from earlier, we have \[f(pq) + p(1-q) + q(1-p) + pq = 1 \implies f(pq) = 1 - p(1-q) - q(1-p) - pq = (1-p)(1-q) = f(p) \cdot f(q)\]
Great! We're almost done now.
Let's actually plug in $2023 = 7 \cdot 17^2$ into the original formula. \[\sum_{d \mid 2023} d \cdot f\left(\frac{2023}{d}\right) = 1\] \[1 \cdot f(2023) + 7 \cdot f(17^2) + 17 \cdot f(7 \cdot 17) + 7 \cdot 17 \cdot f(17) + 17^2 \cdot f(7) + 7 \cdot 17^2 \cdot f(1) = 1\] Let's use our formulas! We know \[f(7) = 1-7 = -6\] \[f(17) = 1-17 = -16\] \[f(7 \cdot 17) = f(7) \cdot f(17) = (-6) \cdot (-16) = 96\] \[f(17^2) = f(17) = -16\]
So plugging ALL that in, we have \[f(2023) = 1 - \left(7 \cdot (-16) + 17 \cdot (-6) \cdot (-16) + 7 \cdot 17 \cdot (-16) + 17^2 \cdot (-6) + 7 \cdot 17^2\right)\] which, be my guest simplifying, is $\boxed{96}$

Solution: First, change the problem into an easier form. \[\sum_{d\mid n}d\cdot f(\frac{n}{d} )=\sum_{d\mid n}\frac{n}{d}f(d)=1\] So now we get \[\frac{1}{n}= \sum_{d\mid n}\frac{f(d)}{d}\] Also, notice that both $\frac{f(d)}{d}$ and $\frac{1}{n}$ are arithmetic functions. Applying Möbius inversion formula, we get \[\frac{f(n)}{n}=\sum_{d\mid n}\frac{ \mu (d) }{\frac{n}{d} }=\frac{1}{n} \sum_{d\mid n}d\cdot \mu (d)\] So \[f(n)=1-p_1-p_2-...+p_1p_2+...=(1-p_1)(1-p_2)...=\prod_{p\mid n}(1-p)\] So the answer should be $f(2023)=\prod_{p\mid 2023}(1-p)=(1-7)(1-17)=\boxed{96}$

Solution: From the problem, we want to find $f(2023)$ . Using the problem, we get $f(2023)+7f(289)+17f(119)+119f(17)+289f(7)+2023f(1)=1$ . By plugging in factors of $2023$ , we get \begin{align} f(7)+7f(1)=1\\ f(17)+17f(1)=1\\ f(119)+7f(17)+17f(7)+119f(1)=1\\ f(289)+17f(17)+289f(1)=1 \end{align} Notice that $(4)-17(2)=f(289)$ , so $f(289)=-16$ . Similarly, notice that $(3)-17(1)=f(119)+7f(17)=-16$ . Now, substituting this all back into our equation to solve for $f(2023)$ , we get \begin{align*} f(2023)+7f(289)+17(f(119)+7f(17))+289(f(7)+7f(1))=1\\ f(2023)+7 \cdot (-16) + 17 \cdot (-16) + 289 \cdot (1) = 1\\ f(2023)=\boxed{96} -PhunsukhWangdu

Solution:
Consider any $n \in \Bbb N$ with prime factorization $n = \Pi_{i=1}^k p_i^{\alpha_i}$ .
Thus, the equation given in this problem can be equivalently written as \[ \sum_{\beta_1 = 0}^{\alpha_1} \sum_{\beta_2 = 0}^{\alpha_2} \cdots \sum_{\beta_k = 0}^{\alpha_k} \Pi_{i=1}^k p_i^{\alpha_i - \beta_i} \cdot f \left( \Pi_{i=1}^k p_i^{\beta_i} \right) = 1 . \]
$\noindent \textbf{Special case 1}$ $n = 1$
We have $f \left( 1 \right) = 1$
$\noindent \textbf{Special case 2}$ $n$ is a prime.
We have \[ 1 \cdot f \left( n \right) + n \cdot f \left( 1 \right) = 1 . \]
Thus, $f \left( n \right) = 1 - n$
$\noindent \textbf{Special case 3}$ $n$ is the square of a prime, $n = p_1^2$
We have \[ 1 \cdot f \left( p_1^2 \right) + p_1 \cdot f \left( p_1 \right) + p_1^2 \cdot f \left( 1 \right) = 1. \]
Thus, $f \left( p_1^2 \right) = 1 - p_1$
$\noindent \textbf{Special case 4}$ $n$ is the product of two distinct primes, $n = p_1 p_2$
We have \[ 1 \cdot f \left( p_1 p_2 \right) + p_1 \cdot f \left( p_2 \right) + p_2 \cdot f \left( p_1 \right) + p_1 p_2 \cdot f \left( 1 \right) = 1. \]
Thus, $f \left( p_1 p_2 \right) = 1 - p_1 - p_2 + p_1 p_2$
$\noindent \textbf{Special case 5}$ $n$ takes the form $n = p_1^2 p_2$ , where $p_1$ and $p_2$ are two distinct primes.
We have \[ 1 \cdot f \left( p_1^2 p_2 \right) + p_1 \cdot f \left( p_1 p_2 \right) + p_1^2 \cdot f \left( p_2 \right) + p_2 \cdot f \left( p_1^2 \right) + p_1 p_2 f \left( p_1 \right) + p_1^2 p_2 f \left( 1 \right) = 1. \]
Thus, $f \left( p_1^2 p_2 \right) = 1 - p_1 - p_2 + p_1 p_2$
The prime factorization of 2023 is $7 \cdot 17^2$ .
Therefore, \begin{align*} f \left( 2023 \right) & = 1 - 7 - 17 + 7 \cdot 17 \\ & = \boxed{96}


Answer: 2.0

Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$ , where $i = \sqrt{-1}$ . What is $f(2022)$
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$

Solution:
Converting both summands to exponential form, \begin{align*} -1 + i\sqrt{3} &= 2e^{\frac{2\pi i}{3}}, \\ -1 - i\sqrt{3} &= 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}. \end{align*} Notice that both are scaled copies of the third roots of unity.
When we replace the summands with their exponential form, we get \[f(n) = \left(e^{\frac{2\pi i}{3}}\right)^n + \left(e^{\frac{4\pi i}{3}}\right)^n.\] When we substitute $n = 2022$ , we get \[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{2022} + \left(e^{\frac{4\pi i}{3}}\right)^{2022}.\] We can rewrite $2022$ as $3 \cdot 674$ , how does that help? \[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{3 \cdot 674} + \left(e^{\frac{4\pi i}{3}}\right)^{3 \cdot 674} = \left(\left(e^{\frac{2\pi i}{3}}\right)^{3}\right)^{674} + \left(\left(e^{\frac{4\pi i}{3}}\right)^{3}\right)^{674} = 1^{674} + 1^{674} = \boxed{2}.\] Since any third root of unity must cube to $1$

Solution: The numbers $\frac{-1+i\sqrt{3}}{2}$ and $\frac{-1-i\sqrt{3}}{2}$ are both $\textbf{Eisenstein Units}$ (along with $1$ ), denoted as $\omega$ and $\omega^2$ , respectively. They have the property that when they are cubed, they equal to $1$ . Thus, we can immediately solve: \[\omega^{2022} + \omega^{2 \cdot 2022} = \omega^{3 \cdot 674} + \omega^{3 \cdot 2 \cdot 674} = 1^{674} + 1^{2 \cdot 674} = \boxed{2}.\] ~mathboy100

Solution: We begin by recognizing this form looks similar to the definition of cosine: \[\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.\] We can convert our two terms into exponential form to find \[f(n) = \left( e^{\frac{2\pi i}{3}} \right ) ^n + \left ( e^{-\frac{2\pi i}{3}} \right ) ^n=e^{\frac{2 \pi i n}{3}} + e^{-\frac{2\pi i n}{3}}.\] This simplifies nicely: \[f(n)=2\cos\left( \frac{2\pi n}{3} \right).\] Thus, \[f(2022)=2\cos \left ( \frac{2\pi (2022) }{3} \right) = 2\cos(1348 \pi) = \boxed{2}.\]

Solution: Notice how this looks like the closed form of the Fibonacci sequence except different roots. This is motivation to turn this closed formula into a recurrence relation. The base of the exponents are the roots of the characteristic equation $r^3-1=0$ . So we have \begin{align*} a_0&=2\\ a_1&=-1\\ a_2&=-1\\ a_n&=a_{n-3} \end{align*} Every time $n$ is multiple of $3$ as is true when $n=2022$ $a_n= \boxed{2}$ ~lopkiloinm

Solution:
Converting the two terms into rectangular form,
\[f(2022)=\left(\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}\right)^{2022}+\left(\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}\right)^{2022}.\]
By DeMoivre's Theorem,
\[f(2022)=\left(\cos{\left(\frac{2\pi}{3}\cdot{2022}\right)}+i\sin{\left(\frac{2\pi}{3}\cdot{2022}\right)}\right)+\left(\cos{\left(\frac{4\pi}{3}\cdot{2022}\right)}+i\sin{\left(\frac{4\pi}{3}\cdot{2022}\right)}\right).\]
Note that $\cos{\pi\cdot{k}}=1$ if $k$ is even and $-1$ if $k$ is odd, and that $\sin{\pi\cdot{k}}=0$ for all integers $k$
All arguments are even in the second equation for $f(2022)$ , so the two $\cos$ terms are equal to $1$ , and the two $\sin$ terms are equal to $0$
Therefore the answer is $1+1=\boxed{2}.$


Answer: 258.0

Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n$ . Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$

Solution:
Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2<n<(k+1)^2$ for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$ . Note that this formula also returns the correct value when $n=(k+1)^2$ , but not when $n=k^2$ . Thus $f(n)=k^2+3k+2-n$ for $k^2<n \leq (k+1)^2$
If $2 \mid (k+1)^2-n$ $g(n)$ returns the same value as $f(n)$ . This is because the recursion once again stops at $(k+1)^2$ . We seek a case in which $f(n)<g(n)$ , so obviously this is not what we want. We want $(k+1)^2,n$ to have a different parity, or $n, k$ have the same parity. When this is the case, $g(n)$ instead returns $(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n$
Write $7f(n)=4g(n)$ , which simplifies to $3k^2+k-10=3n$ . Notice that we want the LHS expression to be divisible by 3; as a result, $k \equiv 1 \pmod{3}$ . We also want n to be strictly greater than $k^2$ , so $k-10>0, k>10$ . The LHS expression is always even (since $3k^2+k-10$ factors to $k(3k+1)-10$ , and one of $k$ and $3k+1$ will be even), so to ensure that $k$ and $n$ share the same parity, $k$ should be even. Then the least $k$ that satisfies these requirements is $k=16$ , giving $n=258$
Indeed - if we check our answer, it works. Therefore, the answer is $\boxed{258}$

Solution: Since $n$ isn't a perfect square, let $n=m^2+k$ with $0<k<2m+1$ . If $k$ is odd, then $f(n)=g(n)$ . If $k$ is even, then \begin{align*} f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\ g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k, \end{align*} from which \begin{align*} 7(3m+2-k)&=4(5m+6-k) \\ m&=3k+10. \end{align*} Since $k$ is even, $m$ is even. Since $k\neq 0$ , the smallest $k$ is $2$ which produces the smallest $n$ \[k=2 \implies m=16 \implies n=16^2+2=\boxed{258}.\] ~Afo

Solution:
To begin, note that if $n$ is a perfect square, $f(n)=g(n)$ , so $f(n)/g(n)=1$ , so we must look at values of $n$ that are not perfect squares (what a surprise). First, let the distance between $n$ and the first perfect square greater than or equal to it be $k$ , making the values of $f(n+k)$ and $g(n+k)$ integers. Using this notation, we see that $f(n)=k+f(n+k)$ , giving us a formula for the numerator of our ratio. However, since the function of $g(n)$ does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of $\sqrt{n+k}$ in $g(n)$ unless $k$ is an even number. However, this is impossible, since if $k$ was an even number, $f(n)=g(n)$ , giving a ratio of one. Thus, $k$ must be an odd number.
Thus, since $k$ must be an odd number, regardless of whether $n$ is even or odd, to get an integral value in $g(n)$ , we must get to the next perfect square after $n+k$ . To make matters easier, let $z^2=n+k$ . Thus, in $g(n)$ , we want to achieve $(z+1)^2$
Expanding $(z+1)^2$ and substituting in the fact that $z=\sqrt{n+k}$ yields:
\[(z+1)^2=z^2+2z+1=n+k+2\sqrt{n+k}+1\]
Thus, we must add the quantity $k+2z+1$ to $n$ to achieve a integral value in the function $g(n)$ . Thus.
\[g(n)=(k+2z+1)+\sqrt{n+k+2\sqrt{n+k}+1}\]
However, note that the quantity within the square root is just $(z+1)^2$ , and so:
\[g(n)=k+3z+2\]
Thus, \[\frac{f(n)}{g(n)}=\frac{k+z}{k+3z+2}\]
Since we want this quantity to equal $\frac{4}{7}$ , we can set the above equation equal to this number and collect all the variables to one side to achieve
\[3k-5z=8\]
Substituting back in that $z=\sqrt{n+k}$ , and then separating variables and squaring yields that
\[9k^2-73k+64=25n\]
Now, if we treat $n$ as a constant, we can use the quadratic formula in respect to $k$ to get an equation for $k$ in terms of $n$ (without all the squares). Doing so yields
\[\frac{73\pm\sqrt{3025+900n}}{18}=k\]
Now, since $n$ and $k$ are integers, we want the quantity within the square root to be a perfect square. Note that $55^2=3025$ . Thus, assume that the quantity within the root is equal to the perfect square, $m^2$ . Thus, after using a difference of squares, we have \[(m-55)(m+55)=900n\] Since we want $n$ to be an integer, we know that the $LHS$ should be divisible by five, so, let's assume that we should have $m$ divisible by five. If so, the quantity $18k-73$ must be divisible by five, meaning that $k$ leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!).
Thus, we see that to achieve integers $n$ and $k$ that could potentially satisfy the problem statement, we must try the values of $k$ congruent to one modulo five. However, if we recall a statement made earlier in the solution, we see that we can skip all even values of $k$ produced by this modulo argument.
Also, note that $k=1,6$ won't work, as they are too small, and will give an erroneous value for $n$ . After trying $k=11,21,31$ , we see that $k=31$ will give a value of $m=485$ , which yields $n=\boxed{258}$ , which, if plugged in to for our equations of $f(n)$ and $g(n)$ , will yield the desired ratio, and we're done.

Solution:
First of all, if $n$ is a perfect square, $f(n)=g(n)=\sqrt{n}$ and their quotient is $1.$ So, for the rest of this solution, assume $n$ is not a perfect square.
Let $a^2$ be the smallest perfect square greater than $n$ and let $b^2$ be the smallest perfect square greater than $n$ with the same parity as $n,$ and note that either $b=a$ or $b=a+1.$ Notice that $(a-1)^2 < n < a^2.$
With a bit of inspection, it becomes clear that $f(n) = a+(a^2-n)$ and $g(n) = b+(b^2-n).$
If $a$ and $n$ have the same parity, we get $a=b$ so $f(n) = g(n)$ and their quotient is $1.$ So, for the rest of this solution, we let $a$ and $n$ have opposite parity. We have two cases to consider.
Case 1: $n$ is odd and $a$ is even
Here, we get $a=2k$ for some positive integer $k.$ Then, $b = 2k+1.$ We let $n = a^2-(2m+1)$ for some positive integer $m$ so $f(n) = 2k+2m+1$ and $g(n) = 2k+1+2m+1+4k+1 = 6k+2m+3.$
We set $\frac{2k+2m+1}{6k+2m+3}=\frac{4}{7},$ cross multiply, and rearrange to get $6m-10k=5.$ Since $k$ and $m$ are integers, the LHS will always be even and the RHS will always be odd, and thus, this case yields no solutions.
Case 2: $n$ is even and $a$ is odd
Here, we get $a=2k+1$ for some positive ineger $k.$ Then, $b=2k+2.$ We let $n = a^2-(2m+1)$ for some positive integer $m$ so $f(n)=2k+1+2m+1=2k+2m+2$ and $g(n)=2k+2+2m+1+4k+3 = 6k+2m+6.$
We set $\frac{2k+2m+2}{6k+2m+6} = \frac{4}{7},$ cross multiply, and rearrange to get $5k=3m-5,$ or $k=\frac{3}{5}m-1.$ Since $k$ and $m$ are integers, $m$ must be a multiple of $5.$ Some possible solutions for $(k,m)$ with the least $k$ and $m$ are $(2,5), (5,10), (8,15),$ and $(11,20).$
We wish to minimize $k$ since $a=2k+1.$ One thing to keep in mind is the initial assumption $(a-1)^2 < n < a^2.$
The pair $(2,5)$ gives $a=2(2)+1=5$ and $n=5^2-(2(5)+1)=14.$ But $4^2<14<5^2$ is clearly false, so we discard this case.
The pair $(5,10)$ gives $a=2(5)+1=11$ and $n=11^2-(2(10)+1)=100,$ which is a perfect square and therefore can be discarded.
The pair $(8,15)$ gives $a=2(8)+1=17$ and $n=17^2-(2(15)+1)=258,$ which is between $16^2$ and $17^2$ so it is our smallest solution.
So, $\boxed{258}$ is the correct answer.

Solution:
Say the answer is in the form n^2-x, then x must be odd or else f(x) = g(x). Say y = n^2-x. f(y) = x+n, g(y) = 3n+2+x. Because f(y)/g(y) = 4*(an integer)/7*(an integer),  f(y) is 4*(an integer) so n must be odd or else f(y) would be odd. Solving for x in terms of n gives integer x = (5/3)n+8/3 which means n is 2 mod 3, because n is also odd, n is 5 mod 6. x must be less than 2n-1 or else the minimum square above y would be (n-1)^2. We set an inequality
(5/3)n+8/3<2n-1 => 5n+8<6n-3 => n>11.
Since n is 5 mod 6, n = 17 and x = 31 giving 17^2-31 = $\boxed{258}$


Answer: 400.0

Let $f(n)$ be the integer closest to $\sqrt[4]{n}.$ Find $\sum_{k=1}^{1995}\frac 1{f(k)}.$

Solution: This is a pretty easy problem just to bash. Since the max number we can get is $7$ , we just need to test $n$ values for $1.5,2.5,3.5,4.5,5.5$ and $6.5$ . Then just do how many numbers there are times $\frac{1}{\lfloor n \rfloor}$ , which should be $5+17+37+65+101+145+30 = \boxed{400}$


Answer: 2047.0

Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation.  For example, $f(4)=6$ because $4$ can be written as $4$ $2+2$ $2+1+1$ $1+2+1$ $1+1+2$ , and $1+1+1+1$ .  Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd.

Solution:
First of all, note that $f(n)$ $\sum_{i=0}^{k} f(n-2^{i})$ where $k$ is the largest integer such that $2^k \le n$ .  We let $f(0) = 1$ for convenience.
From here, we proceed by induction, with our claim being that the only $n$ such that $f(n)$ is odd are $n$ representable of the form $2^{a} - 1, a \in \mathbb{Z}$
We induct on $a$ .  It is trivially true for $a = 0$ and $a = 1$ .  From here, we show that, if the only numbers $n \le 2^{a-1} - 1$ where $f(n)$ is odd are of the form described above, then the only numbers $n \le 2^{a} -1$ that are odd are of that form.  We first consider all numbers $b$ , such that $2^{a-1} \le b \le 2^{a} - 2$ , going from the lower bound to the upper bound (a mini induction, you might say).  We know that $f(b) = \sum_{i=0}^{a-1} f(b-2^{i})$ .  For a number in this summation to be odd, $b - 2^i = 2^m -1 \rightarrow b = 2^i + 2^m - 1$ .  However, we know that $b > 2^{a-1}$ , so $m$ must be equal to $a-1$ , or else $b$ cannot be in that interval.  Now, from this, we know that $i < a-1$ , as $b<2^{a} - 1$ .  Therefore, $i$ and $m$ are distinct, and thus $f(b - 2^i)$ and $f(b- 2^{a-1})$ are odd; since there are just two odd numbers, the ending sum for any $b$ is even.  Finally, considering $2^{a} - 1$ , the only odd number is $f(2^{a} - 1 - 2^{a-1})$ , so the ending sum is odd. $\Box$
The smallest $n$ greater than $2013$ expressible as $2^d - 1, d \in \mathbb{N}$ is $2^{11} -1 = \boxed{2047}$

Solution:
Of course, as with any number theory problem, use actual numbers to start, not variables! By plotting out the first few sums (do it!) and looking for patterns, we observe that $f(n)=\sum_{\textrm{power}=0}^{\textrm{pow}_{\textrm{larg}}} f(n-2^{\textrm{power}})$ , where $\textrm{pow}_{\textrm{larg}}$ represents the largest power of $2$ that is smaller than $n$ . I will call this sum the Divine Sign, or DS.
But wait a minute... we are trying to determine odd/even of $f(n)$ . Why not call all the evens 0 and odds 1, basically using mod 2? Sounds so simple. Draw a small table for the values: as $n$ goes up from $0$ , you get: $1,1,0,1,0,0,0,1,0...$ . We have to set $f(0)=1$ for this to work. Already it looks like $f(n)$ is only odd if $n=2^{\textrm{power}}-1$
The only tool here is induction. The base case is clearly established. Then let's assume we successfully made our claim up to $2^n-1$ . We need to visit numbers from $2^n$ to $2^{n+1}-1$ . Realize that $2^n$ has $0$ for $f$ because there will be two numbers in DS that give a $f$ of one: $2^{n-1}$ and $1$
But to look at whether a value of $f(\textrm{number})$ is 1 or 0, we need to revisit our first equation. We can answer this rather natural question: When will a number to be inducted upon, say $2^n+k$ , ever have a 1 as $f(\textrm{number})$ in the DS equation? Well- because by our assumption of the claim up to $2^n-1$ , we know that the only way for that to happen is if $2^n+k-2^{\textrm{power}}$ in the DS is equal to $2^{\textrm{Some power}} - 1$ . Clearly $1 \leq k \leq 2^n - 1$
Finally, we can simplify. Using our last equation, $2^n+k-2^{\textrm{power}}=2^{\textrm{Some power}}-1$ , regrouping gives $2^n+k=2^{\textrm{power}}+2^{\textrm{Some power}}-1$
Most importantly, realize that $\textrm{power}$ can be from $0$ to $n$ , because of the restraints on $k$ mentioned earlier. Same with $\textrm{Some power}$ . Immediately at least one of $\textrm{power}$ and $\textrm{Some power}$ has to be $n$ . If both were smaller, LHS is greater, contradiction. If both were greater, RHS is greater, contradiction.
Therefore, by setting one of $\textrm{power}$ or $\textrm{Some power}$ to $n$ , we realize $k=2^{\textrm{A certain power}}-1$
The conclusion is clear, right? Each $k$ from $1$ to $2^{n-1}-1$ yields two distinct cases: one of $\textrm{power}$ and $\textrm{Some power}$ is equal to $n$ , while the other is LESS THAN $n$ . But for $k=2^n-1$ , there is ONE CASE: BOTH values have to equal $n$ . Therefore, the only $k$ that has $f(2^n+k)$ as odd must only be $2^n-1$ , because the other ones yield a $f$ of 1+1=0 in our mod. That proves our induction for a new power of 2, namely $n+1$ , meaning that $f(\textrm{number})$ is only odd if $\textrm{number} = 2^{\textrm{Power of two}} - 1$ , and we are almost done...
Thus, the answer is $2^{11}-1=\boxed{2047}$


Answer: 16089.0

Solution:
$g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1$
$h_{1}(x)=g(f(x))\text{ = }g(10^{10x}=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}$
Proof by induction that $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$
For $n=1$ $h_{1}(x)=10x - 1$
Assume $h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$ is true for n:
\begin{align*} h_{n+1}(x)&= h_{1}(h_{n}(x))\\ &=10 h_{n}(x) - 1\\ &=10 (10^n x   - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1\\ &= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1\\ &= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1}) \end{align*}
Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n.
$h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})$ , which is the 2011-digit number 8888...8889
The sum of the digits is 8 times 2010 plus 9, or $\boxed{16089}$

Solution: As before, $h_1(x)=10x-1$ . Compute $h_1(x)$ $h_2(x)$ , and $h_3(x)$ to yield 9, 89, and 889. Notice how this trend will repeat this trend (multiply by 10, subtract 1, repeat). As such, $h_{2011}$ is just 2010 8's followed by a nine. $2010(8)+9=\boxed{16089}$


Answer: 12.0

Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36$

Solution:
The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.
We note that since all of the $\sin$ factors are inside a logarithm, the function is undefined where the inside of the logarithm is less than or equal to $0$
First, let us find the number of zeros of the inside of the logarithm.
\begin{align*}\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x) &= 0\\ \sin(\pi x) &= 0\\ x &= 0, 1\\ \sin(2 \pi x) &= 0\\ x &= 0, \frac{1}{2}, 1\\ \sin(3 \pi x) &= 0\\ x &= 0, \frac{1}{3}, \frac{2}{3}, 1\\ \sin(4 \pi x) &= 0\\ x &= 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1\\ &\cdots\end{align*}
After counting up the number of zeros for each factor and eliminating the excess cases we get $23$ zeros and $22$ intervals.
In order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc.
The first interval $\left(0, \frac{1}{8}\right)$ is obviously positive. This means the next interval $\left(\frac{1}{8}, \frac{1}{7}\right)$ is negative. Continuing the pattern and accounting for doubled roots (which do not flip sign), we realize that there are $5$ negative intervals from $0$ to $\frac{1}{2}$ . Since the function is symmetric, we know that there are also $5$ negative intervals from $\frac{1}{2}$ to $1$
And so, the total number of disjoint open intervals is $22 - 2\cdot{5} = \boxed{12}$

Solution:
Note that the expression $\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)$ must be greater than zero, since logarithm functions are undefined for $0$ and negative numbers. Let $x_1, x_2, x_3, ..., x_8$ temporarily be the dependent variables of the functions $y_1 = \sin(\pi x_1), y_2 = \sin(2\pi x_2), ..., y_8 = \sin(8\pi x_8)$ . It is easy to see that for $y_i$ to be positive for $1\leq i\leq8$ $\lfloor i x_i \rfloor$ must be even for $1 \leq i\leq8$ . Since an even number of positives times an even number of negatives equals a positive, there can be $2, 4, 6,$ or $8$ positive values of $y_i$ for $1 \leq i\leq 8$ for a given value of $x$ . (since $y_1$ is always positive on the range $[0, 1]$ )
Since MAA allows rulers (and you should bring one to the actual exam), use it to your advantage and draw a larged scaled number line from $0$ to $1$ . (I recommend increments of at most $0.1$ .) If you don't have a ruler but have graph paper, you can use that instead. Then, designate rows for $y_1, y_2, ..., y_8$ , respectively. Draw a large bar (label it with $+$ so you know it's positive) for all values of $x_i$ such that $\lfloor i x_i \rfloor$ is even, and do that for all eight rows. Then, use your ruler (or another viable straightedge, such as the edge of another sheet of paper), place the straightedge perpendicular to the vertical line on your digram at $0$ , and slowly work your way to $1$ , marking all disjoint intervals in which your straightedge touches $2, 4, 6,$ or $8$ boxes simultaneously. (If an interval excludes a value in that interval, you still have to count it as two disjoint intervals. Note that this will be important as to not undercounting disjoint intervals. )
If done correctly, you should obtain $\boxed{12}$ as your answer.

Solution:
You should be able to somewhat visualize what the $\sin$ function looks like (if you can't then you should look it up and try to memorize it). To summarize, the graph of $y= \sin{x}$ is positive from the interval $(0, \pi)$ and negative from the interval $(\pi, 2\pi)$ (notice how the intervals use parentheses instead of brackets, as brackets would denote inclusive bounds and that it incorrect).
All of the various $\sin$ functions that we are multiplying are in the form of $k\pi x$ , which means the intervals have now changed to integer values.
Also, the period of the function $y =\sin{k \pi x}$ is equivalent to $\frac{2}{k}$ ; For example the period of $y=  \sin{4\pi x}$ is $\frac{1}{2}$
Let a "wave" be a full period of a $\sin$ curve. Then, a "half-wave" is half of a full period of a $\sin$ curve. Over the interval $[0, 1],$ $y= \sin{\pi x}$ will display $1$ "half-wave", $y= \sin{2\pi x}$ will display $2$ "half-waves", and $y= \sin{k\pi x}$ will display $k$ "half-waves". Also notice that the graph of $y= \sin{k \pi x}$ crosses the $x$ -axis exactly $k-1$ times over the interval $(0, 1)$
The problem asks for the number of disjoint intervals, not the values of the intervals themselves. This means we only want the positive values of
$\prod_{k=1}^8 \sin{k\pi x}$
(the product of all of the various $\sin$ functions)
We don't care about the specific values themselves, we just need to know what the sign (+/-) of the product is. Notice that if an even number of the functions are negative, the product will be positive; conversely, if the number of negative functions is odd, the product will be negative. Therefore we only need to count the number of intervals during which an even number of functions are negative/positive. By definition, the sign of a function changes after it crosses one of its roots (because trig functions don't have multiplicities). Notice that the roots of $y = \sin{k\pi x}$ are in the form
$\frac{1}{k}, \frac{2}{k}, \frac{3}{k},...$
(this is in agreement with the fact that $y= \sin{k \pi x}$ crosses the $x$ -axis exactly $k-1$ times over the interval $(0, 1)$
Therefore, across the interval $[0, 1]$ , the sign will change every time the function hits a root of one of the $\sin$ functions. We must be careful, however, as some roots (such as $\frac{1}{2}$ and $\frac{2}{4}$ ) are duplicates.
Notice that the duplicates always come in pairs. For example, $\frac{1}{2}$ is also $\frac{2}{4}, \frac{3}{6},$ and $\frac{4}{8}$ ; there are four equivalent roots. Across the interval $[0, 1],$ there are no duplicate roots that come in triplets, or in fives. The roots are always single, doubled up, or quadrupled up. So every time the function reaches a single root, it switches signs, and every time it reaches a duplicate root, the sign remains the same (this still counts as 2 distinct intervals, though, as the zeroes "separate" them). Listing out the roots (not including $x=0$ and $x=1$ ):
$1$ : n/a
$2: \frac{1}{2}$
$3: \frac{1}{3}, \frac{2}{3}$
$4: \frac{1}{4}, \frac{2}{4}, \frac{3}{4}$
$5: \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}$
$6: \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}$
$7: \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$
$8: \frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{4}{8}, \frac{5}{8}, \frac{6}{8}, \frac{7}{8}$
Then in order (showing duplicates as well):
$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$
Then creating a sequence of signs based on the roots:
$+,-,+,-,+,+,-,-,+,-,+,+,-,+,-,-,+,+,-,+,-,+$
We see that there are $12$ positive intervals $\Rightarrow \boxed{12}$


Answer: 2.0

Let $f(x) = \sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}}.$ An equivalent form of $f(x)$ is
$\text{(A) }1-\sqrt{2}\sin{x} \qquad \text{(B) }-1+\sqrt{2}\cos{x} \qquad \text{(C) }\cos{\frac{x}{2}} - \sin{\frac{x}{2}} \qquad \text{(D) }\cos{x} - \sin{x} \qquad \text{(E) }\cos{2x}$
Solution

Solution:
By the Pythagorean identity we can rewrite the given expression as follows. \[\sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}} = \sqrt{\sin^4{x} + 4(1 - \sin^2{x})} - \sqrt{\cos^4{x} + 4(1 - \cos^2{x})}\]
Expanding each bracket gives \[\sqrt{\sin^4{x} - 4\sin^2{x} + 4} - \sqrt{\cos^4{x} - 4\cos^2{x} + 4}\]
The expressions under the square roots can be factored to get \[\sqrt{(\sin^2{x} - 2)^2} - \sqrt{(\cos^2{x} - 2)^2}\]
Since $\sin^2{x} < 2$ and $\cos^2{x} < 2$ for all real $x$ , the expression must evaluate to $(2 - \sin^2{x}) - (2 - \cos^2{x})$ , which simplifies to $\cos^2{x} - \sin^2{x} = \boxed{2}$

Solution:
We don't actually have to solve the question. Just let $x$ equal some easy value to calculate $\cos {x}, \cos {2x}, \sin {x}, \sin {\frac{x}{2}},$ and $\cos {\frac{x}{2}}.$ For this solution, let $x=60^\circ.$ This means that the expression in the problem will give $\sqrt{\sin^4{60^\circ} + 4 \cos^2{60^\circ}} - \sqrt{\cos^4{60^\circ} + 4 \sin^2{60^\circ}}=\sqrt{(\frac{\sqrt{3}}{2})^4 + 4(\frac{1}{2})^2} - \sqrt{(\frac{1}{2})^4 + 4(\frac{\sqrt{3}}{2})^2}=\sqrt{\frac{9}{16} +1} - \sqrt{\frac{1}{16} + 3} = \frac{-1}{2}.$ Plugging in $x=60^\circ$ for the rest of the expressions, we get
$\text{(A) }1-\sqrt{2}\sin{60^\circ}=1-\frac{\sqrt{6}}{2} \neq \frac{-1}{2}.$
$\text{(B) }1+\sqrt{2}\cos{60^\circ}=1+\frac{\sqrt{2}}{2} \neq \frac{-1}{2}.$
$\text{(C) }\cos{\frac{60^\circ}{2}} - \sin{\frac{60^\circ}{2}}=\frac{\sqrt{3}}{2}-1 \neq \frac{-1}{2}.$
$\text{(D) }\cos{60^\circ} - \sin{60^\circ}=\frac{1-\sqrt{3}}{2} \neq \frac{-1}{2}.$
$\text{(E) }\cos {120^\circ}=\frac{-1}{2}.$
Therefore, our answer is $\boxed{2}$


Answer: 25.0

Let $f(x) = x^2 + 6x + 1$ , and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that \[f(x) + f(y) \le 0 \qquad \text{and} \qquad f(x)-f(y) \le 0\] The area of $R$ is closest to
$\textbf{(A) } 21 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 23 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 25$

Solution:
The first condition gives us that \[x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16\]
which is a circle centered at $(-3,-3)$ with radius $4$ . The second condition gives us that
\[x^2 + 6x + 1 - y^2 - 6y - 1 \le 0 \Longrightarrow (x^2 - y^2) + 6(x-y) \le 0 \Longrightarrow (x-y)(x+y+6) \le 0\]
Thus either
\[x - y \ge 0,\quad x+y+6 \le 0\]
or
\[x - y \le 0,\quad x+y+6 \ge 0\]
Each of those lines passes through $(-3,-3)$ and has slope $\pm 1$ , as shown above. Therefore, the area of $R$ is half of the area of the circle, which is $\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx \boxed{25}$

Solution:
Similar to Solution 1, we proceed to get the area of the circle satisfying $f(x)+f(y) \le 0$ , or $16 \pi$
Since $f(x)-f(y) \le 0 \implies f(x) \le f(y)$ , we have that by symmetry, if $(x,y)$ is in $R$ , then $(y,x)$ is not, and vice versa. Therefore, the shaded part of the circle above the line $y=x$ has the same area as the unshaded part below $y=x$ , and the unshaded part above $y=x$ has the same area as the shaded part below $y=x$ . This means that exactly half the circle is shaded, allowing us to divide by two to get $\frac{16 \pi }{2} = 8\pi \approx \boxed{25}$ . ~samrocksnature + ddot1


Answer: 0.0

Solution: First, note that $f(x) = f(1-x)$ . We can see this since \[f(x) = x^2(1-x)^2 = (1-x)^2x^2 = (1-x)^{2}\left(1-\left(1-x\right)\right)^{2} = f(1-x)\] Using this result, we regroup the terms accordingly: \[\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) +  \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2017}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1010}{2019} \right) \right)\] \[= \left( f \left(\frac{1}{2019} \right) - f \left(\frac{1}{2019} \right) \right) +  \left( f \left(\frac{2}{2019} \right) - f \left(\frac{2}{2019} \right) \right) + \cdots + \left( f \left(\frac{1009}{2019} \right) - f \left(\frac{1009}{2019} \right) \right)\] Now it is clear that all the terms will cancel out (the series telescopes), so the answer is $\boxed{0}$


Answer: 676.0

Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$

Solution:
If the leading term of $f(x)$ is $ax^m$ , then the leading term of $f(x)f(2x^2) = ax^m \cdot a(2x^2)^m = 2^ma^2x^{3m}$ , and the leading term of $f(2x^3 + x) = 2^max^{3m}$ . Hence $2^ma^2 = 2^ma$ , and $a = 1$ . Because $f(0) = 1$ , the product of all the roots
of $f(x)$ is $\pm 1$ . If $f(\lambda) = 0$ , then $f(2\lambda^3 + \lambda) = 0$ . Assume that there exists a root $\lambda$ with $|\lambda| \neq 1$ . Then there must be such a root $\lambda_1$ with $|\lambda_1| > 1$ . Then \[|2\lambda^3 + \lambda| \ge 2|\lambda|^3 - |\lambda| > 2|\lambda| - |\lambda| = |\lambda|.\] But then $f(x)$ would have infinitely many roots, given by $\lambda_{k+1} = 2\lambda_k^3 + \lambda_k$ , for $k \ge 1$ .
Therefore $|\lambda| = 1$ for all of the roots of the polynomial. Thus $\lambda\overline{\lambda}=1$ , and $(2\lambda^3 + \lambda)\overline{(2\lambda^3 + \lambda)} = 1$ .
Solving these equations simultaneously for $\lambda = a + bi$ yields $a = 0$ $b^2 = 1$ , and so $\lambda^2 = -1$ . Because the polynomial has real coefficients, the polynomial must have the form $f(x) = (1 + x^2)^n$ for some integer $n \ge 1$ . The condition $f(2) + f(3) = 125$ implies $n = 2$ , giving $f(5) = \boxed{676}$

Solution:
Let $r$ be a root of $f(x)$ . Then we have $f(r)f(2r^2)=f(2r^3+r)$ ; since $r$ is a root, we have $f(r)=0$ ; therefore $2r^3+r$ is also a root. Thus, if $r$ is real and non-zero, $|2r^3+r|>r$ , so $f(x)$ has infinitely many roots. Since $f(x)$ is a polynomial (thus of finite degree) and $f(0)$ is nonzero, $f(x)$ has no real roots.
Note that $f(x)$ is not constant. We then find two complex roots: $r = \pm i$ . We find that $f(i)f(-2) = f(-i)$ , and that $f(-i)f(-2) = f(i)$ . This means that $f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2)^2 - 1) = 0$ . Thus, $\pm i$ are roots of the polynomial, and so $(x - i)(x + i) = x^2 + 1$ will be a factor of the polynomial. (Note: This requires the assumption that $f(-2)\neq1$ . Clearly, $f(-2)\neq-1$ , because that would imply the existence of a real root.)
The polynomial is thus in the form of $f(x) = (x^2 + 1)g(x)$ . Substituting into the given expression, we have
\[(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)\] \[(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)\]
Thus either $4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)$ is 0 for any $x$ , or $g(x)$ satisfies the same constraints as $f(x)$ . Continuing, by infinite descent, $f(x) = (x^2 + 1)^n$ for some $n$
Since $f(2)+f(3)=125=5^n+10^n$ for some $n$ , we have $n=2$ ; so $f(5) = \boxed{676}$

Solution:
Let $r$ be a root of $f(x).$ This means that $f(r)f(2r^2)=f(2r^3+r).$ In other words, $2r^3+r$ is a root of $f(x)$ too. Since $f(x)$ can't have infinitely many roots, \[Q(x)=P(P(\dotsb P(P(r)) \dotsb))\] is cyclic, where $P(x)=2x^3+x.$ Now, we will do casework.
Case 1: $\deg f\geq1$
Subcase 1: $|r|>1$
This means that \[|2r^3+r|\geq|2r^3|-|r|=|r|(2|r|^2-1)>|r|(2\cdot1^2-1)=|r|.\] It follows that $|2r^3+r|>|r|$ for all $r.$ This implies that $Q(r)$ can't be cyclic. Thus, it is impossible for $|r|>1$ to be true.
Subcase 2: $|r|<1$
This means that $|2r^3+r|\geq2|r^3|-|r|=|r|(|2r^2|-1)<|r|.$ It follows that $|2r^3+r|<|r|$ for all $r.$ This implies that $Q(r)$ can't be cyclic. Thus, it is impossible for $|r|>1$ to be true.
Subcase 3: $|r|=1.$
Since $|r|$ is not greater than or less than 1, $|r|=1.$ This means that all the roots of the polynomial have a magnitude of $1.$ More specifically, $|2r^3+r|$ has a magnitude of one. Since this would mean an equality condition from the triangle inequality, $2r^3$ and $r$ are collinear with the origin in the complex plane. In other words, $\frac{2r^3}{r}=\pm c\Leftrightarrow cr=2r^3\Leftrightarrow 2r^2=c\Leftrightarrow r=\pm\sqrt{\pm\frac{c}{2}},$ for some real constant $c.$ Now, from $|r|=1,$ we find that $\left|\pm\sqrt{\pm\frac{c}{2}}\right|=1\Leftrightarrow \sqrt{\pm\frac{c}{2}}=1\Leftrightarrow \pm\frac{c}{2}=1\Leftrightarrow c=\pm2.$ Putting this back into the equation, we find that $r=1,-1,i,-i.$ Now, this means that $2r^3+r=3,-3,i,-i.$ $3$ and $-3$ obviously doesn't have a magnitude of $1.$ Thus, $i,-i$ are the only possible roots of the polynomial. Since roots come in conjugate pairs, $f(x)=[(x-i)(x+i)]^n=(x^2+1)^n,$ works for all constants $n\neq0.$
Case 2: $\deg f=0.$
This means that $f(x)=c,$ for some constant $c.$ In other words, $c^2=c.$ We can easily find that this means that $c=0,1.$ Combining all the cases, we conclude that $f(x)=(x^2+1)^n,0,1$ are the only polynomials that satisfy this equation.
Now, we can test! $f(x)=0,1$ obviously don't satisfy $f(2)+f(3)=125.$ Thus, $f(x)=(x^2+1)^n.$ Substituting, we find that $5^n+10^n=125\Leftrightarrow n=2.$ We conclude that $f(5)=(5^2+1)^2=26^2=\boxed{676}.$


Answer: 72.0

Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$

Solution:
Let $f(x)$ $ax^3+bx^2+cx+d$ .
Since $f(x)$ is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up.
By drawing a coordinate axis, and two lines representing $12$ and $-12$ , it is easy to see that $f(1)=f(5)=f(6)$ , and $f(2)=f(3)=f(7)$ ; otherwise more bends would be required in the graph. Since only the absolute value of $f(0)$ is required, there is no loss of generalization by stating that $f(1)=12$ , and $f(2)=-12$ . This provides the following system of equations. \[a +     b +   c +   d =  12\] \[8a +   4b + 2c +   d = -12\] \[27a +   9b + 3c +   d = -12\] \[125a + 25b + 5c +   d =  12\] \[216a + 36b + 6c +   d =  12\] \[343a + 49b + 7c +   d = -12\] Using any four of these functions as a system of equations yields $d = |f(0)| = \boxed{072}$

Solution:
By drawing the function, and similar to Solution 1, WLOG let $f(1)=f(5)=f(6)=12$ . Then, $f(2)=f(3)=f(7)$ . Set $g(x)+12=f(x)$ . Then the roots of $g(x)$ are $1,5,6$ . So, $g(x)=a(x-1)(x-5)(x-6)$ . Plug in $x=2$ to find a. We know \[-24=-12-12=f(2)-12=g(2)=a(1)(-3)(-4)=12a\] .
So, $a=-2$ . Thus, $f(x)=g(x)+12=-2(x-1)(x-5)(x-6)+12$ , and then $|f(0)|=60+12=\boxed{072}$

Solution: Without loss of generality, let $f(1) = 12$ . (If $f(1) = -12$ , then take $-f(x)$ as the polynomial, which leaves $|f(0)|$ unchanged.) Because $f$ is third-degree, write \[f(x) - 12 = a(x - 1)(x - b)(x - c)\] \[f(x) + 12 = a(x - d)(x - e)(x - f)\] where $\{b, c, d, e, f \}$ clearly must be a permutation of $\{2, 3, 5, 6, 7\}$ from the given condition. Thus $b + c + d + e + f = 2 + 3 + 5 + 6 + 7 = 23.$ However, subtracting the two equations gives $-24 = a[(x - 1)(x - b)(x - c) - (x - d)(x - e)(x - f)]$ , so comparing $x^2$ coefficients gives $1 + b + c = d + e + f$ and thus both values equal to $\dfrac{24}{2} = 12$ . As a result, $\{b, c \} = \{5, 6 \}$ . As a result, $-24 = a (12)$ and so $a = -2$ . Now, we easily deduce that $f(0) = (-2) \cdot (-1) \cdot (-5) \cdot (-6) + 12 = 72,$ and so removing the without loss of generality gives $|f(0)| = \boxed{072}$ , which is our answer.

Solution: The following solution is similar to solution 3, but assumes nothing. Let $g(x)=(f(x))^2-144$ . Since $f$ has degree 3, $g$ has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, $g(x)=k(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ for some $k$ . Hence $|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}$ . Note that $g(x)=(f(x)+12)(f(x)-12)$ . Since $f$ has degree 3, so do $f(x)+12$ and $f(x)-12$ ; and both have the same leading coefficient. Hence $f(x)+12=a(x-q)(x-r)(x-s)$ and $f(x)-12=a(x-t)(x-u)(x-v)$ for some $a\neq 0$ (else $f$ is not cubic) where $\{q,r,s,t,u,v\}$ is the same as the set $\{1,2,3,5,6,7\}$ . Subtracting the second equation from the first, expanding, and collecting like terms, we have that \[24=a((t+u+v-(q+r+s))x^2-a(tu+uv+tv-(qr+qs+rs))x+a(tuv-qrs)\] which must hold for all $x$ . Since $a\neq 0$ we have that (1) $t+u+v=q+r+s$ , (2) $tu+uv+tv=qr+qs+rs$ and (3) $a(tuv-qrs)=24$ . Since $q+r+s+t+u+v$ is the sum of 1,2,3,5,6, and 7, we have $q+r+s+t+u+v=24$ so that by (1) we have $q+r+s=12$ and $t+u+v=12$ . We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be $\geq 12$ with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be $\{2,3,7\}$ and $\{1,5,6\}$ . Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that $|a|=\frac{24}{|tuv-qrs|}=\frac{24}{12}=2$ . Since $a$ is the leading coefficient of $f(x)$ , the leading coefficient of $(f(x))^2$ is $a^2=|a|^2=2^2=4$ . Thus the leading coefficient of $g(x)$ is 4, i.e. $k=4$ . Then from earlier, $|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}=\sqrt{1260\cdot4+144}=\sqrt{5184}=72$ so that the answer is $\boxed{072}$

Solution: Express $f(x)$ in terms of powers of $(x-4)$ \[f(x) = a(x-4)^3 + b(x-4)^2 + c(x-4) + d\] By the same argument as in the first Solution, we see that $f(x)$ is an odd function about the line $x=4$ , so its coefficients $b$ and $d$ are 0. From there it is relatively simple to solve $f(2)=f(3)=-12$ (as in the above solution, but with a smaller system of equations): \[a(1)^3 + c(1) = -12\] \[a(2)^3 + c(2) = -12\] $a=2$ and $c=-14$ \[|f(0)| = |2(-4)^3 - 14(-4)| = \boxed{072}\]

Solution: Because a cubic must come in a "wave form" with two points of inflection, we can see that $f(1)=f(5)=f(6)$ , and $f(2)=f(3)=f(7)$ . By symmetry, $f(4)=0$ . Now, WLOG let $f(1)=12$ , and $f(2)=f(3)=-12$ . Then, we can use finite differences to get that the third (constant) difference is $12$ , and therefore $f(0)=12+(24+(24+12))=\boxed{072}$

Solution: We can rewrite our function as two different cubics, $f(x)=k(x-a)(x-b)(x-c)\pm12=k(x-d)(x-e)(x-f)\mp12$ . Note that $k$ is the same in both because they must be equal, so their leading terms must be. We must then, following Vieta's, choose our roots such that $a+b+c=d+e+f$ and verify that the other Vieta's formulas hold. Additionally, a cubic must only cross the x-axis thrice, restricting our choices for roots further. Choosing $a=1$ $b=5$ $c=6$ $d=2$ $e=3$ $f=7$ yields: \[kx^3-12kx^2+41kx-30k\pm12=kx^3-12kx^2+41kx-42k\mp12\] For the constant terms to have a difference of 24 ( $|\pm12-\mp12|$ ), we must have $k=\pm2$ , so the constant term of our polynomial is $\pm72$ , the absolute value of which is $\boxed{072}$ . -- Solution by eiis1000


Answer: 2.0

Let $f(x)= \sqrt{ax^2+bx}$ .  For how many real values of $a$ is there at least one positive value of $b$ for which the domain of $f$ and the range of $f$ are the same set
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} }$

Solution:
If $f(x)=y$ , then squaring both sides of the given equation and subtracting $ax^2$ and $bx$ yields $y^2-ax^2-bx=0$ . Completing the square, we get $(x+\frac{b}{2a})^2-\frac{y^2}{a}=\frac{b^2}{4a^2}$ where $y\geq 0$ . Divide out by $\frac{b^2}{4a^2}$ to put the equation in the standard form of an ellipse or hyperbola (depending on the sign of $a$ ) to get $\frac{(x+\frac{b}{2a})^2}{(\frac{b}{2a})^2}-\frac{y^2}{(\frac{b}{2\sqrt {\pm a}})^2}=1$
Before continuing, it is important to note that because $f(x)=\sqrt{ax^2+bx}=\sqrt{ax(x+\frac{b}{a})}$ $f(x)$ has roots 0 and $-\frac{b}{a}$ . Now, we can use the function we deduced to figure out some of its properties when:
$a>0$ : A semi-hyperbola above or on the x-axis. Therefore, no positive value of $a$ allows the domain and range to be the same set because the range will always be $[0, \infty)$ and the domain will always be undefined on some finite range between some value and zero.
$a=0$ : We must refer back to the original function; this results in a horizontal semiparabola in the first quadrant, which satisfies that the domain and range of the function are equal. Specifically, both sets are $[0, \infty )$ $a=0$ is the only case where this happens.
$a<0$ : A semi-ellipse in quadrant one. Since its roots are 0 and $-\frac{b}{a}$ , its domain must be $[0, -\frac{b}{a}]$ . To make its domain and range equal, the maximum value of the ellipse must then be $-\frac{b}{a}$ . But we have another expression for the maximum value of the ellipse, which is $0+\frac{b}{2\sqrt {\pm a}}$ . Setting these two expressions equal to each other will find us the final value of $a$ that satisfies the question.
$\frac{b}{-a}=\frac{b}{2\sqrt {\pm a}}$
$-a=2\sqrt {\pm a}$
$a^2=\pm 4a$
$a=0,\pm 4$
We already knew 0 was a solution from earlier, so -4 is our only new solution (we already ruled out any positive value of $a$ as a solution, so 4 does not work). Thus there are $\boxed{2}$ values of $a$ that make the domain and range of $f(x)$ the same set.


Answer: 21.0

Let $f(x)=(x^2+3x+2)^{\cos(\pi x)}$ . Find the sum of all positive integers $n$ for which \[\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.\]

Solution:
First, let's split it into two cases to get rid of the absolute value sign
$\left |\sum_{k=1}^n\log_{10}f(k)\right|=1 \iff \sum_{k=1}^n\log_{10}f(k)=1,-1$
Now we simplify using product-sum logarithmic identites:
$\log_{10}{f(1)}+\log_{10}{f(2)}+...+\log_{10}{f(n)}=\log_{10}{f(1)\cdot f(2) \cdot ... \cdot f(n)}=1,-1$
Note that the exponent $\cos{\pi(x)}$ is either $-1$ if $x$ is odd or $1$ if $x$ is even.
Writing out the first terms we have
$\frac{1}{(2)(3)}(3)(4)\frac{1}{(4)(5)} \ldots$
This product clearly telescopes (i.e. most terms cancel) and equals either $10$ or $\frac{1}{10}$ . But the resulting term after telescoping depends on parity (odd/evenness), so we split it two cases, one where $n$ is odd and another where $n$ is even.
$\textbf{Case 1: Odd n}$
For odd $n$ , it telescopes to $\frac{1}{2(n+2)}$ where $n$ is clearly $3$
$\textbf{Case 2: Even n}$
For even $n$ , it telescopes to $\frac{n+2}{2}$ where $18$ is the only possible $n$ value. Thus the answer is $\boxed{021}$

Solution:
Note that $\cos(\pi x)$ is $-1$ when $x$ is odd and $1$ when $x$ is even.  Also note that $x^2+3x+2=(x+1)(x+2)$ for all $x$ .  Therefore \[\log_{10}f(x)=\log_{10}(x+1)+\log_{10}(x+2)\ \ \ \text{if }x \text{ is even}\] \[\log_{10}f(x)=-\log_{10}(x+1)-\log_{10}(x+2)\ \ \ \text{if }x \text{ is odd}\] Because of this, $\sum_{k=1}^n\log_{10}f(k)$ is a telescoping series of logs, and we have \[\sum_{k=1}^n\log_{10}f(k)= \log_{10}(n+2)-\log_{10}2=\log_{10}\frac{n+2}{2}\ \ \ \text{if }n \text{ is even}\] \[\sum_{k=1}^n\log_{10}f(k)= -\log_{10}(n+2)-\log_{10}2=-\log_{10}2(n+2)\ \ \ \text{if }n \text{ is odd}\] Setting each of the above quantities to $1$ and $-1$ and solving for $n$ ,
we get possible values of $n=3$ and $n=18$ so our desired answer is $3+18=\boxed{021}$


Answer: 3.0

Let $f(x)=ax^2+bx+c$ , where $a$ $b$ , and $c$ are integers. Suppose that $f(1)=0$ $50<f(7)<60$ $70<f(8)<80$ $5000k<f(100)<5000(k+1)$ for some integer $k$ . What is $k$
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution:
From $f(1) = 0$ , we know that $a+b+c = 0$
From the first inequality, we get $50 < 49a+7b+c < 60$ . Subtracting $a+b+c = 0$ from this gives us $50 < 48a+6b < 60$ , and thus $\frac{25}{3} < 8a+b < 10$ . Since $8a+b$ must be an integer, it follows that $8a+b = 9$
Similarly, from the second inequality, we get $70 < 64a+8b+c < 80$ . Again subtracting $a+b+c = 0$ from this gives us $70 < 63a+7b < 80$ , or $10 < 9a+b < \frac{80}{7}$ . It follows from this that $9a+b = 11$
We now have a system of three equations: $a+b+c = 0$ $8a+b = 9$ , and $9a+b = 11$ . Solving gives us $(a, b, c) = (2, -7, 5)$ and from this we find that $f(100) = 2(100)^2-7(100)+5 = 19305$
Since $15000 < 19305 < 20000 \to 5000(3) < 19305 < 5000(4)$ , we find that $k = 3 \rightarrow \boxed{3}$

Solution:
$f(x)$ is some non-monic quadratic with a root at $x=1$ . Knowing this, we'll forget their silly $a$ $b$ , and $c$ and instead write it as $f(x)=p(x-1)(x-r)$
$f(7)=6p(7-r)$ , so $f(7)$ is a multiple of 6. They say $f(7)$ is between 50 and 60, exclusive. Notice that the only multiple of 6 in that range is 54. Thus, $f(7)=6p(7-r)=54$
$f(8)=7p(8-r)$ , so $f(8)$ is a multiple of 7. They say $f(8)$ is between 70 and 80, exclusive. Notice that the only multiple of 7 in that range is 77. Thus, $f(8)=7p(8-r)=77$
Now, we solve a system of equations in two variables.
\begin{align*} 6p(7-r)&=54 \\ 7p(8-r)&=77 \\ \\ p(7-r)&=9 \\ p(8-r)&=11 \\ \\ 7p-pr&=9 \\ 8p-pr&=11 \\ \\ (8p-pr)-(7p-pr)&=11-9 \\ \\ p&=2 \\ \\ 2(7-r)&=9 \\ \\ r&=2.5 \end{align*}
$f(100)=2(100-1)(100-2.5)=19305 \implies k=3 \implies \boxed{3}$

Solution:
So we know that $a,b,c$ are integers so we can use this to our advantage
$\quad$
Using $f(1)=0$ , we get the equation $a+b+c=0$ and $f(7)=49a+7b+c=5X$ where $X$ is a decimal digit placeholder. (Ex. $X=2$ provides the value $52$
$\quad$
Solving for $b$ using the system of equations, we get $48a+6b=5X$ $\implies$ $b=-8a+ \frac{5X}{6}$
$\quad$
Since we know that $a$ and $b$ are both integers, we know that $\frac{5X}{6}$ $\in$ $\mathbb{Z}$ $\implies$ $X=4$ and by extension $b=-8a+9$
$\quad$
Attempting to solve for $b$ again using the system $f(8)=64a+8b+c=7Y$ $Y$ is another decimal digit placeholder), $f(1)=a+b+c=0$ gives us $b=-9a+ \frac{7Y}{7}$ $\implies$ $Y=7$ $\implies$ $b=-9a+11$
$\quad$
This leads to $-8a+9=-9a+11$ $\implies$ $a=2$ $\implies$ $b=-7$
$\quad$
Plugging in the values of $a$ and $b$ into $f(1)=a+b+c=0$ , we get $c=5$
$\quad$
Substituting the values of $a,b,c$ into $f(100)=10000a+100b+c$ , we get $f(100)=19305$ and $5000k<19305<5000(k+1)$ $\implies$ $k=3$ $\implies$ $\boxed{3}$


Answer: 17.0

Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0, 1, 2, \dots , 25 \}$ .
The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:
$\textbf{(A)}\ 25\qquad \textbf{(B)}\ 22\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 17$

Solution:
Note that for all polynomials $f(x)$ $f(x + 6) \equiv f(x) \pmod 6$
Proof:
If $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ , then $f(x+6) = a_n(x+6)^n + a_{n-1}(x+6)^{n-1} +...+ a_0$ .  In the second equation, we can use the binomial expansion to expand every term, and then subtract off all terms that have a factor of $6^1$ or higher, since subtracting a multiple of $6$ will not change congruence $\pmod 6$ .  This leaves $a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ , which is $f(x)$ , so $f(x+6) \equiv f(x) \pmod 6$
So, we only need to test when $f(x)$ has a remainder of $0$ for $0, 1, 2, 3, 4, 5$ .  The set of numbers $6, 7, 8, 9, 10, 11$ will repeat remainders, as will all other sets.  The remainders are $2, 0, 0, 2, 0, 0$
This means for $s=1, 2, 4, 5$ $f(s)$ is divisible by $6$ .  Since $f(1)$ is divisible, so is $f(s)$ for $s=7, 13, 19, 25$ , which is $5$ values of $s$ that work.  Since $f(2)$ is divisible, so is $f(s)$ for $s=8, 14, 20$ , which is $4$ more values of $s$ that work.  The values of $s=4, 5$ will also generate $4$ solutions each, just like $f(2)$ .  This is a total of $17$ values of $s$ , for an answer of $\boxed{17}$

Solution: $f(x)$ $x^2$ $3x$ $2$ is $0$ congruent modulo 6 that implies $x+1$ or/and $x+2$ is $0$ congruent modulo $6$ .The numbers are of the form $6k+5$ and $6k+4$ for some integer $k$ and due to restrictions of number of elements in the set $S$ we get the inequality $k<4$ .Then we calculate the possible combinations to get $17$ as answer i.e. option $\boxed{17}$


Answer: 2.0

Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$ . The sum of the largest and smallest values of $f(x)$ is
$\textbf {(A)}\ 1 \qquad  \textbf {(B)}\ 2 \qquad  \textbf {(C)}\ 4 \qquad  \textbf {(D)}\ 6 \qquad  \textbf {(E)}\ \text{none of these}$

Solution:
Note that at $x=2,3,4,$ one of the three absolute values is equal to $0.$
Without using absolute values, we rewrite $f(x)$ as a piecewise function: \[f(x) = \begin{cases} (x-2)+(4-x)-(6-2x) & \mathrm{if} \ 2\leq x<3 \\ (x-2)+(4-x)-(2x-6) & \mathrm{if} \ 3\leq x<4 \\ (x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8 \end{cases},\] which simplifies to \[f(x) = \begin{cases} 2x-4 & \mathrm{if} \ 2\leq x<3 \\ -2x+8 & \mathrm{if} \ 3\leq x<4 \\ 0 & \mathrm{if} \ 4\leq x\leq8 \end{cases}.\] The graph of $y=f(x)$ is shown below.  The largest value of $f(x)$ is $2,$ and the smallest value of $f(x)$ is $0.$ So, their sum is $\boxed{2}.$


Answer: 15.0

Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$

Solution:
It is best to get rid of the absolute values first.
Under the given circumstances, we notice that $|x-p|=x-p$ $|x-15|=15-x$ , and $|x-p-15|=15+p-x$
Adding these together, we find that the sum is equal to $30-x$ , which attains its minimum value (on the given interval $p \leq x \leq 15$ ) when $x=15$ , giving a minimum of $\boxed{015}$

Solution: Let $p$ be equal to $15 - \varepsilon$ , where $\varepsilon$ is an almost neglectable value. Because of the small value $\varepsilon$ , the domain of $f(x)$ is basically the set ${15}$ . plugging in $15$ gives $\varepsilon + 0 + 15 - \varepsilon$ , or $15$ , so the answer is $\boxed{15}$


Answer: 301.0

Let $f_0(x)=x+|x-100|-|x+100|$ , and for $n\geq 1$ , let $f_n(x)=|f_{n-1}(x)|-1$ .  For how many values of $x$ is $f_{100}(x)=0$
$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$

Solution:
1. Draw the graph of $f_0(x)$ by dividing the domain into three parts.
2. Apply the recursive rule a few times to find the pattern.
Note: $f_n(x) = |f_{n-1}(x)| - 10$ is used to enlarge the difference, but the reasoning is the same.
3. Extrapolate to $f_{100}$ . Notice that the summits start $100$ away from $0$ and get $1$ closer each iteration, so they reach $0$ exactly at $f_{100}$
$f_{100}(x)$ reaches $0$ at $x = -300$ , then zigzags between $0$ and $-1$ , hitting $0$ at every even $x$ , before leaving $0$ at $x = 300$
This means that $f_{100}(x) = 0$ at all even $x$ where $-300 \le x \le 300$ . This is a $601$ -integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or $\frac{601+1}{2} = \boxed{301}$ .
(Revised by Flamedragon & Jason,C & emerald_block

Solution:
First, notice that the recursive rule moves the current value $1$ closer to $0$ . Upon reaching $0$ , it alternates between $-1$ and $0$ . This means that $f_{100}(x) = 0$ exactly when $|f_0(x)| \le 100$ (to reach $0$ in time) and $f_0(x)$ is even (so $f_{100}(x) \ne -1$ ).
Casework each part of $f_0(x)$ (where the expressions in the absolute values do not change sign): \[x \le -100 \implies f_0(x) = x-(x-100)+(x+100) = x+200\] so even $-300 \le x \le -100$ work. \[-100 \le x \le 100 \implies f_0(x) = x-(x-100)-(x+100) = -x\] so even $-100 \le x \le 100$ work. \[100 \le x \implies f_0(x) = x+(x-100)-(x+100) = x-200\] so even $100 \le x \le 300$ work.
Putting these together, all even $x$ where $-300 \le x \le 300$ work. So, the answer is $2\cdot150+1 = \boxed{301}$ .
~revised by emerald_block


Answer: 8.0

Let $f_1(x) = \frac23 - \frac3{3x+1}$ , and for $n \ge 2$ , define $f_n(x) = f_1(f_{n-1}(x))$ . The value of $x$ that satisfies $f_{1001}(x) = x-3$ can be expressed in the form $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
After evaluating the first few values of $f_k (x)$ , we obtain $f_4(x) = f_1(x) = \frac{2}{3} - \frac{3}{3x+1} = \frac{6x-7}{9x+3}$ . Since $1001 \equiv 2 \mod 3$ $f_{1001}(x) = f_2(x) = \frac{3x+7}{6-9x}$ . We set this equal to $x-3$ , i.e.
$\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}$ . The answer is thus $5+3 = \boxed{008}$


Answer: 8.0

Let $f_n (x) = \text{sin}^n x + \text{cos}^n x.$ For how many $x$ in $[0,\pi]$ is it true that
\[6f_{4}(x)-4f_{6}(x)=2f_{2}(x)?\]
$\text{(A) }2 \qquad \text{(B) }4  \qquad \text{(C) }6 \qquad \text{(D) }8 \qquad \text{(E) more than }8$

Solution:
Divide by 2 on both sides to get \[3f_{4}(x)-2f_{6}(x)=f_{2}(x)\] Substituting the definitions of $f_{2}(x)$ $f_{4}(x)$ , and $f_{6}(x)$ , we may rewrite the expression as \[3(\text{sin}^4{x} + \text{cos}^4{x}) - 2(\text{sin}^6{x} + \text{cos}^6{x}) = 1\] We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.
We can rewrite $3(\text{sin}^4 x + \text{cos}^4 x)$ as $3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x$ , which is equivalent to $3 - 6\text{sin}^2 x \text{cos}^2 x$
As for $2(\text{sin}^6 x + \text{cos}^6 x)$ , we may factor it as $2(\text{sin}^2 x + \text{cos}^2 x)(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$ which can be rewritten as $2(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$ , and then as $2(\text{sin}^2 x + \text{cos}^2)^2) - 6\text{sin}^2 x \text{cos}^2 x$ , which is equivalent to $2 - 6\text{sin}^2 x \text{cos}^2 x$
Putting everything together, we have $(3 - 6\text{sin}^2 {x} \text{cos}^2 {x}) - (2 - 6\text{sin}^2 {x} \text{cos}^2 {x}) = 1$ or $1 = 1$ . Therefore, the given equation $3f_{4}(x)-2f_{6}(x)=f_{2}(x)$ is true for all real $x$ , meaning that there are more than 8 values of $x$ that satisfy the given equation and so the answer is $\boxed{8}$


Answer: 226.0

Let $f_{1}(x)=\sqrt{1-x}$ , and for integers $n \geq 2$ , let $f_{n}(x)=f_{n-1}(\sqrt{n^2 - x})$ . If $N$ is the largest value of $n$ for which the domain of $f_{n}$ is nonempty, the domain of $f_{N}$ is $[c]$ . What is $N+c$
$\textbf{(A)}\ -226 \qquad \textbf{(B)}\ -144 \qquad \textbf{(C)}\ -20 \qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ 144$

Solution:
The domain of $f_{1}(x)=\sqrt{1-x}$ is defined when $x\leq1$ \[f_{2}(x)=f_{1}\left(\sqrt{4-x}\right)=\sqrt{1-\sqrt{4-x}}\]
Applying the domain of $f_{1}(x)$ and the fact that square roots must be positive, we get $0\leq\sqrt{4-x}\leq1$ . Simplifying, the domain of $f_{2}(x)$ becomes $3\leq x\leq4$
Repeat this process for $f_{3}(x)=\sqrt{1-\sqrt{4-\sqrt{9-x}}}$ to get a domain of $-7\leq x\leq0$
For $f_{4}(x)$ , since square roots must be nonnegative, we can see that the negative values of the previous domain will not work, so $\sqrt{16-x}=0$ . Thus we now arrive at $16$ being the only number in the of domain of $f_4 x$ that defines $x$ . However, since we are looking for the largest value for $n$ for which the domain of $f_{n}$ is nonempty, we must continue checking until we arrive at a domain that is empty.
We continue with $f_{5}(x)$ to get a domain of $\sqrt{25-x}=16 \implies x=-231$ . Since square roots cannot be negative, this is the last nonempty domain. We add to get $5-231=\boxed{226}$


Answer: 399.0

Let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = z^2 + iz + 1$ . How many complex numbers $z$ are there such that $\text{Im}(z) > 0$ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $10$
$\textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D)} \ 431 \qquad \textbf{(E)} \ 441$

Solution:
Suppose $f(z)=z^2+iz+1=c=a+bi$ . We look for $z$ with $\operatorname{Im}(z)>0$ such that $a,b$ are integers where $|a|, |b|\leq 10$
First, use the quadratic formula:
$z = \frac{1}{2} (-i \pm \sqrt{-1-4(1-c)}) = -\frac{i}{2} \pm \sqrt{ -\frac{5}{4} + c }$
Generally, consider the imaginary part of a radical of a complex number: $\sqrt{u}$ , where $u = v+wi = r e^{i\theta}$
$\operatorname{Im}(\sqrt{u}) = \operatorname{Im}(\pm \sqrt{r} e^{i\theta/2}) = \pm \sqrt{r} \sin(\theta/2) = \pm \sqrt{r}\sqrt{\frac{1-\cos\theta}{2}} = \pm \sqrt{\frac{r-v}{2}}$
Now let $u= -5/4 + c$ , then $v = -5/4 + a$ $w=b$ $r=\sqrt{v^2 + w^2}$
Note that $\operatorname{Im}(z)>0$ if and only if $\pm \sqrt{\frac{r-v}{2}}>\frac{1}{2}$ . The latter is true only when we take the positive sign, and that $r-v > 1/2$
or $v^2 + w^2 > (1/2 + v)^2 = 1/4 + v + v^2$ $w^2 > 1/4 + v$ , or $b^2 > a-1$
In other words, when $b^2 > a-1$ , the equation $f(z)=a+bi$ has unique solution $z$ in the region $\operatorname{Im}(z)>0$ ; and when $b^2 \leq a-1$ there is no solution. Therefore the number of desired solution $z$ is the same as the number of ordered pairs $(a,b)$ such that integers $|a|, |b|\leq 10$ , and that $b^2 \geq a$
When $a\leq 0$ , there is no restriction on $b$ so there are $11\cdot 21 = 231$ pairs;
when $a > 0$ , there are $2(1+4+9+10+10+10+10+10+10+10)=2(84)=168$ pairs.
So there are $231+168=\boxed{399}$ in total.

Solution:
We consider the function $f(z)$ as a mapping from the 2-D complex plane onto itself. We complete the square of $f(z)=z^2+iz+1=(z+\frac{i}{2})^2+\frac{5}{4}$
Now, we must decide the range of $f(z)$ based on the domain of $z$ $\operatorname{Im}(z)>0$ . To do this, we are interested in mapping the boundary line $\operatorname{Im}(z)=0$ . To make the mapping simpler, let $f(z)=g(z)+\frac{5}{4}$ , or $g(z)=(z+\frac{i}{2})^2$
We intend to map of the line $\operatorname{Im}(z)=0$ using the function $g(z)$ . This transformation is equivalent to the polar equation $r=(\frac{1}{2}\csc(\frac{\theta}{2}))^2$ . Using polar and trig identities, we can restate this equation as the rectangular form of a parabola,
$x=y^2-\frac{1}{4}$
where $x=\operatorname{Re}(z)$ and $y=\operatorname{Im}(z)$ . So, we conclude that $f(z)$ maps the line $\operatorname{Im}(z)=0$ to the parabola
$x=y^2-\frac{1}{4}+\frac{5}{4}=y^2+1$
A quick check reveals that the range of $f(z)$ is to the left of the parabola, meaning that any point on or to the right of parabola cannot be reached.
Since the problem requires $|\operatorname{Re}(z)|$ and $|\operatorname{Im}(z)|$ to both be integers and at most 10, all that remains is counting all points with integer coordinates in the range of $f(z), \operatorname{Im}(z)>0$ . To do this, we employ complementary counting.
The points of interest are $|\operatorname{Re}(z)|\leq 10$ and $|\operatorname{Im}(z)|\leq 10$ , resulting in a total of $441$ points. For lattice points on or to the right of the parabola, there are $10$ points for $x=0$ $9$ points for $x=\pm 1$ $6$ points for $x=\pm 2$ , and $1$ point for $x=\pm 3$ . Summing it all together, our answer is $441-(10+2*9+2*6+2*1)=\boxed{399}$


Answer: 1.0

Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1<a<b<c.$ What is $g(1)$ in terms of $a,b,$ and $c?$
$\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}$

Solution: Note that $f(1/x)$ has the same roots as $g(x)$ , if it is multiplied by some monomial so that the leading term is $x^3$ they will be equal. We have \[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\] so we can see that \[g(x) = \frac{x^3}{c}f(1/x)\] Therefore \[g(1) = \frac{1}{c}f(1) = \boxed{1}\]

Solution:
Let the three roots of $f(x)$ be $d$ $e$ , and $f$ . (Here e does NOT mean 2.7182818...)
We know that $a=-(d+e+f)$ $b=de+ef+df$ , and $c=-def$ , and that $g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}$ (Vieta's). This is equal to $\frac{def-de-df-ef+d+e+f-1}{def}$ , which equals $\boxed{1}$ . -dstanz5

Solution: If we let $p, q,$ and $r$ be the roots of $f(x)$ $f(x) = (x-p)(x-q)(x-r)$ and $g(x) = \left(x-\frac{1}{p}\right)\left(x-\frac{1}{q}\right)\left(x-\frac{1}{r}\right)$ . The requested value, $g(1)$ , is then \[\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)\left(1-\frac{1}{r}\right) = \frac{(p-1)(q-1)(r-1)}{pqr}\] The numerator is $-f(1)$ (using the product form of $f(x)$ ) and the denominator is $-c$ , so the answer is \[\frac{f(1)}{c} = \boxed{1}\]

Solution: It is well known that reversing the order of the coefficients of a polynomial turns each root into its corresponding reciprocal. Thus, a polynomial with the desired roots may be written as $cx^3+bx^2+a+1$ . As the problem statement asks for a monic polynomial, our answer is \[\boxed{1}\]


Answer: 6.0

Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$

Solution:
Let $r(x)$ be the remainder when $g(x^{12})$ is divided by $g(x)$ . Then $r(x)$ is the unique polynomial such that \[g(x^{12}) - r(x) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 - r(x)\] is divisible by $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1$ , and $\deg r(x) < 5$
Note that $(x - 1)(x^5 + x^4 + x^3 + x^2 + 1) = x^6 - 1$ is a multiple of $g(x)$ . Also, \[g(x^{12}) - 6 = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} - 5 \\ = (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).\] Each term is a multiple of $x^6 - 1$ . For example, \[x^{60} - 1 = (x^6 - 1)(x^{54} + x^{48} + \dots + x^6 + 1).\] Hence, $g(x^{12}) - 6$ is a multiple of $x^6 - 1$ , which means that $g(x^{12}) - 6$ is a multiple of $g(x)$ . Therefore, the remainder is $\boxed{6}$ . The answer is (A).

Solution: We express the quotient and remainder as follows. \[g(x^{12}) = Q(x) g(x) + R(x)\] Note that the solutions to $g(x)$ correspond to the 6th roots of unity, excluding $1$ . Hence, we have $x^6 = 1$ , allowing us to set: \[g(x^{12}) = 6\] \[g(x) = 0\] We have $5$ values of $x$ that return $R(x) = 6$ . However, $g(x)$ is quintic, implying the remainder is of degree at most $4$ . Since there are $5$ solutions, the only possibility is that the remainder is a constant $\boxed{6}$

Solution: We can use the Chinese remainder theorem over $\mathbb{Q}[x].$ Since $g(x)=(x+1)(x^2-x+1)(x^2+x+1)$ $\mathbb{Q}[x]/g\cong \mathbb{Q}[x]/(x+1)\times \mathbb{Q}[x]/(x^2-x+1)\times \mathbb{Q}[x]/(x^2+x+1).$ This means that if we can find the remainder of $g(x^{12})$ modulo $x+1,x^2-x+1,x^2+x+1$ , we can reconstruct the remainder modulo $g.$ We can further use that each factor is irreducible and that if $p(x)$ is an irreducible polynomial over $\mathbb{Q}$ with root $\alpha$ $\mathbb{Q}[x]/p\cong \mathbb{Q}(\alpha)$ so to evaluate the remainders of $g(x^{12})$ , we just need to evaluate it on one of the roots of the irreducible factors. The first factor has root $-1$ , the second has roots the primitive sixth roots of unity, and the third as roots the primitive cube roots of unity (this is easily seen as $g(x)(x-1)=x^6-1$ ). Evaluating $g(x^{12})$ on each of these values yields $g(1)=6$ so the remainder is $6$ on each factor on the right of the isomorphism. Hence, by the Chinese remainder theorem, the remainder modulo $g$ must be $\boxed{6}$ as well.


Answer: 8.0

Let $h_n$ and $k_n$ be the unique relatively prime positive integers such that \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{h_n}{k_n}.\] Let $L_n$ denote the least common multiple of the numbers $1, 2, 3, \ldots, n$ . For how many integers with $1\le{n}\le{22}$ is $k_n<L_n$
$\textbf{(A) }0 \qquad\textbf{(B) }3 \qquad\textbf{(C) }7 \qquad\textbf{(D) }8\qquad\textbf{(E) }10$

Solution:
We are given that \[\sum_{i=1}^{n}\frac1i = \frac{1}{L_n}\sum_{i=1}^{n}\frac{L_n}{i} = \frac{h_n}{k_n}.\] Since $k_n < L_n,$ we need $\gcd\left(\sum_{i=1}^{n}\frac{L_n}{i}, L_n\right)>1.$
For all primes $p$ such that $p\leq n,$ let $v_p(L_n)=e\geq1$ be the largest power of $p$ that is a factor of $L_n.$
It is clear that $L_n\equiv0\pmod{p},$ so we test whether $\sum_{i=1}^{n}\frac{L_n}{i}\equiv0\pmod{p}.$ Note that \[\sum_{i=1}^{n}\frac{L_n}{i} \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p^e) \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p).\] We construct the following table for $v_p(L_n)=e:$ \[\begin{array}{c|c|l|c}  \textbf{Case of }\boldsymbol{(p,e)} & \textbf{Interval of }\boldsymbol{n} & \hspace{22.75mm}\textbf{Sum} & \boldsymbol{\stackrel{?}{\equiv}0\ (\operatorname{mod} \ p)} \\ [0.5ex] \hline\hline  & & & \\ [-2ex] (2,1) & [2,3] & L_n/2 &  \\      (2,2) & [4,7] & L_n/4 &  \\     (2,3) & [8,15] & L_n/8 & \\     (2,4) & [16,22] & L_n/16 &  \\ [0.5ex] \hline   & & & \\ [-2ex] (3,1) & [3,5] & L_n/3 & \\     & [6,8] & L_n/3 + L_n/6 & \checkmark \\    (3,2) & [9,17] & L_n/9 & \\  & [18,22] & L_n/9 + L_n/18 & \checkmark \\ [0.5ex] \hline   & & & \\ [-2ex] (5,1) & [5,9] & L_n/5 & \\     & [10,14] & L_n/5 + L_n/10 & \\ & [15,19] & L_n/5 + L_n/10 + L_n/15 & \\   & [20,22] & L_n/5 + L_n/10 + L_n/15 + L_n/20 & \checkmark \\ [0.5ex] \hline   & & & \\ [-2ex] (7,1) & [7,13] & L_n/7 & \\     & [14,20] & L_n/7 + L_n/14 & \\ & [21,22] & L_n/7 + L_n/14 + L_n/21 & \\ [0.5ex] \hline   & & & \\ [-2ex] (11,1) & [11,21] & L_n/11 &  \\ & \{22\} & L_n/11 + L_n/22 & \\ [0.5ex] \hline & & & \\ [-2ex] (13,1) & [13,22] & L_n/13 &  \\ [0.5ex] \hline & & & \\ [-2ex] (17,1) & [17,22] & L_n/17 &  \\ [0.5ex] \hline & & & \\ [-2ex] (19,1) & [19,22] & L_n/19 & \\ [0.5ex] \end{array}\] Note that:
Together, there are $\boxed{8}$ such integers $n,$ namely \[6,7,8,18,19,20,21,22.\] ~MRENTHUSIASM


Answer: 8064.0

Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$ , then the numbers that Bernardo writes are $16, 25, 36, 49, 64$ , and the numbers showing on the board after Silvia erases are $1, 2, 3, 4,$ and $6$ , and thus $f(1) = 5$ . What is the sum of the digits of $f(2) + f(4)+ f(6) + \dots + f(2016)$
$\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064$

Solution:
Consider $f(2)$ . The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$ . Since $100\le (x+1)^2-x^2=2x+1$ , this first happens at $x\ge \lfloor 99/2\rfloor = 50$ . The perfect squares from here go: $2500,  2601, 2704, 2809\dots$ . Note that the ones and tens also make the perfect squares, $1^2,2^2,3^2\dots$ . After the ones and tens make $100$ , the hundreds place will go up by $2$ , thus reaching our goal. Since $10^2=100$ , the last perfect square to be written will be $\left(50+10\right)^2=60^2=3600$ . The missing number is one less than the number of hundreds $(k=2)$ of $3600$ , or $35$
Now consider $f(4)$ . Instead of the difference between two squares needing to be $100$ , the difference must now be $10000$ . This first happens at $x\ge 5000$ . After this point, similarly, $\sqrt{10000}=100$ more numbers are needed to make the $10^4$ th's place go up by $2$ . This will take place at $\left(5000+100\right)^2=5100^2= 26010000$ . Removing the last four digits (the zeros) and subtracting one yields $2600$ for the skipped value.
In general, each new value of $f(k+2)$ will add two digits to the " $5$ " and one digit to the " $1$ ". This means that the last number Bernardo writes for $k=6$ is $\left(500000+1000\right)^2$ , the last for $k = 8$ will be $\left(50000000+10000\right)^2$ , and so on until $k=2016$ . Removing the last $k$ digits as Silvia does will be the same as removing $k/2$ trailing zeroes on the number to be squared. This means that the last number on the board for $k=6$ is $5001^2$ $k=8$ is $50001^2$ , and so on. So the first missing number is $5001^2-1,50001^2-1\text{ etc.}$ The squaring will make a " $25$ " with two more digits than the last number, a " $10$ " with one more digit, and a " $1$ ". The missing number is one less than that, so the "1" will be subtracted from $f(k)$ . In other words, $f(k) = 25\cdot 10^{k-2}+1\cdot 10^{k/2}$
Therefore:
\[f(2) =35 =25 +10\] \[f(4) =2600 =2500 +100\] \[f(6) =251000 =250000 +1000\] \[f(8) = 25010000 = 25000000 + 10000\]
And so on. The sum $f(2) + f(4) + f(6) +\dots + f(2016)$ is:
$2.52525252525\dots 2525\cdot 10^{2015}$ $1.11111\dots 110\cdot 10^{1008}$ , with $2016$ repetitions each of " $25$ " and " $1$ ".
There is no carrying in this addition. Therefore each $f(k)$ adds $2 + 5 + 1 = 8$ to the sum of the digits.
Since $2n = 2016$ $n = 1008$ , and $8n = 8064$ , or $\boxed{8064}$

Solution:
We assume $n \geq 1$ for all claims.
We will let $g_k(x) = \left \lfloor \frac{x^2}{10^k}\right \rfloor$ . This is the result when the last k digits are truncated off $x^2$
Let $x_n$ = the smallest a, such that $g_{2n}(a) - g_{2n}(a-1) \geq 2$
We then have $\left \lfloor \frac{a^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(a-1)^2}{10^{2n}} \right \rfloor \geq 2$
Claim 1: $x_n > 5\cdot10^{2n-1}$
Proof of Claim 1:
Assume for the sake of contradiction, we have $x_n \leq 5 \cdot 10^{2n-1}$
Let $c = \frac{2x_n - 1}{10^{2n}}$ $d = \frac{(x_n-1)^2}{10^{2n}}$
Note that since $x_n \leq 5 \cdot 10^{2n-1}$ , we have $c < 1$
It is well known that $\lfloor i+j \rfloor \leq \lfloor i \rfloor + \lfloor j \rfloor + 1$ for any $i$ and $j$
Since $x_n$ satisfies the condition, we have:
$1 = \lfloor c \rfloor + 1 \geq  \lfloor{c+d} \rfloor - \lfloor d \rfloor \geq 2$
This is a contradiction. $\blacksquare$
Claim 2: $x_n = 5 \cdot 10^{2n-1} + 10^n$
Proof of Claim 2:
We will show our choice of $x_n = 5 \cdot 10^{2n-1} + 10^n$ satisfies the criteria above.
$\left \lfloor \frac{{x_n}^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(x_n-1)^2}{10^{2n}} \right \rfloor = \left \lfloor \frac{25 \cdot 10^{4n - 2} + 10^{3n} + 10^{2n}}{10^{2n}} \right \rfloor - \left \lfloor \frac{25 \cdot 10^{4n - 2} + 10^{3n}   - 2 \cdot 10^n + 1}{10^{2n}} \right \rfloor$
$= (25 \cdot 10^{2n-2} + 10^{n} + 1) - (25 \cdot 10^{2n-2} + 10^{n} - 1) = 2$
Now, we will show all values smaller than $5 \cdot 10^{2n-1} + 10^n$ don’t satisfy the criteria.
We will assume for the sake of contradiction, there exists an $x'_n$ which satisfies the criteria.
We will write $x'_n$ as $5 \cdot 10^{2n - 1} + k$ . By our hypothesis, we have $k < 10^n$ . We also have $k > 0$ by claim 1.
We have $\left \lfloor \frac{{x'}_n^2}{10^{2n}} \right \rfloor -  \left \lfloor \frac{(x'_n-1)^2}{10^{2n}} \right \rfloor = \left \lfloor \frac{25 \cdot 10^{4n - 2} + k \cdot 10^{2n} + k^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{25 \cdot 10^{4n - 2} + (k-1) \cdot 10^{2n} + (k-1)^2   }{10^{2n}} \right \rfloor$
$= (25 \cdot 10^{2n-2} + k) - (25 \cdot 10^{2n-2} + k - 1) = 1 \geq 2.$ This gets a contradiction. $\blacksquare$
Claim 3: $f(2n) = 25 \cdot10^{2n - 2} + 10^{n}$
Proof of Claim 3: Because of claim 2, $x_n$ is $5 \cdot 10^{2n-1} + 10^n$ . Note that $g_n(x_n)$ and $g_n(x_n - 1)$ are the first numbers written that will differ by at least 2. Thus, $f(2n) = g_n(x_n - 1) + 1 = 25 \cdot 10^{2n-2} + 10^n$ $\blacksquare$
Thus, $f(2) + f(4) \dots f(2016) = \underbrace{(2525 \dots 25)}_{1008 \ 25s} + \underbrace{(111111 \dots 1111)}_{1008 \ 1s}$ . This addition has no regroups/carry overs, so we can just take the sums of the digits of each of the addends, and sum them together to get 8064. $\boxed{8064}$


Answer: 6.0

Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$

Solution:
$k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$
So, $k^2 \equiv 0 \pmod{10}$ . Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$
Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$ . So the units digit is $6 \Rightarrow \boxed{6}$

Solution:
I am going to share another approach to this problem.
A units digit $k$ for an integer $n$ implies $n \equiv k \pmod{10}$
Let us take this step by step. First, we consider $k^2.$
Note that $k^2 = \left(2008^2 + 2^{2008}\right)^2 = 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016}.$ Now we calculate $k^2 \pmod{10}$
Before continuing, though, we must take note of the following:
\begin{align*} 2^1 &\equiv 2 \pmod {10} \\ 2^2 &\equiv 4 \pmod {10} \\ 2^3 &\equiv 8 \pmod {10} \\ 2^4 &\equiv 6 \pmod {10} \end{align*}
Now, we continue with the calculation.
\begin{align*} 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016} &\equiv 8^4 + 2 \cdot 8^2 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^1 \cdot 2^6 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^{1+6+2008} + 2^{5016} \pmod{10} \\ &\equiv 6 + 2^{2015} + 6 \pmod{10} \\ &\equiv 6 + 2^{3} + 6 \pmod{10} \\ &\equiv 6 + 8 + 6 \pmod{10} \\ &\equiv 20 \pmod{10} \\ &\equiv 0 \pmod{10} \\ \end{align*}
We do the same with $2^k.$ However, we just need to find $k \pmod 4$ in order to do this calculation since we have the table of $2^k \pmod 10.$
\begin{align*} 2008^2 + 2^{2008} &\equiv 8^2 \pmod{4} \\ &\equiv 64 \pmod 4\\ &\equiv 0 \pmod 4 \end{align*}
This implies that \begin{align*} 2^k &\equiv 2^{4} \pmod{10} \\ &\equiv 6 \pmod{10} \end{align*}
Thus, \begin{align*} k^2 + 2^k &\equiv 6+0 \pmod{10}\\ &\equiv \boxed{6}


Answer: 6.0

Solution:
$k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$
So, $k^2 \equiv 0 \pmod{10}$ . Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$
Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$ . So the units digit is $6 \Rightarrow \boxed{6}$

Solution:
I am going to share another approach to this problem.
A units digit $k$ for an integer $n$ implies $n \equiv k \pmod{10}$
Let us take this step by step. First, we consider $k^2.$
Note that $k^2 = \left(2008^2 + 2^{2008}\right)^2 = 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016}.$ Now we calculate $k^2 \pmod{10}$
Before continuing, though, we must take note of the following:
\begin{align*} 2^1 &\equiv 2 \pmod {10} \\ 2^2 &\equiv 4 \pmod {10} \\ 2^3 &\equiv 8 \pmod {10} \\ 2^4 &\equiv 6 \pmod {10} \end{align*}
Now, we continue with the calculation.
\begin{align*} 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016} &\equiv 8^4 + 2 \cdot 8^2 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^1 \cdot 2^6 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^{1+6+2008} + 2^{5016} \pmod{10} \\ &\equiv 6 + 2^{2015} + 6 \pmod{10} \\ &\equiv 6 + 2^{3} + 6 \pmod{10} \\ &\equiv 6 + 8 + 6 \pmod{10} \\ &\equiv 20 \pmod{10} \\ &\equiv 0 \pmod{10} \\ \end{align*}
We do the same with $2^k.$ However, we just need to find $k \pmod 4$ in order to do this calculation since we have the table of $2^k \pmod 10.$
\begin{align*} 2008^2 + 2^{2008} &\equiv 8^2 \pmod{4} \\ &\equiv 64 \pmod 4\\ &\equiv 0 \pmod 4 \end{align*}
This implies that \begin{align*} 2^k &\equiv 2^{4} \pmod{10} \\ &\equiv 6 \pmod{10} \end{align*}
Thus, \begin{align*} k^2 + 2^k &\equiv 6+0 \pmod{10}\\ &\equiv \boxed{6}


Answer: 243.0

Let $m \ge 3$ be an integer and let $S = \{3,4,5,\ldots,m\}$ . Find the smallest value of $m$ such that for every partition of $S$ into two subsets, at least one of the subsets contains integers $a$ $b$ , and $c$ (not necessarily distinct) such that $ab = c$
Note : a partition of $S$ is a pair of sets $A$ $B$ such that $A \cap B = \emptyset$ $A \cup B = S$

Solution:
We claim that $243$ is the minimal value of $m$ . Let the two partitioned sets be $A$ and $B$ ; we will try to partition $3, 9, 27, 81,$ and $243$ such that the $ab=c$ condition is not satisfied. Without loss of generality , we place $3$ in $A$ . Then $9$ must be placed in $B$ , so $81$ must be placed in $A$ , and $27$ must be placed in $B$ . Then $243$ cannot be placed in any set, so we know $m$ is less than or equal to $243$
For $m \le 242$ , we can partition $S$ into $S \cap \{3, 4, 5, 6, 7, 8, 81, 82, 83, 84 ... 242\}$ and $S \cap \{9, 10, 11 ... 80\}$ , and in neither set are there values where $ab=c$ (since $8 < (3\text{ to }8)^2 < 81$ and $81^2>242$ and $(9\text{ to }80)^2 > 80$ ). Thus $m = \boxed{243}$


Answer: 8.0

Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$ .
The largest integer which divides all possible numbers of the form $m^2-n^2$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$

Solution:
First, factor the difference of squares \[(m+n)(m-n)\] Since $m$ and $n$ are odd numbers, let $m=2a+1$ and $n=2b+1$ , where $a$ and $b$ can be any integer. \[(2a+2b+2)(2a-2b)\] Factor the resulting expression. \[4(a+b+1)(a-b)\] If $a$ and $b$ are both even, then $a-b$ is even.  If $a$ and $b$ are both odd, then $a-b$ is even as well.  If $a$ is odd and $b$ is even (or vise versa), then $a+b+1$ is even.  Therefore, in all cases, $8$ can be divided into all numbers with the form $m^2-n^2$
This can be confirmed by setting $m=3$ and $n=1$ , making $m^2-n^2=9-1=8$ .  Since $8$ is not a multiple of $3$ and is less than $16$ , we can confirm that the answer is $\boxed{8}$


Answer: 248.0

Let $m$ and $n$ be odd integers greater than $1.$ An $m\times n$ rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers $1$ through $n$ , those in the second row are numbered left to right with the integers $n + 1$ through $2n$ , and so on. Square $200$ is in the top row, and square $2000$ is in the bottom row. Find the number of ordered pairs $(m,n)$ of odd integers greater than $1$ with the property that, in the $m\times n$ rectangle, the line through the centers of squares $200$ and $2000$ intersects the interior of square $1099$

Solution:
Let us take some cases. Since $m$ and $n$ are odds, and $200$ is in the top row and $2000$ in the bottom, $m$ has to be $3$ $5$ $7$ , or $9$ . Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of $< 1$ . Therefore, $m < 1800 \mod n < 1800-m$
If $m=3$ $n$ can range from $667$ to $999$ . However, $900$ divides $1800$ , so looking at mods, we can easily eliminate $899$ and $901$ . Now, counting these odd integers, we get $167 - 2 = 165$
Similarly, let $m=5$ . Then $n$ can range from $401$ to $499$ . However, $450|1800$ , so one can remove $449$ and $451$ . Counting odd integers, we get $50 - 2 = 48$
Take $m=7$ . Then, $n$ can range from $287$ to $333$ . However, $300|1800$ , so one can verify and eliminate $299$ and $301$ . Counting odd integers, we get $24 - 2 = 22$
Let $m = 9$ . Then $n$ can vary from $223$ to $249$ . However, $225|1800$ . Checking that value and the values around it, we can eliminate $225$ . Counting odd integers, we get $14 - 1 = 13$
Add all of our cases to get \[165+48+22+13 = \boxed{248}\]


Answer: 407.0

Let $m$ and $n$ be positive integers satisfying the conditions
$\quad\bullet\ \gcd(m+n,210)=1,$
$\quad\bullet\ m^m$ is a multiple of $n^n,$ and
$\quad\bullet\ m$ is not a multiple of $n.$
Find the least possible value of $m+n.$

Solution: Taking inspiration from $4^4 \mid 10^{10}$ we are inspired to take $n$ to be $p^2$ , the lowest prime not dividing $210$ , or $11 \implies n = 121$ . Now, there are $242$ factors of $11$ , so $11^{242} \mid m^m$ , and then $m = 11k$ for $k \geq 22$ . Now, $\gcd(m+n, 210) = \gcd(11+k,210) = 1$ . Noting $k = 26$ is the minimal that satisfies this, we get $(n,m) = (121,286)$ . Thus, it is easy to verify this is minimal and we get $\boxed{407}$ . ~awang11

Solution:
Assume for the sake of contradiction that $n$ is a multiple of a single digit prime number, then $m$ must also be a multiple of that single digit prime number to accommodate for $n^n | m^m$ . However that means that $m+n$ is divisible by that single digit prime number, which violates $\gcd(m+n,210) = 1$ , so contradiction.
$n$ is also not 1 because then $m$ would be a multiple of it.
Thus, $n$ is a multiple of 11 and/or 13 and/or 17 and/or...
Assume for the sake of contradiction that $n$ has at most 1 power of 11, at most 1 power of 13...and so on...
Then, for $n^n | m^m$ to be satisfied, $m$ must contain at least the same prime factors that $n$ has. This tells us that for the primes where $n$ has one power of, $m$ also has at least one power, and since this holds true for all the primes of $n$ $n|m$ . Contradiction.
Thus $n$ needs more than one power of some prime.
The obvious smallest possible value of $n$ now is $11^2 =121$ .
Since $121^{121}=11^{242}$ , we need $m$ to be a multiple of 11 at least $242$ that is not divisible by $121$ and most importantly, $\gcd(m+n,210) = 1$ $242$ is divisible by $121$ , out. $253+121$ is divisible by 2, out. $264+121$ is divisible by 5, out. $275+121$ is divisible by 2, out. $286+121=37\cdot 11$ and satisfies all the conditions in the given problem, and the next case $n=169$ will give us at least $169\cdot 3$ , so we get $\boxed{407}$


Answer: 889.0

Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$

Solution:
For $0 < k < m$ , we have
Thus the product $a_{k}a_{k+1}$ is a monovariant : it decreases by 3 each time $k$ increases by 1.  For $k = 0$ we have $a_{k}a_{k+1} = 37\cdot 72$ , so when $k = \frac{37 \cdot 72}{3} = 888$ $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m = \boxed{889}$ , our answer.

Solution: Plugging in $k = m-1$ to the given relation, we get $0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}$ . Inspecting the value of $a_{k}a_{k+1}$ for small values of $k$ , we see that $a_{k}a_{k+1} = 37\cdot 72 - 3k$ . Setting the RHS of this equation equal to $3$ , we find that $m$ must be $\boxed{889}$

Solution:
Note that $a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3$ . Then, we can generate a sum of series of equations such that $\sum_{k=1}^{m-1} a_{k+1}\cdot a_{k} - a_{k}\cdot a_{k-1} = - 3(m-1)$ . Then, note that all but the first and last terms on the LHS cancel out, leaving us with $a_m\cdot a_{m-1} - a_1\cdot a_0 = -3(m-1)$ . Plugging in $a_m=0$ $a_0=37$ $a_1=72$ , we have $-37\cdot 72 = -3(m-1)$ . Solving for $m$ gives $m=\boxed{889}$ .
~sigma


Answer: 263.0

Let $m$ be the largest real solution to the equation
$\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4$
There are positive integers $a$ $b$ , and $c$ such that $m=a+\sqrt{b+\sqrt{c}}$ . Find $a+b+c$

Solution:
The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $\frac{3}{x-3}$ , then the fraction becomes of the form $\frac{x}{x - 3}$ . A similar cancellation happens with the other four terms. If we assume $x = 0$ is not the highest solution (which is true given the answer format) we can cancel the common factor of $x$ from both sides of the equation.
$\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x - 17} + \frac{1}{x - 19} = x - 11$
Then, if we make the substitution $y = x - 11$ , we can further simplify.
$\frac{1}{y + 8} + \frac{1}{y + 6} + \frac{1}{y - 6} + \frac{1}{y - 8} =y$
If we group and combine the terms of the form $y - n$ and $y + n$ , we get this equation:
$\frac{2y}{y^2 - 64} + \frac{2y}{y^2 - 36} = y$
Then, we can cancel out a $y$ from both sides, knowing that $x = 11$ is not a possible solution given the answer format.
After we do that, we can make the final substitution $z = y^2$
$\frac{2}{z - 64} + \frac{2}{z - 36} = 1$
$2z - 128 + 2z - 72 = (z - 64)(z - 36)$
$4z -  200 = z^2 - 100z + 64(36)$
$z^2 - 104z + 2504 = 0$
Using the quadratic formula, we get that the largest solution for $z$ is $z = 52 + 10\sqrt{2}$ . Then, repeatedly substituting backwards, we find that the largest value of $x$ is $11 + \sqrt{52 + \sqrt{200}}$ . The answer is thus $11 + 52 + 200 = \boxed{263}$


Answer: 476.0

Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$

Solution:
The three-digit integers divisible by $17$ , and their digit sum: \[\begin{array}{c|c} m & s(m)\\ \hline 102 & 3 \\ 119 & 11\\ 136 & 10\\ 153 & 9\\ 170 & 8\\ 187 & 16\\ 204 & 6\\ 221 & 5\\ 238 & 13\\ 255 & 12\\ 272 & 11\\ 289 & 19\\ 306 & 9\\ 323 & 8\\ 340 & 7\\ 357 & 15\\ 374 & 14\\ 391 & 13\\ 408 & 12\\ 425 & 11\\ 442 & 10\\ 459 & 18\\ 476 & 17 \end{array}\]
Thus the answer is $\boxed{476}$

Solution:
We can do the same thing as solution 1, except note the following fact: $102$ is a multiple of $17$ and its digits sum to $3$
Therefore, we can add it onto an existing multiple of $17$ that we know of to have $s(m) = 14$ , shown in the right-hand column, provided that its units digit is less than $8$ and its hundreds digit is less than $9$ . Unfortunately, $68$ does not fit the criteria, but $374$ does, meaning that, instead of continually adding multiples of $17$ , we can stop here and simply add $102$ to reach our final answer of $\boxed{476}$

Solution:
The digit sum of a base $10$ integer $m$ is just $m\pmod{9}$ . In this problem, we know $17\mid m$ , or $m=17k$ for a positive integer $k$
Also, we know that $m\equiv 17\equiv -1\pmod{9}$ , or $17k\equiv -k\equiv -1\pmod{9}$
Obviously $k=1$ is a solution. This means in general, $k=9x+1$ is a solution for non-negative integer $x$
Checking the first few possible solutions, we find that $m=\boxed{476}$ is the first solution that has $s(m)=17$ , and we're done.

Solution:
Since the sum of the digits in the base-10 representation of $m$ is $17$ , we must have $m\equiv 17 \pmod{9}$ or $m\equiv -1\pmod{9}$ .
We also know that since $m$ is divisible by 17, $m\equiv 0 \pmod{17}$
To solve this system of linear congruences, we can use the Chinese Remainder Theorem. If we set $m\equiv (-1)(17)(8)\pmod {153}$ , we find that $m\equiv 0\pmod{17}$ and $m\equiv -1\pmod{9}$ , because $17\cdot 8\equiv 136 \equiv 1\pmod{9}$ . The trick to getting here was to find the number $x$ such that $17x\equiv 1\pmod{9}$ , so that when we take things $\pmod{9}$ , the $17$ goes away. We can do this using the Extended Euclidean Algorithm or by guess and check to find that $x\equiv 8\pmod{9}$
Finally, since $m\equiv 17\pmod{153}$ , we repeatedly add multiples of $153$ until we get a number in which its digits sum to 17, which first happens when $m=\boxed{476}$

Solution:
We proceed by casework on the number of digits. Clearly the answer must have at least two digits, seeing as the maximum digit sum for a one-digit number is 9. The answer must also have less than 4 digits, because this is the AIME.
Case 1: The answer is a 2-digit number.
Represent the number as $10a + b$ , where $0 < a \leq 9$ and $0 \leq b \leq 9$ . The conditions of the problem translated into algebra are: \[17|10a+b\] \[a+b=17\] By the Euclidean Algorithm, this is equivalent to: \[17|9a\] 9 is not a factor of 17, so $17|a$ . So $a$ must be a multiple of 17, but this is impossible because of the conditions we placed on $a$ and $b$ .
(Alternatively, note that the only possible options are 89 and 98, and neither works.)
Case 2: The answer is a 3-digit number.
Represent the number as $100a+10b+c$ , where $0 < a \leq 9$ and $0 \leq b,c \leq 9$ . Translating the conditions again: \[17|100a+10b+c\] \[a+b+c=17\] \[17|99a+9b\] \[17|9(11a+b)\] \[17|11a+b\] Testing multiples of 17 yields $(4, 7, 6)$ as the minimal solution for $(a, b, c)$ and thus the answer is $\boxed{476}$


Answer: 750.0

Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$

Solution:
We can use complementary counting. We can choose a five-element subset in ${14\choose 5}$ ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 $A$ s and 5 $B$ s, thereby showing that there are ${10\choose 5}$ such sets.
Given a five-element subset $S$ of $\{1,2,\dots,14\}$ in which no two numbers are consecutive, we can start by writing down a string of length 14, in which the $i$ -th character is $A$ if $i\in S$ and $B$ otherwise. Now we got a string with 5 $A$ s and 9 $B$ s. As no two numbers were consecutive, we know that in our string no two $A$ s are consecutive. We can now remove exactly one $B$ from between each pair of $A$ s to get a string with 5 $A$ s and 5 $B$ s. And clearly this is a bijection, as from each string with 5 $A$ s and 5 $B$ s we can reconstruct one original set by reversing the construction.
Hence we have $m = {14\choose 5} - {10\choose 5} = 2002 - 252 = 1750$ , and the answer is $1750 \bmod 1000 = \boxed{750}$

Solution:
We will approach this problem using complementary counting. First it is obvious, there are $\binom{14}{5}$ total sets. To find the number of sets with NO consecutive elements, we do a little experimentation. Consider the following:
Start of with any of the $14$ numbers, say WLOG, $1$ . Then we cannot have $2$ . So again WLOG, we may pick $3$ . Then we cannot have $4$ , so again WLOG, we may pick $5$ . Then not $6$ , but $7$ , then not $8$ , but $9$ . Now we have the set ${1,3,5,7,9}$ , and we had to remove $4$ digits from the $14$ total amount, so there wasn't any consecutive numbers. So we have that the number of non-consecutive cardinality $5$ sets, is $\binom{14-4}{5}=\binom{10}{5}$ . Now you already read the solutions above, and if you are here, you are either confused, or looking for more. So now the answer is simply $1750$ , which is $\boxed{750}$ (mod $1000$ ).

Solution:
We will proceed by complementary counting. We can easily see that there are $\binom{14}{5}$ ways to select $5$ element subsets from the original set. Now, we must find the number of subsets part of that group that do NOT have any consecutive integers.
We can think of this situation as stars and bars. We have $9$ stars and $5$ bars, where the stars represent the integers NOT in the subset while the bars ARE the integers in the chosen subset. We want the amount of combinations where there is at least one star between each bar (meaning that none of the integers in the subset are consecutive). To do this, we remove $4$ stars from the total $9$ stars ( $4$ because that is the amount that is between each bar). Now, we have $5$ stars and $5$ bars with no restrictions, so we calculate this as $\binom{10}{5}$ , which is $252$ . Hence, $\binom{14}{15} - \binom{10}{5} = 2002 - 252 = 1750$ , which becomes $\boxed{750}$