Answer: 448.0

For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+1=5$ and for $\{5\}$ it is simply $5$ . Find the sum of all such alternating sums for $n=7$

Solution:
Let $S$ be a non- empty subset of $\{1,2,3,4,5,6\}$
Then the alternating sum of $S$ , plus the alternating sum of $S \cup \{7\}$ , is $7$ . This is because, since $7$ is the largest element, when we take an alternating sum, each number in $S$ ends up with the opposite sign of each corresponding element of $S\cup \{7\}$
Because there are $2^{6}=64$ of these pairs of sets, the sum of all possible subsets of our given set is $64 \cdot 7$ , giving an answer of $\boxed{448}$

Solution: Consider a given subset $T$ of $S$ that contains $7$ ; then there is a subset $T'$ which contains all the elements of $T$ except for $7$ , and only those elements . Since each element of $T'$ has one fewer element preceding it than it does in $T$ , their signs are opposite. Thus the sum of the alternating sums of $T$ and $T'$ is equal to 7. There are $2^6$ subsets containing 7, so our answer is $7 \cdot 2^6 = \boxed{448}$

Solution: Denote the desired total of all alternating sums of an $n$ -element set as $S_n$ . We are looking for $S_7$ . Notice that all alternating sums of an $n$ -element set are also alternating sums of an $n+1$ -element set. However, when we go from an $n$ to $n+1$ element set, for each subset with the new element, we are adding the new element and subtracting one of the alternating sums of the $n$ -element set. There are $2^n$ subsets of an $n+1$ -element set that includes the new element, giving us the relationship $S_{n+1} = S_n + 2^n(n+1) - S_n = 2^n(n+1)$ . When $n = 6$ , we therefore get $S_ 7 = 2^6(7) = \boxed{448}$

Solution:
We analyze all the numbers from 1 to 7 separately to see where the number contributes its positive or negative to the sum of the alternating sums. Whenever 7 appears, which it does 64 times, it contributes a positive because it is always first. This gives a net gain of $7 \cdot 64=448$
If we look at when 6 appears, which it also does 64 times, whether it comes as positive or negative depends on the presence of 7. Half of the subsets with 6 have 7 resulting in subtracting 6 each time, while the other half does not have 7 adding 6 each time, so these contributions of sixes cancel each other out giving a net gain of 0.
The same thing happens to any positive integer less than 7. This is because the determination of a positive or negative contribution is dependent on the number of larger numbers in front of it(For example, the sign of 3 is dependent on the presence of 4, 5, 6, and 7 in the subset). If the number of larger numbers is even, it gives in a positive copy while odd produces its negative. We know that the frequencies of these two cases occurring are the same because $0=(1-1)^{n}=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...\binom{n}{n}$ via the Binomial Theorem. Therefore, all positive integers less than 7 will not have any effect and our sum will be $\boxed{448}$

Solution: Let $\mathbb{N}_n := \{1, 2, 3, \dots n\}$ . Let the alternating sum of a certain subset of $S$ of $\mathbb{N}_n$ be $\xi(S),$ and let \[\mathcal{A}(\mathbb{N}_n) := \sum_{S \subseteq \mathbb{N}_n} \xi(S).\] We see that \[\mathcal{A}(\mathbb{N}_n) = \sum_{S \subseteq \mathbb{N}_n} \xi(S) = \sum_{n \in S, S \subseteq \mathbb{N}_n} \xi(S) + \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = \mathcal{A}(\mathbb{N}_{n-1}) + \sum_{n \in S, S \subseteq \mathbb{N}_n} \left( n - \xi(S - \{n\}) \right),\] as if $n \in S,$ $n$ is the largest element in $S.$ Now, we know that \[\sum_{n \in S, S \subseteq \mathbb{N}_n} n - \xi(S - \{n\}) = \sum_{S \subseteq \mathbb{N}_{n-1}} n - \xi(S) = n \cdot 2^{n-1} - \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = n \cdot 2^{n-1} - \mathcal{A}(\mathbb{N}_{n-1}),\] so \[\mathcal{A}(\mathbb{N}_{n}) = n \cdot 2^{n-1}.\] Thus, our answer (which is the $n = 7$ case) is $\mathcal{A}(\mathbb{N}_{7}) = 7 \cdot 2^6 = \boxed{448}.$


Answer: 7.0

For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ .  Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ .  What is the maximum value of $N(k)$
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$

Solution:
Notice that $2$ is a prime factor of an integer $n$ if and only if $n$ is even. Therefore, given any sufficiently high positive integral value of $k$ , dividing $I_k$ by $2^6$ yields a terminal digit of zero, and dividing by 2 again leaves us with $2^7 \cdot a = I_k$ where $a$ is an odd integer.
Observe then that $\boxed{7}$ must be the maximum value for $N(k)$ because whatever value we choose for $k$ $N(k)$ must be less than or equal to $7$

Solution:
Similar to the other solutions, notice that $I_k$ can be written as $10^{k+2}+64 \Rightarrow 2^{k+2}5^{k+2}+2^6$ . Factoring out $2^6$ we see that
$I_k = 2^6(2^{k-4}5^{k+2}+1)$
Notice that for $k < 4$ $2^{k-4}$ will not be an integer, and will "steal" some $2$ 's from the $2^6$ . We don't want this to happen, since we want to maximize the exponent of $2$ . We start by considering $k = 4$ . Then
$I_k = 2^6(*5^6+1) \Rightarrow 2^6(5^6+1)$
$5^6$ is an odd number; more specifically, it ends in $25$ (all powers of $5$ after $5^1$ end in $25$ ). Therefore the value in the parentheses will be some large number that ends in $26$ . Considering the rules of divisibility, we find that $5^6+1$ is even, so it is divisible by $2$ . Now our exponent of $2$ is at $7$ . But the divisibility rule for $4$ is the last $2$ digits of the number must be divisible by $4$ . We know the last digits: $26$ , which is not divisible by $4$ . Therefore $5^6 + 1$ is divisible by $2$ , but not $4$ . Testing more values of $k$ , we find that for $k \ge 5$ , the last digit becomes $1$ , which means it is not even divisible by $2$ . Therefore the highest possible exponent of $2$ that we can reach is $7 \Rightarrow \boxed{7}$

Solution:
Let $m=k+2$ $v_2(10^m+2^6)=6$ if $m>6$ and $v_2(10^m+2^6)=m$ if $m<6$ .
However, if $m=6$ , then $v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)$ . By LTE, $v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3$ . Since $v_2(5^6-1)=3$ $v_2(5^6+1)$ must equal $1$ . So, the answer is $6+1=7 \Rightarrow \boxed{7}$


Answer: 7.0

Solution:
The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$
For $k\in\{1,2,3\}$ we have $I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)$ . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\leq 5$
For $k>4$ we have $I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)$ . For $k>4$ the value in the parentheses is odd, hence $N(k)=6$
This leaves the case $k=4$ . We have $I_4 = 2^6 \left( 5^6 + 1 \right)$ . The value $5^6 + 1$ is obviously even. And as $5\equiv 1 \pmod 4$ , we have $5^6 \equiv 1 \pmod 4$ , and therefore $5^6 + 1 \equiv 2 \pmod 4$ . Hence the largest power of $2$ that divides $5^6+1$ is $2^1$ , and this gives us the desired maximum of the function $N$ $N(4) = \boxed{7}$


Answer: 7.0

Solution:
Notice that $2$ is a prime factor of an integer $n$ if and only if $n$ is even. Therefore, given any sufficiently high positive integral value of $k$ , dividing $I_k$ by $2^6$ yields a terminal digit of zero, and dividing by 2 again leaves us with $2^7 \cdot a = I_k$ where $a$ is an odd integer.
Observe then that $\boxed{7}$ must be the maximum value for $N(k)$ because whatever value we choose for $k$ $N(k)$ must be less than or equal to $7$

Solution:
Similar to the other solutions, notice that $I_k$ can be written as $10^{k+2}+64 \Rightarrow 2^{k+2}5^{k+2}+2^6$ . Factoring out $2^6$ we see that
$I_k = 2^6(2^{k-4}5^{k+2}+1)$
Notice that for $k < 4$ $2^{k-4}$ will not be an integer, and will "steal" some $2$ 's from the $2^6$ . We don't want this to happen, since we want to maximize the exponent of $2$ . We start by considering $k = 4$ . Then
$I_k = 2^6(*5^6+1) \Rightarrow 2^6(5^6+1)$
$5^6$ is an odd number; more specifically, it ends in $25$ (all powers of $5$ after $5^1$ end in $25$ ). Therefore the value in the parentheses will be some large number that ends in $26$ . Considering the rules of divisibility, we find that $5^6+1$ is even, so it is divisible by $2$ . Now our exponent of $2$ is at $7$ . But the divisibility rule for $4$ is the last $2$ digits of the number must be divisible by $4$ . We know the last digits: $26$ , which is not divisible by $4$ . Therefore $5^6 + 1$ is divisible by $2$ , but not $4$ . Testing more values of $k$ , we find that for $k \ge 5$ , the last digit becomes $1$ , which means it is not even divisible by $2$ . Therefore the highest possible exponent of $2$ that we can reach is $7 \Rightarrow \boxed{7}$

Solution:
Let $m=k+2$ $v_2(10^m+2^6)=6$ if $m>6$ and $v_2(10^m+2^6)=m$ if $m<6$ .
However, if $m=6$ , then $v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)$ . By LTE, $v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3$ . Since $v_2(5^6-1)=3$ $v_2(5^6+1)$ must equal $1$ . So, the answer is $6+1=7 \Rightarrow \boxed{7}$


Answer: 7.0

Solution:
The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$
For $k\in\{1,2,3\}$ we have $I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)$ . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\leq 5$
For $k>4$ we have $I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)$ . For $k>4$ the value in the parentheses is odd, hence $N(k)=6$
This leaves the case $k=4$ . We have $I_4 = 2^6 \left( 5^6 + 1 \right)$ . The value $5^6 + 1$ is obviously even. And as $5\equiv 1 \pmod 4$ , we have $5^6 \equiv 1 \pmod 4$ , and therefore $5^6 + 1 \equiv 2 \pmod 4$ . Hence the largest power of $2$ that divides $5^6+1$ is $2^1$ , and this gives us the desired maximum of the function $N$ $N(4) = \boxed{7}$


Answer: 47.0

For $n \ge 1$ call a finite sequence $(a_1, a_2 \ldots a_n)$ of positive integers progressive if $a_i < a_{i+1}$ and $a_i$ divides $a_{i+1}$ for all $1 \le i \le n-1$ . Find the number of progressive sequences such that the sum of the terms in the sequence is equal to $360$

Solution:
If the first term is $x$ , then dividing through by $x$ , we see that we can find the number of progressive sequences whose sum is $\frac{360}{x} - 1$ , and whose first term is not 1. If $a(k)$ denotes the number of progressive sequences whose sum is $k$ and whose first term is not 1, then we can express the answer $N$ as follows:
\begin{align*}N &= a(359) + a(179) + a(119) + a(89) + a(71) + a(59) + a(44) + a(39) \\ &+ a(35) + a(29) + a(23) + a(19) + a(17) + a(14) + a(11) + a(9) \\ &+ a(8) + a(7) + a(5) + a(4) + a(3) + a(2) + a(1) + 1 \end{align*}
The $+1$ at the end accounts for the sequence whose only term is 360. Fortunately, many of these numbers are prime; we have $a(p) = 1$ for primes $p$ as the only such sequence is " $p$ " itself. Also, $a(1) = 0$ . So we have
\[N = 15 + a(119) + a(44) + a(39) + a(35) + a(14) + a(9) + a(8) + a(4)\]
For small $k$ $a(k)$ is easy to compute: $a(4) = 1$ $a(8) = 2$ $a(9) = 2$ . For intermediate $k$ (e.g. $k=21$ below), $a(k)$ can be computed recursively using previously-computed values of $a(\cdot)$ , similar to dynamic programming. Then we have \begin{align*} a(14) &= 1+a(6) = 1+2 = 3\\ a(35) &= 1+a(6)+a(4) = 1 + 2 + 1 = 4 \\ a(39) &= 2 + a(12) = 2+4 = 6 \\ a(44) &= 2 + a(21) + a(10) = 2+4+2=8 \\ a(119) &= 1 + a(16) + a(6) = 1 + 3 + 2 = 6 \end{align*} Thus the answer is $N = 15+6+8+6+4+3+2+2+1 = \boxed{47}$


Answer: 2.0

For $n$ a positive integer, let $R(n)$ be the sum of the remainders when $n$ is divided by $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , and $10$ . For example, $R(15) = 1+0+3+0+3+1+7+6+5=26$ . How many two-digit positive integers $n$ satisfy $R(n) = R(n+1)\,?$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution: Note that we can add $9$ to $R(n)$ to get $R(n+1)$ , but must subtract $k$ for all $k|n+1$ . Hence, we see that there are four ways to do that because $9=7+2=6+3=5+4=4+3+2$ . Note that only $7+2$ is a plausible option, since $4+3+2$ indicates $n+1$ is divisible by $6$ $5+4$ indicates that $n+1$ is divisible by $2$ $6+3$ indicates $n+1$ is divisible by $2$ , and $9$ itself indicates divisibility by $3$ , too. So, $14|n+1$ and $n+1$ is not divisible by any positive integers from $2$ to $10$ , inclusive, except $2$ and $7$ . We check and get that only $n+1=14 \cdot 1$ and $n+1=14 \cdot 7$ give possible solutions so our answer is $\boxed{2}$

Solution:
Denote by ${\rm Rem} \ \left( n, k \right)$ the remainder of $n$ divided by $k$ .
Define $\Delta \left( n, k \right) = {\rm Rem} \ \left( n + 1, k \right) - {\rm Rem} \ \left( n, k \right)$
Hence, \[ \Delta \left( n, k \right) = \left\{ \begin{array}{ll} 1 & \mbox{ if } n \not\equiv -1 \pmod{k} \\ - \left( k  -1 \right) & \mbox{ if } n \equiv -1 \pmod{k} \end{array} \right.. \]
Hence, this problem asks us to find all $n \in \left\{ 10 , 11, \cdots , 99 \right\}$ , such that $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$
$\textbf{Case 1}$ $\Delta \left( n, 10 \right) = - 9$
We have $\sum_{k = 2}^9 \Delta \left( n, k \right) \leq \sum_{k = 2}^9 1 = 8$
Therefore, there is no $n$ in this case.
$\textbf{Case 2}$ $\Delta \left( n, 10 \right) = 1$ and $\Delta \left( n, 9 \right) = -8$
The condition $\Delta \left( n, 9 \right) = -8$ implies $n \equiv - 1 \pmod{9}$ .
This further implies $n \equiv - 1 \pmod{3}$ .
Hence, $\Delta \left( n, 3 \right) = -2$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}} \Delta \left( n, k \right) = 9$
However, we have $\sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 2 , 4 , 5 , 6, 7, 8\right\}}  1 = 6$
Therefore, there is no $n$ in this case.
$\textbf{Case 3}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 9 , 10 \right\}$ and $\Delta \left( n, 8 \right) = -7$
The condition $\Delta \left( n, 8 \right) = -7$ implies $n \equiv - 1 \pmod{k}$ with $k \in \left\{ 2, 4 \right\}$ .
Hence, $\Delta \left( n, 2 \right) = -1$ and $\Delta \left( n, 4 \right) = -3$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 3, 5, 6, 7 \right\}} \Delta \left( n, k \right) = 9$
However, we have $\sum_{k \in \left\{ 3, 5, 6, 7 \right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 3, 5, 6, 7 \right\}}  1 = 4$
Therefore, there is no $n$ in this case.
$\textbf{Case 4}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 8, \cdots , 10 \right\}$ and $\Delta \left( n, 7 \right) = -6$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 2, 3, 4, 5, 6 \right\}} \Delta \left( n, k \right) = 3$
Hence, we must have $\Delta \left( n, 2 \right) = -1$ and $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 3 , 4 , 5 , 6 \right\}$
Therefore, $n = 13, 97$
$\textbf{Case 5}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 7 , \cdots , 10 \right\}$ and $\Delta \left( n, 6 \right) = -5$
The condition $\Delta \left( n, 6 \right) = -5$ implies $n \equiv - 1 \pmod{k}$ with $k \in \left\{ 2, 3 \right\}$ .
Hence, $\Delta \left( n, 2 \right) = -1$ and $\Delta \left( n, 3 \right) = -2$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 4, 5 \right\}} \Delta \left( n, k \right) = 4$
However, we have $\sum_{k \in \left\{ 4, 5 \right\}} \Delta \left( n, k \right) \leq \sum_{k \in \left\{ 4, 5 \right\}}  1 = 2$
Therefore, there is no $n$ in this case.
$\textbf{Case 6}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 6 , \cdots , 10 \right\}$ and $\Delta \left( n, 5 \right) = -4$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\sum_{k \in \left\{ 2, 3, 4 \right\}} \Delta \left( n, k \right) = -1$
This can be achieved if $\Delta \left( n, 2 \right) = 1$ $\Delta \left( n, 3 \right) = 1$ $\Delta \left( n, 4 \right) = -3$
However, $\Delta \left( n, 4 \right) = -3$ implies $n \equiv - 1 \pmod{4}$ . This implies $n \equiv -1 \pmod{2}$ . Hence, $\Delta \left( n, 2 \right) = -1$ .
We get a contradiction.
Therefore, there is no $n$ in this case.
$\textbf{Case 7}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 5 , \cdots , 10 \right\}$ and $\Delta \left( n, 4 \right) = -3$
The condition $\Delta \left( n, 4 \right) = -3$ implies $n \equiv - 1 \pmod{k}$ with $k = 2$ .
Hence, $\Delta \left( n, 2 \right) = -1$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\Delta \left( n, 3 \right) = - 2$ . This implies $n \equiv - 1 \pmod{3}$
Because $n \equiv - 1 \pmod{2}$ and $n \equiv - 1 \pmod{3}$ , we have $n \equiv - 1 \pmod{6}$ .
Hence, $\Delta \left( n, 6 \right) = - 5$ .
However, in this case, we assume $\Delta \left( n, 6 \right) = 1$ .
We get a contradiction.
Therefore, there is no $n$ in this case.
$\textbf{Case 8}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{ 4 , \cdots , 10 \right\}$ and $\Delta \left( n, 3 \right) = -2$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\Delta \left( n, 2 \right) = - 5$ . This is infeasible.
Therefore, there is no $n$ in this case.
$\textbf{Case 9}$ $\Delta \left( n, k \right) = 1$ for $k \in \left\{3 , \cdots , 10 \right\}$
To get $\sum_{k = 2}^{10} \Delta \left( n, k \right) = 0$ , we have $\Delta \left( n, 2 \right) = - 8$ . This is infeasible.
Therefore, there is no $n$ in this case.
Putting all cases together, the answer is $\boxed{2}$


Answer: 80200.0

For $p=1, 2, \cdots, 10$ let $S_p$ be the sum of the first $40$ terms of the arithmetic progression whose first term is $p$ and whose common difference is $2p-1$ ; then $S_1+S_2+\cdots+S_{10}$ is
$\mathrm{(A)\ } 80000 \qquad \mathrm{(B) \ }80200 \qquad \mathrm{(C) \  } 80400 \qquad \mathrm{(D) \  } 80600 \qquad \mathrm{(E) \  }80800$

Solution:
The $40\text{th}$ term of an arithmetic progression with a first term $p$ and a common difference $2p-1$ is $p+39(2p-1)=79p-39$ . Therefore, the sum of the first $40$ terms of such a progression is $\frac{40}{2}(79p-39+p)=1600p-780$
We now want to evaluate $\sum_{p=1}^{10}(1600p-780)$ \[\sum_{p=1}^{10}(1600p-780)=1600\sum_{p=1}^{10}(p)-\sum_{p=1}^{10}(780)\] \[=(1600)\left(\frac{10\cdot11}{2}\right)-(780)(10)=88000-7800=80200, \boxed{80200}.\]


Answer: 905.0

Solution:
Because the order of the $a$ 's doesn't matter, we simply need to find the number of $1$ $2$ $3$ s and $4$ s that minimize $S_2$ . So let $w, x, y,$ and $z$ represent the number of $1$ s, $2$ s, $3$ s, and $4$ s respectively. Then we can write three equations based on these variables. Since there are a total of $350$ $a$ s, we know that $w + x + y + z = 350$ . We also know that $w + 2x + 3y + 4z = 513$ and $w + 16x + 81y + 256z = 4745$ . We can now solve these down to two variables: \[w = 350 - x - y - z\] Substituting this into the second and third equations, we get \[x + 2y + 3z = 163\] and \[15x + 80y + 255z = 4395.\] The second of these can be reduced to \[3x + 16y + 51z = 879.\] Now we substitute $x$ from the first new equation into the other new equation. \[x = 163 - 2y - 3z\] \[3(163 - 2y - 3z) + 16y + 51z = 879\] \[489 + 10y + 42z = 879\] \[5y + 21z = 195\] Since $y$ and $z$ are integers, the two solutions to this are $(y,z) = (39,0)$ or $(18,5)$ .
If you plug both these solutions in to $S_2$ it is apparent that the second one returns a smaller value. It turns out that $w = 215$ $x = 112$ $y = 18$ , and $z = 5$ , so $S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = \boxed{905}$


Answer: 10.0

For a certain complex number $c$ , the polynomial \[P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)\] has exactly 4 distinct roots. What is $|c|$
$\textbf{(A) } 2 \qquad \textbf{(B) } \sqrt{6} \qquad \textbf{(C) } 2\sqrt{2} \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \sqrt{10}$

Solution:
The polynomial can be factored further broken down into
$P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$
by using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of two (not necessarily distinct) factors from the set: $(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),$ and $(x - [2 + 2i])$ . We need the two factors to yield a constant term of $4$ when multiplied together. The only combinations that work are $(x - [1 - i])$ and $(x - [2 + 2i])$ , or $(x - [1+i])$ and $(x - [2-2i])$ . When multiplied together, the polynomial is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$ . Therefore, $c = 3 \pm i$ and $|c| = \boxed{10}$


Answer: 201400.0

For a certain positive integer $n$ less than $1000$ , the decimal equivalent of $\frac{1}{n}$ is $0.\overline{abcdef}$ , a repeating decimal of period of $6$ , and the decimal equivalent of $\frac{1}{n+6}$ is $0.\overline{wxyz}$ , a repeating decimal of period $4$ . In which interval does $n$ lie?
$\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]$

Solution:
Solution by e_power_pi_times_i
If $\frac{1}{n} = 0.\overline{abcdef}$ $n$ must be a factor of $999999$ . Also, by the same procedure, $n+6$ must be a factor of $9999$ . Checking through all the factors of $999999$ and $9999$ that are less than $1000$ , we see that $n = 297$ is a solution, so the answer is $\boxed{201,400}$

Solution:
Notice that the repeating fraction $0.\overline{abcdef}$ can be represented as $\frac{abcdef}{999999},$ and thereby, $n|999999.$ Also, notice that $0.\overline{wxyz} = \frac{wxyz}{9999},$ so $(n+6)|9999.$ However, we have to make some restrictions here. For instance, if $n|99999,$ then $\frac{1}{n}$ could be expressed as $\frac{a’b’c’d’e’}{99999} = .\overline{a’b’c’d’e’}$ which cannot happen. Therefore, from this, we see that the smallest $m$ such that $n|\underbrace{999\cdots999}_{m \text{ nines}}$ is $m = 6.$ Also, the smallest number $m$ such that $(n+6)|\underbrace{999\cdots999}_{m \text{ nines}}$ is $m = 4$ by similar reasoning.
Proceeding, we can factorize $9999 = 99 \times 101,$ after which we see that $n+6$ must contain a prime factor of $101$ as it cannot divide $99$ but must divide $9999.$ However, $101$ is prime, so $101|(n+6)$ ! Looking at the answer choices, all of the intervals are less than $1000,$ so we know that (the minimum value of) $n+6$ is thereby either $101, 101 \times 3,$ or $101 \times 9.$ Testing, we see that $n+6 = 303$ gives $n = 297 = 3^3 \times 11,$ which in fact is a divisor of $999 \times 1001$ while not being a divisor of $999.$ Therefore, the answer is $\boxed{201,400}.$


Answer: 991.0

For a finite sequence $A=(a_1,a_2,...,a_n)$ of numbers, the Cesáro sum of A is defined to be $\frac{S_1+\cdots+S_n}{n}$ , where $S_k=a_1+\cdots+a_k$ and $1\leq k\leq n$ . If the Cesáro sum of
the 99-term sequence $(a_1,...,a_{99})$ is 1000,  what is the Cesáro sum of the 100-term sequence $(1,a_1,...,a_{99})$
$\text{(A) } 991\quad \text{(B) } 999\quad \text{(C) } 1000\quad \text{(D) } 1001\quad \text{(E) } 1009$

Solution: Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is \[S_1 + S_2 + S_3 + ... + S_n\] \[= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)\] \[= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.\] If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is $1000 * 99 = 99,000,$ than the Cesáro total of the second sequence is $n \cdot a_1 + 99,000 = 100 \cdot 1 + 99,000 = 99,100.$ Thus the Cesáro sum of the second sequence is $\frac{99,100}{100} = \boxed{991,}\, .$


Answer: 20.5

For a given arithmetic series the sum of the first $50$ terms is $200$ , and the sum of the next $50$ terms is $2700$ .
The first term in the series is:
$\textbf{(A)}\ -1221 \qquad \textbf{(B)}\ -21.5 \qquad \textbf{(C)}\ -20.5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 3.5$

Solution:
Let the first term of the arithmetic sequence be $a$ and the common difference be $d$
The $50^{\text{th}}$ term of the sequence is $a+49d$ , so the sum of the first $50$ terms is $\frac{50(a + a + 49d)}{2}$
The $51^{\text{th}}$ term of the sequence is $a+50d$ and the $100^{\text{th}}$ term of the sequence is $a+99d$ , so the sum of the next $50$ terms is $\frac{50(a+50d+a+99d)}{2}$
Substituting in values results in this system of equations \[2a+49d=8\] \[2a+149d=108\] Solving for $a$ yields $a = \frac{-41}{2}$ .  The first term is $-20.5$ , which is answer choice $\boxed{20.5}$


Answer: 69.0

For a positive integer $n$ , let $d_n$ be the units digit of $1 + 2 + \dots + n$ . Find the remainder when \[\sum_{n=1}^{2017} d_n\] is divided by $1000$

Solution:
We see that $d_n$ appears in cycles of $20$ and the cycles are \[1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\] adding a total of $70$ each cycle.
Since $\left\lfloor\frac{2017}{20}\right\rfloor=100$ , we know that by $2017$ , there have been $100$ cycles and $7000$ has been added. This can be discarded as we're just looking for the last three digits.
Adding up the first $17$ of the cycle of $20$ , we can see that the answer is $\boxed{069}$


Answer: 10.0

For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$

Solution: We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$ , and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$ . Therefore, the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$ , and so $12 \cdot 10! = 12 \cdot N!$ . Cancelling the $12$ s, it is clear that $N=\boxed{10}$

Solution: Since $5! = 120$ , we obtain $120\cdot 9!=12\cdot N!$ , which becomes $12\cdot 10\cdot 9!=12\cdot N!$ and thus $12 \cdot 10!=12\cdot N!$ . We therefore deduce $N=\boxed{10}$

Solution:
We notice that $5! \cdot 9! = (5!)^2 \cdot (9 \cdot 8 \cdot 7 \cdot 6).$
We know that $5! = 120,$ so we have $120(5! \cdot 9 \cdot 8 \cdot 7 \cdot 6) = 12 \cdot N!$
Isolating $N!$ we have $N! = 10 \cdot 5! \cdot 9 \cdot 8 \cdot 7 \cdot 6 \Rightarrow N! = 10! \Rightarrow N = \boxed{10}.$


Answer: 18.0

For a positive integer $n$ and nonzero digits $a$ $b$ , and $c$ , let $A_n$ be the $n$ -digit integer each of whose digits is equal to $a$ ; let $B_n$ be the $n$ -digit integer each of whose digits is equal to $b$ , and let $C_n$ be the $2n$ -digit (not $n$ -digit) integer each of whose digits is equal to $c$ . What is the greatest possible value of $a + b + c$ for which there are at least two values of $n$ such that $C_n - B_n = A_n^2$
$\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20$

Solution:
By geometric series, we have \begin{alignat*}{8} A_n&=a\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=a\left(1+10+10^2+\cdots+10^{n-1}\right)&&=a\cdot\frac{10^n-1}{9}, \\ B_n&=b\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=b\left(1+10+10^2+\cdots+10^{n-1}\right)&&=b\cdot\frac{10^n-1}{9}, \\ C_n&=c\bigl(\phantom{ }\underbrace{111\cdots1}_{2n\text{ digits}}\phantom{ }\bigr)&&=c\left(1+10+10^2+\cdots+10^{2n-1}\right)&&=c\cdot\frac{10^{2n}-1}{9}. \end{alignat*} By substitution, we rewrite the given equation $C_n - B_n = A_n^2$ as \[c\cdot\frac{10^{2n}-1}{9} - b\cdot\frac{10^n-1}{9} = a^2\cdot\left(\frac{10^n-1}{9}\right)^2.\] Since $n > 0,$ it follows that $10^n > 1.$ We divide both sides by $\frac{10^n-1}{9}$ and then rearrange: \begin{align*} c\left(10^n+1\right) - b &= a^2\cdot\frac{10^n-1}{9} \\ 9c\left(10^n+1\right) - 9b &= a^2\left(10^n-1\right) \\ \left(9c-a^2\right)10^n &= 9b-9c-a^2. &&(\bigstar) \end{align*} Let $y=10^n.$ Note that $(\bigstar)$ is a linear equation with $y,$ and $y$ is a one-to-one function of $n.$ Since $(\bigstar)$ has at least two solutions of $n,$ it has at least two solutions of $y.$ We conclude that $(\bigstar)$ must be an identity, so we get the following system of equations: \begin{align*} 9c-a^2&=0, \\ 9b-9c-a^2&=0. \end{align*} The first equation implies that $c=\frac{a^2}{9}.$ Substituting this into the second equation gives $b=\frac{2a^2}{9}.$
To maximize $a + b + c = a + \frac{a^2}{3},$ we need to maximize $a.$ Clearly, $a$ must be divisible by $3.$ The possibilities for $(a,b,c)$ are $(9,18,9),(6,8,4),$ or $(3,2,1),$ but $(9,18,9)$ is invalid. Therefore, the greatest possible value of $a + b + c$ is $6+8+4=\boxed{18}.$

Solution:
Immediately start trying $n = 1$ and $n = 2$ . These give the system of equations $11c - b = a^2$ and $1111c - 11b = (11a)^2$ (which simplifies to $101c - b = 11a^2$ ). These imply that $a^2 = 9c$ , so the possible $(a, c)$ pairs are $(9, 9)$ $(6, 4)$ , and $(3, 1)$ . The first puts $b$ out of range but the second makes $b = 8$ . We now know the answer is at least $6 + 8 + 4 = 18$
We now only need to know whether $a + b + c = 20$ might work for any larger $n$ . We will always get equations like $100001c - b = 11111a^2$ where the $c$ coefficient is very close to being nine times the $a$ coefficient. Since the $b$ term will be quite insignificant, we know that once again $a^2$ must equal $9c$ , and thus $a = 9, c = 9$ is our only hope to reach $20$ . Substituting and dividing through by $9$ , we will have something like $100001 - \frac{b}{9} = 99999$ . No matter what $n$ really was, $b$ is out of range (and certainly isn't $2$ as we would have needed).
The answer then is $\boxed{18}$

Solution: The given equation can be written as \[c \cdot (\phantom{ } \overbrace{1111 \ldots 1111}^{2n\text{ digits}}\phantom{ }) - b \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ })^2.\] Divide by $\overbrace{11 \ldots 11}^{n\text{ digits}}$ on both sides: \[c \cdot (\phantom{ } \overbrace{1000 \ldots 0001}^{n+1\text{ digits}}\phantom{ }) - b = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }).\] Next, split the first term to make it easier to deal with: \begin{align*} 2c + c \cdot (\phantom{ }\overbrace{99 \ldots 99}^{n\text{ digits}}\phantom{ }) - b &= a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) \\ 2c - b &= (a^2 - 9c) \cdot (\phantom{ }\overbrace{11 \ldots 11}^{n\text{ digits}}\phantom{ }). \end{align*} Because $2c - b$ and $a^2 - 9c$ are constants and because there must be at least two distinct values of $n$ that satisfy, $2c - b = a^2 - 9c = 0.$ Thus, we have \begin{align*} 2c&=b, \\ a^2&=9c. \end{align*} Knowing that $a,b,$ and $c$ are single digit positive integers and that $9c$ must be a perfect square, the values of $(a,b,c)$ that satisfy both equations are $(3,2,1)$ and $(6,8,4).$ Finally, $6 + 8 + 4 = \boxed{18}.$

Solution: By PaperMath’s sum , the answer is $6+8+4=\boxed{18}$


Answer: 18.0

Solution:
By geometric series, we have \begin{alignat*}{8} A_n&=a\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=a\left(1+10+10^2+\cdots+10^{n-1}\right)&&=a\cdot\frac{10^n-1}{9}, \\ B_n&=b\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=b\left(1+10+10^2+\cdots+10^{n-1}\right)&&=b\cdot\frac{10^n-1}{9}, \\ C_n&=c\bigl(\phantom{ }\underbrace{111\cdots1}_{2n\text{ digits}}\phantom{ }\bigr)&&=c\left(1+10+10^2+\cdots+10^{2n-1}\right)&&=c\cdot\frac{10^{2n}-1}{9}. \end{alignat*} By substitution, we rewrite the given equation $C_n - B_n = A_n^2$ as \[c\cdot\frac{10^{2n}-1}{9} - b\cdot\frac{10^n-1}{9} = a^2\cdot\left(\frac{10^n-1}{9}\right)^2.\] Since $n > 0,$ it follows that $10^n > 1.$ We divide both sides by $\frac{10^n-1}{9}$ and then rearrange: \begin{align*} c\left(10^n+1\right) - b &= a^2\cdot\frac{10^n-1}{9} \\ 9c\left(10^n+1\right) - 9b &= a^2\left(10^n-1\right) \\ \left(9c-a^2\right)10^n &= 9b-9c-a^2. &&(\bigstar) \end{align*} Let $y=10^n.$ Note that $(\bigstar)$ is a linear equation with $y,$ and $y$ is a one-to-one function of $n.$ Since $(\bigstar)$ has at least two solutions of $n,$ it has at least two solutions of $y.$ We conclude that $(\bigstar)$ must be an identity, so we get the following system of equations: \begin{align*} 9c-a^2&=0, \\ 9b-9c-a^2&=0. \end{align*} The first equation implies that $c=\frac{a^2}{9}.$ Substituting this into the second equation gives $b=\frac{2a^2}{9}.$
To maximize $a + b + c = a + \frac{a^2}{3},$ we need to maximize $a.$ Clearly, $a$ must be divisible by $3.$ The possibilities for $(a,b,c)$ are $(9,18,9),(6,8,4),$ or $(3,2,1),$ but $(9,18,9)$ is invalid. Therefore, the greatest possible value of $a + b + c$ is $6+8+4=\boxed{18}.$

Solution:
Immediately start trying $n = 1$ and $n = 2$ . These give the system of equations $11c - b = a^2$ and $1111c - 11b = (11a)^2$ (which simplifies to $101c - b = 11a^2$ ). These imply that $a^2 = 9c$ , so the possible $(a, c)$ pairs are $(9, 9)$ $(6, 4)$ , and $(3, 1)$ . The first puts $b$ out of range but the second makes $b = 8$ . We now know the answer is at least $6 + 8 + 4 = 18$
We now only need to know whether $a + b + c = 20$ might work for any larger $n$ . We will always get equations like $100001c - b = 11111a^2$ where the $c$ coefficient is very close to being nine times the $a$ coefficient. Since the $b$ term will be quite insignificant, we know that once again $a^2$ must equal $9c$ , and thus $a = 9, c = 9$ is our only hope to reach $20$ . Substituting and dividing through by $9$ , we will have something like $100001 - \frac{b}{9} = 99999$ . No matter what $n$ really was, $b$ is out of range (and certainly isn't $2$ as we would have needed).
The answer then is $\boxed{18}$

Solution: The given equation can be written as \[c \cdot (\phantom{ } \overbrace{1111 \ldots 1111}^{2n\text{ digits}}\phantom{ }) - b \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ })^2.\] Divide by $\overbrace{11 \ldots 11}^{n\text{ digits}}$ on both sides: \[c \cdot (\phantom{ } \overbrace{1000 \ldots 0001}^{n+1\text{ digits}}\phantom{ }) - b = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }).\] Next, split the first term to make it easier to deal with: \begin{align*} 2c + c \cdot (\phantom{ }\overbrace{99 \ldots 99}^{n\text{ digits}}\phantom{ }) - b &= a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) \\ 2c - b &= (a^2 - 9c) \cdot (\phantom{ }\overbrace{11 \ldots 11}^{n\text{ digits}}\phantom{ }). \end{align*} Because $2c - b$ and $a^2 - 9c$ are constants and because there must be at least two distinct values of $n$ that satisfy, $2c - b = a^2 - 9c = 0.$ Thus, we have \begin{align*} 2c&=b, \\ a^2&=9c. \end{align*} Knowing that $a,b,$ and $c$ are single digit positive integers and that $9c$ must be a perfect square, the values of $(a,b,c)$ that satisfy both equations are $(3,2,1)$ and $(6,8,4).$ Finally, $6 + 8 + 4 = \boxed{18}.$

Solution: By PaperMath’s sum , the answer is $6+8+4=\boxed{18}$


Answer: 132.0

For a real number $a$ , let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$ . Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$ . The region $\mathcal{R}$ is completely contained in a disk of radius $r$ (a disk is the union of a circle and its interior). The minimum value of $r$ can be written as $\frac {\sqrt {m}}{n}$ , where $m$ and $n$ are integers and $m$ is not divisible by the square of any prime. Find $m + n$

Solution: The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$ , namely $(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).$ Since the points themselves are symmetric about $(0,0)$ , the boxes are symmetric about $\left(\frac12,\frac12\right)$ . The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box that lays on an axis, for instance $(6,1)$ , is $\sqrt {\frac {11}2^2 + \frac12^2} = \sqrt {\frac {122}4}.$ The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box in the middle of a quadrant, for instance $(5,4)$ , is $\sqrt {\frac92^2 + \frac72^2} = \sqrt {\frac {130}4}.$ The latter is the larger, and is $\frac {\sqrt {130}}2$ , giving an answer of $130 + 2 = \boxed{132}$


Answer: 10.0

For a real number $x$ , define $\heartsuit(x)$ to be the average of $x$ and $x^2$ . What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 20$

Solution:
The average of two numbers, $a$ and $b$ , is defined as $\frac{a+b}{2}$ . Thus the average of $x$ and $x^2$ would be $\frac{x(x+1)}{2}$ . With that said,  we need to find the sum when we plug, $1$ $2$ and $3$ into that equation. So:
\[\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{10}\]


Answer: 10.0

For a real number $x$ let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$ , and define $\{x\} = x - \lfloor x \rfloor$ to be the fractional part of $x$ . For example, $\{3\} = 0$ and $\{4.56\} = 0.56$ . Define $f(x)=x\{x\}$ , and let $N$ be the number of real-valued solutions to the equation $f(f(f(x)))=17$ for $0\leq x\leq 2020$ . Find the remainder when $N$ is divided by $1000$

Solution:
To solve $f(f(f(x)))=17$ , we need to solve $f(x) = y$ where $f(f(y))=17$ , and to solve that we need to solve $f(y) = z$ where $f(z) = 17$
It is clear to see for some integer $a \geq 17$ there is exactly one value of $z$ in the interval $[a, a+1)$ where $f(z) = 17$ . To understand this, imagine the graph of $f(z)$ on the interval $[a, a+1)$ The graph starts at $0$ , is continuous and increasing, and approaches $a+1$ . So as long as $a+1 > 17$ , there will be a solution for $z$ in the interval.
Using this logic, we can find the number of solutions to $f(x) = y$ . For every interval $[a, a+1)$ where $a \geq \left \lfloor{y}\right \rfloor$ there will be one solution for $x$ in that interval. However, the question states $0 \leq x \leq 2020$ , but because $x=2020$ doesn't work we can change it to $0 \leq x < 2020$ . Therefore, $\left \lfloor{y}\right \rfloor  \leq a \leq 2019$ , and there are $2020 - \left \lfloor{y}\right \rfloor$ solutions to $f(x) = y$
We can solve $f(y) = z$ similarly. $0 \leq y < 2020$ to satisfy the bounds of $x$ , so there are $2020 - \left \lfloor{z}\right \rfloor$ solutions to $f(y) = z$ , and $0 \leq z < 2020$ to satisfy the bounds of $y$
Going back to $f(z) = 17$ , there is a single solution for z in the interval $[a, a+1)$ , where $17 \leq a \leq 2019$ . (We now have an upper bound for $a$ because we know $z < 2020$ .) There are $2003$ solutions for $z$ , and the floors of these solutions create the sequence $17, 18, 19, ..., 2018, 2019$
Lets first look at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 17$ . Then $f(y) = z$ would have $2003$ solutions, and the floors of these solutions would also create the sequence $17, 18, 19, ..., 2018, 2019$
If we used the solution of $y$ where $\left \lfloor{y}\right \rfloor = 17$ , there would be $2003$ solutions for $f(x) = y$ . If we used the solution of $y$ where $\left \lfloor{y}\right \rfloor = 18$ , there would be $2002$ solutions for $x$ , and so on. So for the solution of $z$ where $\left \lfloor{z}\right \rfloor = 17$ , there will be $2003 + 2002 + 2001 + ... + 2 + 1 = \binom{2004}{2}$ solutions for $x$
If we now look at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 18$ , there would be $\binom{2003}{2}$ solutions for $x$ . If we looked at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 19$ , there would be $\binom{2002}{2}$ solutions for $x$ , and so on.
The total number of solutions to $x$ is $\binom{2004}{2} + \binom{2003}{2} + \binom{2002}{2} + ... + \binom{3}{2} + \binom{2}{2}$ . Using the hockey stick theorem, we see this equals $\binom{2005}{3}$ , and when we take the remainder of that number when divided by $1000$ , we get the answer, $\boxed{10}$


Answer: 4.0

For a sale, a store owner reduces the price of a $$10$ scarf by $20\%$ .  Later the price is lowered again, this time by one-half the reduced price.  The price is now
$\text{(A)}\ 2.00\text{ dollars} \qquad \text{(B)}\ 3.75\text{ dollars} \qquad \text{(C)}\ 4.00\text{ dollars} \qquad \text{(D)}\ 4.90\text{ dollars} \qquad \text{(E)}\ 6.40\text{ dollars}$

Solution:
$100\%-20\%=80\%$
$10\times80\%=10\times0.8$
$10\times0.8=8$
$\frac{8}{2}=4=\boxed{4.00}$

Solution:
The first discount has percentage 20, which is then discounted again for half of the already discounted price.
$100-20=80$
$\frac{80}{2}=40$
$40\%\times10=10\times0.4=4=\boxed{4.00}$


Answer: 48.0

For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54$

Solution:
Let $x$ be the number of acorns that both animals had.
So by the info in the problem:
$\frac{x}{3}=\left( \frac{x}{4} \right)+4$
Subtracting $\frac{x}{4}$ from both sides leaves
$\frac{x}{12}=4$
$\boxed{48}$


Answer: 48.0

For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54$

Solution: If $x$ is the number of holes that the chipmunk dug, then $3x=4(x-4)$ , so $3x=4x-16$ , and $x=16$ . The number of acorns hidden by the chipmunk is equal to $3x = \boxed{48}$

Solution: Trying answer choices, we see that $\boxed{48}$ works. ~Extremelysupercooldude


Answer: 19.0

For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$
$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$

Solution:
It is possible to obtain $0$ $1$ $3$ $4$ $5$ , and $6$ points of intersection, as demonstrated in the following figures:
It is clear that the maximum number of possible intersections is ${4 \choose 2} = 6$ , since each pair of lines can intersect at most once.  We now prove that it is impossible to obtain two intersections.
We proceed by contradiction.  Assume a configuration of four lines exists such that there exist only two intersection points.  Let these intersection points be $A$ and $B$ .  Consider two cases:
Case 1 :  No line passes through both $A$ and $B$
Then, since an intersection point is obtained by an intersection between at least two lines, two lines pass through each of $A$ and $B$ .  Then, since there can be no additional intersections, the 2 lines that pass through $A$ cant intersect the 2 lines that pass through $B$ , and so 2 lines passing through $A$ must be parallel to 2 lines passing through $B$ . Then the two lines passing through $B$ are parallel to each other by transitivity of parallelism, so they coincide, contradiction.
Case 2 :  There is a line passing through $A$ and $B$
Then there must be a line $l_a$ passing through $A$ , and a line $l_b$ passing through $B$ .  These lines must be parallel.  The fourth line $l$ must pass through either $A$ or $B$ .  Without loss of generality, suppose $l$ passes through $A$ .  Then since $l$ and $l_a$ cannot coincide, they cannot be parallel.  Then $l$ and $l_b$ cannot be parallel either, so they intersect, contradiction.
All possibilities have been exhausted, and thus we can conclude that two intersections is impossible.  Our answer is given by the sum $0+1+3+4+5+6=\boxed{19}$


Answer: 19.0

Solution:
It is possible to obtain $0$ $1$ $3$ $4$ $5$ , and $6$ points of intersection, as demonstrated in the following figures:
It is clear that the maximum number of possible intersections is ${4 \choose 2} = 6$ , since each pair of lines can intersect at most once.  We now prove that it is impossible to obtain two intersections.
We proceed by contradiction.  Assume a configuration of four lines exists such that there exist only two intersection points.  Let these intersection points be $A$ and $B$ .  Consider two cases:
Case 1 :  No line passes through both $A$ and $B$
Then, since an intersection point is obtained by an intersection between at least two lines, two lines pass through each of $A$ and $B$ .  Then, since there can be no additional intersections, the 2 lines that pass through $A$ cant intersect the 2 lines that pass through $B$ , and so 2 lines passing through $A$ must be parallel to 2 lines passing through $B$ . Then the two lines passing through $B$ are parallel to each other by transitivity of parallelism, so they coincide, contradiction.
Case 2 :  There is a line passing through $A$ and $B$
Then there must be a line $l_a$ passing through $A$ , and a line $l_b$ passing through $B$ .  These lines must be parallel.  The fourth line $l$ must pass through either $A$ or $B$ .  Without loss of generality, suppose $l$ passes through $A$ .  Then since $l$ and $l_a$ cannot coincide, they cannot be parallel.  Then $l$ and $l_b$ cannot be parallel either, so they intersect, contradiction.
All possibilities have been exhausted, and thus we can conclude that two intersections is impossible.  Our answer is given by the sum $0+1+3+4+5+6=\boxed{19}$


Answer: 1.0

For all integers $n$ greater than $1$ , define $a_n = \frac{1}{\log_n 2002}$ . Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$ . Then $b- c$ equals
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1  \qquad\mathrm{(C)}\ \frac{1}{2002} \qquad\mathrm{(D)}\ \frac{1}{1001} \qquad\mathrm{(E)}\ \frac 12$

Solution:
Note that $\frac{1}{\log_a b}=\log_b a$ . Thus $a_n=\log_{2002} n$ . Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following:
\[\log_{2002}\left(\frac{2*3*4*5}{10*11*12*13*14}=\frac{1}{11*13*14}=\frac{1}{2002}\right)=\boxed{1}\]


Answer: 1.0

Solution:
Note that $a_2 = \frac{1}{\log_2 2002}$ $1$ is also equal to $\log_2 2$ . So $a_2 = \frac{\log_2 2}{\log_2 2002}$ . By the change of bases formula, $a_2 = \log_{2002} 2$ . Following the same reasoning, $a_3 = \log_{2002} 3$ $a_4 = \log_{2002} 4$ and so on.
\[b = \log_{2002} 2 + \log_{2002} 3 + .....+ \log_{2002} 5 = \log_{2002} 5! = \log_{2002} 120\] Now solving for $c$ , we see that it equals $\log_{2002} (10\cdot 11 \cdot 12 \cdot 13 \cdot 14)$ \[b-c = \log_{2002} 120 - \log_{2002} 240240 \rightarrow \log_{2002} \frac{1}{2002} = \boxed{1}\]


Answer: 1.0

For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals
$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$

Solution: Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into \[(x+1)(y-1) = -1\] Plugging in $-1$ and $1$ as the $x$ and $y$ sides respectively, we get $x = -2$ and $y = 2$ . Plugging this in to $\frac{1}{x}-\frac{1}{y}$ gives us $\boxed{1}$ as our final answer.

Solution: Notice that we can do $\frac{x-y}{xy} = \frac{xy}{xy}$ . We are left with $\frac{1}{y} - \frac{1}{x} = 1$ . Multiply by $-1$ to achieve $\frac{1}{x} - \frac{1}{y} = \boxed{1}$


Answer: 2.0

For all positive integers $n$ , let $f(n)=\log_{2002} n^2$ . Let $N=f(11)+f(13)+f(14)$ . Which of the following relations is true?
$\text{(A) }N<1 \qquad \text{(B) }N=1 \qquad \text{(C) }1<N<2 \qquad \text{(D) }N=2 \qquad \text{(E) }N>2$

Solution:
First, note that $2002 = 11 \cdot 13 \cdot 14$
Using the fact that for any base we have $\log a + \log b = \log ab$ , we get that $N = \log_{2002} (11^2 \cdot 13^2 \cdot 14^2) = \log_{2002} 2002^2 = \boxed{2}$


Answer: 448.0

For all positive integers $n$ less than $2002$ , let
\begin{eqnarray*} a_n =\left\{ \begin{array}{lr} 11, & \text{if\ }n\ \text{is\ divisible\ by\ }13\ \text{and\ }14;\\ 13, & \text{if\ }n\ \text{is\ divisible\ by\ }14\ \text{and\ }11;\\ 14, & \text{if\ }n\ \text{is\ divisible\ by\ }11\ \text{and\ }13;\\ 0, & \text{otherwise}. \end{array} \right. \end{eqnarray*}
Calculate $\sum_{n=1}^{2001} a_n$
$\mathrm{(A)}\ 448 \qquad\mathrm{(B)}\ 486 \qquad\mathrm{(C)}\ 1560 \qquad\mathrm{(D)}\ 2001 \qquad\mathrm{(E)}\ 2002$

Solution:
Find the LCMs of the groups of the numbers.
Notice that the groups are relatively prime.
So $a_n=$
11 if $n$ is a multiple of 182.
13 if $n$ is a multiple of 154.
14 if $n$ is a multiple of 143.
When do we see ambiguities (for example: $n$ is a multiple of 11, 13, and 14)? This is only done when $n$ is a multiple of $\operatorname{lcm}(11,13,14)=2002$ . However, since $n<2002$ , this can never happen.
So we have 10 multiples of 182 we have to count (1 to 10 $*182$ ), and similarly, 12 multiples of 154, and 13 multiples of 143. The sum is $10*11+12*13+13*14=448$ . Select $\boxed{448}$


Answer: 511.0

For all positive integers $x$ , let \[f(x)=\begin{cases}1 & \text{if }x = 1\\ \frac x{10} & \text{if }x\text{ is divisible by 10}\\ x+1 & \text{otherwise}\end{cases}\] and define a sequence as follows: $x_1=x$ and $x_{n+1}=f(x_n)$ for all positive integers $n$ . Let $d(x)$ be the smallest $n$ such that $x_n=1$ . (For example, $d(100)=3$ and $d(87)=7$ .) Let $m$ be the number of positive integers $x$ such that $d(x)=20$ . Find the sum of the distinct prime factors of $m$

Solution:
We backcount the number of ways. Namely, we start at $x_{20} = 1$ , which can only be reached if $x_{19} = 10$ , and then we perform $18$ operations that either consist of $A: (-1)$ or $B: (\times 10)$ . We represent these operations in a string format, starting with the operation that sends $f(x_{18}) = x_{19}$ and so forth downwards. There are $2^9$ ways to pick the first $9$ operations; however, not all $9$ of them may be $A: (-1)$ otherwise we return back to $x_{10} = 1$ , contradicting our assumption that $20$ was the smallest value of $n$ . Using complementary counting , we see that there are only $2^9 - 1$ ways.
Since we performed the operation $B: (\times 10)$ at least once in the first $9$ operations, it follows that $x_{10} \ge 20$ , so that we no longer have to worry about reaching $1$ again.
However, we must also account for a sequence of $10$ or more $A: (-1)$ s in a row, because that implies that at least one of those numbers was divisible by $10$ , yet the $\frac{x}{10}$ was never used, contradiction. We must use complementary counting again by determining the number of strings of $A,B$ s of length $18$ such that there are $10$ $A$ s in a row. The first ten are not included since we already accounted for that scenario above, so our string of $10$ $A$ s must be preceded by a $B$ . There are no other restrictions on the remaining seven characters. Letting $\square$ to denote either of the functions, and $^{[k]}$ to indicate that the character appears $k$ times in a row, our bad strings can take the forms: \begin{align*} &\underbrace{BA^{[10]}}\square \square \square \square \square \square \square\\ &\square \underbrace{BA^{[10]}}\square \square \square \square \square \square \\ &\qquad \quad \cdots \quad \cdots \\ &\square \square \square \square \square \square \underbrace{BA^{[10]}} \square \\ &\square \square \square \square \square \square \square \underbrace{BA^{[10]}} \\ \end{align*}
There are $2^7$ ways to select the operations for the $7$ $\square$ s, and $8$ places to place our $BA^{[10]}$ block. Thus, our answer is $2^9(2^9-1)-8\cdot 2^7 = 2^{18} - 2^9 - 8 \cdot 2^7 = 2^9 \times 509$ , and the answer is $\boxed{511}$


Answer: 454.0

For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]

Solution: By PIE, $|A|+|B|-|A \cap B| = |A \cup B|$ . Substituting into the equation and factoring, we get that $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$ , so therefore $A \subseteq B$ or $B \subseteq A$ . WLOG $A\subseteq B$ , then for each element there are $3$ possibilities, either it is in both $A$ and $B$ , it is in $B$ but not $A$ , or it is in neither $A$ nor $B$ . This gives us $3^{5}$ possibilities, and we multiply by $2$ since it could have also been the other way around. Now we need to subtract the overlaps where $A=B$ , and this case has $2^{5}=32$ ways that could happen. It is $32$ because each number could be in the subset or it could not be in the subset. So the final answer is $2\cdot 3^5 - 2^5 = \boxed{454}$

Solution:
We denote $\Omega = \left\{ 1 , 2 , 3 , 4 , 5 \right\}$ .
We denote $X = A \cap B$ $Y = A \backslash \left( A \cap B \right)$ $Z = B \backslash \left( A \cap B \right)$ $W = \Omega \backslash \left( A \cup B \right)$
Therefore, $X \cup Y \cup Z \cup W = \Omega$ and the intersection of any two out of sets $X$ $Y$ $Z$ $W$ is an empty set.
Therefore, $\left( X , Y , Z , W \right)$ is a partition of $\Omega$
Following from our definition of $X$ $Y$ $Z$ , we have $A \cup B = X \cup Y \cup Z$
Therefore, the equation
\[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]
can be equivalently written as
\[\left( | X | + | Y | \right) \left( | X | + | Z | \right) = | X | \left( | X | + | Y | + | Z | \right) .\]
This equality can be simplified as
\[| Y | \cdot | Z | = 0 .\]
Therefore, we have the following three cases: (1) $|Y| = 0$ and $|Z| \neq 0$ , (2) $|Z| = 0$ and $|Y| \neq 0$ , (3) $|Y| = |Z| = 0$ .
Next, we analyze each of these cases, separately.
Case 1: $|Y| = 0$ and $|Z| \neq 0$
In this case, to count the number of solutions, we do the complementary counting.
First, we count the number of solutions that satisfy $|Y| = 0$
Hence, each number in $\Omega$ falls into exactly one out of these three sets: $X$ $Z$ $W$ .
Following from the rule of product, the number of solutions is $3^5$
Second, we count the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$
Hence, each number in $\Omega$ falls into exactly one out of these two sets: $X$ $W$ .
Following from the rule of product, the number of solutions is $2^5$
Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy $|Y| = 0$ minus the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$ , i.e., $3^5 - 2^5$
Case 2: $|Z| = 0$ and $|Y| \neq 0$
This case is symmetric to Case 1. Therefore, the number of solutions in this case is the same as the number of solutions in Case 1, i.e., $3^5 - 2^5$
Case 3: $|Y| = 0$ and $|Z| = 0$
Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is $2^5$
By putting all cases together, following from the rule of sum, the total number of solutions is equal to
\begin{align*} \left( 3^5 - 2^5 \right) + \left( 3^5 - 2^5 \right) + 2^5 & = 2 \cdot 3^5 - 2^5 \\ & = \boxed{454}

Solution: The answer is \begin{align*} \sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] &= 2\sum_{k=0}^{5}\binom{5}{k}2^{5-k}-\sum_{k=0}^{5}\binom{5}{k} \\ &=2(2+1)^5-(1+1)^5 \\ &=2(243)-32 \\ &=\boxed{454} ~MRENTHUSIASM

Solution: The answer is \begin{align*} &\hspace{5.125mm}\sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] \\ &=\left[2\binom{5}{0}2^{5-0}-\binom{5}{0}\right] + \left[2\binom{5}{1}2^{5-1}-\binom{5}{1}\right] + \left[2\binom{5}{2}2^{5-2}-\binom{5}{2}\right] + \left[2\binom{5}{3}2^{5-3}-\binom{5}{3}\right] + \left[2\binom{5}{4}2^{5-4}-\binom{5}{4}\right] + \left[2\binom{5}{5}2^{5-5}-\binom{5}{5}\right] \\ &=\left[2\left(1\right)2^5-1\right] + \left[2\left(5\right)2^4-5\right] + \left[2\left(10\right)2^3-10\right] + \left[2\left(10\right)2^2-10\right] + \left[2\left(5\right)2^1-5\right] + \left[2\left(1\right)2^0-1\right] \\ &=63+155+150+70+15+1 \\ &=\boxed{454} ~MRENTHUSIASM

Solution:
Proceed with Solution 1 to get $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$ . WLOG, assume $|A| = |A \cap B|$ . Thus, $A \subseteq B$
Since $A \subseteq B$ , if $|B| = n$ , there are $2^n$ possible sets $A$ , and there are also ${5 \choose n}$ ways of choosing such $B$
Therefore, the number of possible pairs of sets $(A, B)$ is
\[\sum_{k=0}^{5} 2^n {5 \choose n}\]
We can compute this manually since it's only from $k=0$ to $5$ , and computing gives us $243$ . We can double this result for $B \subseteq A$ , and we get $2(243) = 486$
However, we have double counted the cases where $A$ and $B$ are the same sets. There are $32$ possible such cases, so we subtract $32$ from $486$ to get $\boxed{454}$


Answer: 245.0

For any finite set $X$ , let $| X |$ denote the number of elements in $X$ . Define \[S_n = \sum | A \cap B | ,\] where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3,  \cdots , n \right\}$ with $|A| = |B|$ .
For example, $S_2 = 4$ because the sum is taken over the pairs of subsets \[(A, B) \in \left\{ (\emptyset, \emptyset) , ( \{1\} , \{1\} ), ( \{1\} , \{2\} ) , ( \{2\} , \{1\} ) , ( \{2\} , \{2\} ) , ( \{1 , 2\} , \{1 , 2\} ) \right\} ,\] giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4$ .
Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by
1000.

Solution:
Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$ . The problem states that for $n=2, S_n$ is $4$ . Let's try it out for $n=3$
Let's perform casework on the number of elements in $A, B$
$\textbf{Case 1:} |A| = |B| = 1$
In this case, the only possible equivalencies will be if they are the exact same element, which happens $3$ times.
$\textbf{Case 2:} |A| = |B| = 2$
In this case, if they share both elements, which happens $3$ times, we will get $2$ for each time, and if they share only one element, which also happens $6$ times, we will get $1$ for each time, for a total of $12$ for this case.
$\textbf{Case 3:} |A| = |B| = 3$
In this case, the only possible scenario is that they both are the set $\{1,2,3\}$ , and we have $3$ for this case.
In total, $S_3 = 18$
Now notice, the number of intersections by each element $1 \ldots 3$ , or in general, $1 \ldots n$ is equal for each element because of symmetry - each element when $n=3$ adds $6$ to the answer. Notice that $6 = \binom{4}{2}$ - let's prove that $S_n = n \cdot \binom{2n-2}{n-1}$ (note that you can assume this and answer the problem if you're running short on time in the real test).
Let's analyze the element $k$ - to find a general solution, we must count the number of these subsets that $k$ appears in. For $k$ to be in both $A$ and $B$ , we need both sets to contain $k$ and another subset of $1$ through $n$ not including $k$ . ( $A = \{k\} \cup A'| A' \subset \{1,2,\ldots,n\} \land A' \not \subset \{k\}$ and $B = \{k\} \cup B'| B' \subset \{1,2,\ldots,n\} \land B' \not \subset \{k\}$
For any $0\leq l \leq n-1$ that is the size of both $A'$ and $B'$ , the number of ways to choose the subsets $A'$ and $B'$ is $\binom{n-1}{l}$ for both subsets, so the total number of ways to choose the subsets are $\binom{n-1}{l}^2$ .
Now we sum this over all possible $l$ 's to find the total number of ways to form sets $A$ and $B$ that contain $k$ . This is equal to $\sum_{l=0}^{n-1} \binom{n-1}{l}^2$ . This is a simplification of Vandermonde's identity, which states that $\sum_{k=0}^{r} \binom{m}{k} \cdot \binom{n}{r-k} = \binom{m+n}{r}$ . Here, $m$ $n$ and $r$ are all $n-1$ , so this sum is equal to $\binom{2n-2}{n-1}$ . Finally, since we are iterating over all $k$ 's for $n$ values of $k$ , we have $S_n = n \cdot \binom{2n-2}{n-1}$ , proving our claim.
We now plug in $S_n$ to the expression we want to find. This turns out to be $\frac{2022 \cdot \binom{4042}{2021}}{2021 \cdot \binom{4040}{2020}}$ . Expanding produces $\frac{2022 \cdot 4042!\cdot 2020! \cdot 2020!}{2021 \cdot 4040! \cdot 2021! \cdot 2021!}$
After cancellation, we have \[\frac{2022 \cdot 4042 \cdot 4041}{2021 \cdot 2021 \cdot 2021} \implies \frac{4044\cdot 4041}{2021 \cdot 2021}\]
$4044$ and $4041$ don't have any common factors with $2021$ , so we're done with the simplification. We want to find $4044 \cdot 4041 + 2021^2 \pmod{1000} \equiv 44 \cdot 41 + 21^2 \pmod{1000} \equiv 1804+441 \pmod{1000} \equiv 2245 \pmod{1000} \equiv \boxed{245}$

Solution: We take cases based on the number of values in each of the subsets in the pair. Suppose we have $k$ elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be $n \cdot \frac{k}{n} \cdot \frac{k}{n}$ by linearity of expectation because for each of the $n$ elements, there is a $\frac{k}{n}$ probability that the element will be chosen. To find the sum over all such values, we multiply this quantity by $\binom{n}{k}^2$ . Summing, we get \[\sum_{k=1}^{n} \frac{k^2}{n} \binom{n}{k}^2\] Notice that we can rewrite this as \[\sum_{k=1}^{n} \frac{1}{n} \left(\frac{k \cdot n!}{(k)!(n - k)!}\right)^2 = \sum_{k=1}^{n} \frac{1}{n} n^2 \left(\frac{(n-1)!}{(k - 1)!(n - k)!}\right)^2 = n \sum_{k=1}^{n} \binom{n - 1}{k - 1}^2 = n \sum_{k=1}^{n} \binom{n - 1}{k - 1}\binom{n - 1}{n - k}\] We can simplify this using Vandermonde's identity to get $n \binom{2n - 2}{n - 1}$ . Evaluating this for $2022$ and $2021$ gives \[\frac{2022\binom{4042}{2021}}{2021\binom{4040}{2020}} = \frac{2022 \cdot 4042 \cdot 4041}{2021^3} = \frac{2022 \cdot 2 \cdot 4041}{2021^2}\] Evaluating the numerators and denominators mod $1000$ gives $804 + 441 = 1\boxed{245}$

Solution:
For each element $i$ , denote $x_i = \left( x_{i, A}, x_{i, B} \right) \in \left\{ 0 , 1 \right\}^2$ , where $x_{i, A} = \Bbb I \left\{ i \in A \right\}$ (resp. $x_{i, B} = \Bbb I \left\{ i \in B \right\}$ ).
Denote $\Omega = \left\{ (x_1, \cdots , x_n): \sum_{i = 1}^n x_{i, A} = \sum_{i = 1}^n x_{i, B} \right\}$
Denote $\Omega_{-j} = \left\{ (x_1, \cdots , x_{j-1} , x_{j+1} , \cdots , x_n): \sum_{i \neq j} x_{i, A} = \sum_{i \neq j} x_{i, B} \right\}$
Hence, \begin{align*} S_n & = \sum_{(x_1, \cdots , x_n) \in \Omega} \sum_{i = 1}^n \Bbb I \left\{ x_{i, A} = x_{i, B} = 1 \right\} \\ & = \sum_{i = 1}^n \sum_{(x_1, \cdots , x_n) \in \Omega} \Bbb I \left\{ x_{i, A} = x_{i, B} = 1 \right\} \\ & = \sum_{i = 1}^n \sum_{(x_1, \cdots , x_{i-1} , x_{i+1} , \cdots , x_n) \in \Omega_{-i}} 1  \\ & = \sum_{i = 1}^n \sum_{j=0}^{n-1} \left( \binom{n-1}{j} \right)^2 \\ & = n \sum_{j=0}^{n-1} \left( \binom{n-1}{j} \right)^2 \\ & = n \sum_{j=0}^{n-1} \binom{n-1}{j} \binom{n-1}{n-1-j} \\ & = n \binom{2n-2}{n-1} . \end{align*}
Therefore, \begin{align*} \frac{S_{2022}}{S_{2021}} & = \frac{2022 \binom{4042}{2021}}{2021 \binom{4040}{2020}} \\ & = \frac{4044 \cdot 4041}{2021^2} . \end{align*}
This is in the lowest term.
Therefore, modulo 1000, \begin{align*} p + q  & \equiv 4044 \cdot 4041 + 2021^2 \\ & \equiv 44 \cdot 41 + 21^2 \\ & \equiv \boxed{245}


Answer: 149.0

For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the smallest positive integer $t$ such that $x_t=2090.$

Solution:
Note that for any $x_i$ , for any prime $p$ $p^2 \nmid x_i$ .  This provides motivation to translate $x_i$ into a binary sequence $y_i$
Let the prime factorization of $x_i$ be written as $p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots$ , where $p_i$ is the $i$ th prime number. Then, for every $p_{a_k}$ in the prime factorization of $x_i$ , place a $1$ in the $a_k$ th digit of $y_i$ . This will result in the conversion $x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \cdots$
Multiplication for the sequence $x_i$ will translate to addition for the sequence $y_i$ . Thus, we see that $x_{n+1}X(x_n) = x_np(x_n)$ translates into $y_{n+1} = y_n+1$ . Since $x_0=1$ , and $y_0=0$ $x_i$ corresponds to $y_i$ , which is $i$ in binary. Since $x_{10010101_2} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090$ $t = 10010101_2$ $\boxed{149}$

Solution:
We go through the terms and look for a pattern. We find that
$x_0 = 1$ $x_8 = 7$
$x_1 = 2$ $x_9 = 14$
$x_2 = 3$ $x_{10} = 21$
$x_3 = 6$ $x_{11} = 42$
$x_4 = 5$ $x_{12} = 35$
$x_5 = 10$ $x_{13} = 70$
$x_6 = 15$ $x_{14} = 105$
$x_7 = 30$ $x_{15} = 210$
Commit to the bash. Eventually, you will receive that $x_{149} = 2090$ , so $\boxed{149}$ is the answer. Trust me, this is worth the 10 index points.

Solution:
Let $P_k$ denote the $k$ th prime.
Lemma: $x_{n+2^{k-1}} = P_k \cdot x_{n}$ for all $0 \leq n \leq 2^{k-1} - 1.$
$\mathbf{\mathrm{Proof:}}$
We can prove this using induction. Assume that $x_{2^{k-1}-1} = \prod_{j=1}^{k-1} P_j.$ Then, using the given recursion $x_{k+1} = \frac{x_np(x_n)}{X(x_n)}$ , we would “start fresh” for $x_{2^{k-1}} = P_k.$ It is then easy to see that then $\frac{x_n}{P_k}$ just cycles through the previous $x_{2^{k-1}}$ terms of $\{ x_n \},$ since the recursion process is the same and $P_k$ being a factor of $x_n$ is not affected until $n = 2 \cdot {2^{k-1}} = 2^k,$ when given our assumption $x_{2^k - 1} = \prod_{j=1}^{k} P_j$ and $n = 2^k$ is now the least $n$ such that \[P_{k+1} = p(x_{2^{n-1}}),\] in which $P_a = p(x_n)$ where $a > k$ is the only way that the aforementioned cycle would be affected. Specifically, by cancellation according to our recursion, $x_{2^k} = P_{k+1},$ and the values of $x_n$ just starts cycling through the previous $x_{2^k}$ terms again until $x_{2^{k+1}}$ when a new prime shows up in the prime factorization of $x_n,$ when it starts cycling again, and so on. Using our base cases of $x_0$ and $x_1,$ our induction is complete.
Now, it is easy to see that $2090 = 2 \cdot 5 \cdot 11 \cdot 19 = P_1 \cdot P_3 \cdot P_5 \cdot P_8,$ and by Lemma #1, the least positive integer $n$ such that $19 | x_n$ is $2^7.$ By repeatedly applying our obtained recursion from Lemma #1, it is easy to see that our answer is just $2^7 + 2^4 + 2^2 + 2^0,$ or $10010101_2 = \boxed{149}.$


Answer: 26.0

For any positive integer $M$ , the notation $M!$ denotes the product of the integers $1$ through $M$ . What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$

Solution: Factoring out $98!+99!+100!$ , we have $98! (1+99+99*100)$ , which is $98! (10000)$ . Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$ . The $19$ is because of all the multiples of $5$ .The $3$ is because of all the multiples of $25$ . Now, $10,000$ has $4$ factors of $5$ , so there are a total of $22 + 4 = \boxed{26}$ factors of $5$

Solution:
We can first factor a $98!$ out of the $98! + 99! + 100!$ to get $98! ( 1 + 99 + 100*99 ),$ Simplify to get $98! (10,000)$
Let's first find how many factors of $5 10,000$ has. $10,000$ is $(2*5)^4$ because $10,000$ is $(10)^4$ . After we remove the brackets, we get $2^4$ , and $5^4$ . We only care about the latter (second one), because the problem only ask's for the power of $5$ . We get $4$
Next, we can look at the multiples of 5 in $98!$ $98/5 = 19$ so there is 19 multiples of 5. We get $19$
But we cannot forget the multiples of $5$ with $2$ fives in it. Multiples of $25$ . How many multiples of $25$ are between $1$ and $98$ $3$ $25,50,75,$ and that's it. We get $3$
Finally, we add all of the numbers (powers of $5$ ) up. That is $4 + 19 + 3$ , which is just $26$
So the answer is $26$ . Which is answer choice D $\boxed{26}$


Answer: 26.0

Solution: Also, keep in mind that the number of $5$ ’s in $98! (10,000)$ is the same as the number of trailing zeros. The number of zeros is $98!$ , which means we need pairs of $5$ ’s and $2$ ’s; we know there will be many more $2$ ’s, so we seek to find the number of $5$ ’s in $98!$ , which the solution tells us. And, that is $22$ factors of $5$ $10,000$ has $4$ trailing zeros, so it has $4$ factors of $5$ and $22 + 4 = \boxed{26}$


Answer: 169.0

For any positive integer $k$ , let $f_1(k)$ denote the square of the sum of the digits of $k$ .  For $n \ge 2$ , let $f_n(k) = f_1(f_{n - 1}(k))$ .  Find $f_{1988}(11)$

Solution:
We see that $f_{1}(11)=4$
$f_2(11) = f_1(4)=16$
$f_3(11) = f_1(16)=49$
$f_4(11) = f_1(49)=169$
$f_5(11) = f_1(169)=256$
$f_6(11) = f_1(256)=169$
Note that this revolves between the two numbers. Since $1988$ is even, we thus have $f_{1988}(11) = f_{4}(11) = \boxed{169}$


Answer: 223.0

For any positive integer $x_{}$ , let $S(x)$ be the sum of the digits of $x_{}$ , and let $T(x)$ be $|S(x+2)-S(x)|.$ For example, $T(199)=|S(201)-S(199)|=|3-19|=16.$ How many values of $T(x)$ do not exceed 1999?

Solution: For most values of $x$ $T(x)$ will equal $2$ . For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take $T(a999)$ as an example, \[|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|\] And in general, the values of $T(x)$ will then be in the form of $|2 - 9n| = 9n - 2$ . From $7$ to $1999$ , there are $\left\lceil \frac{1999 - 7}{9}\right\rceil = 222$ solutions; including $2$ and there are a total of $\boxed{223}$ solutions.


Answer: 199.0

For any real number a and positive integer k, define
${a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}$
What is
${-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}$
$\textbf{(A)}\ -199\qquad \textbf{(B)}\ -197\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 197\qquad \textbf{(E)}\ 199$

Solution:
We expand both the numerator and the denominator.
\begin{align*} \binom{-\frac{1}{2}}{100}\div\binom{\frac{1}{2}}{100} &= \frac{   \dfrac{     (-\frac{1}{2})     (-\frac{1}{2} - 1)     (-\frac{1}{2} - 2)     \cdots     (-\frac{1}{2} - (100 - 1))   }{\cancel{(100)(99)\cdots(1)}} }{   \dfrac{     (\frac{1}{2})     (\frac{1}{2} - 1)     (\frac{1}{2} - 2)     \cdots     (\frac{1}{2} - (100 - 1))   }{\cancel{(100)(99)\cdots(1)}} } \\ &= \frac{     (-\frac{1}{2})     (-\frac{1}{2} - 1)     (-\frac{1}{2} - 2)     \cdots     (-\frac{1}{2} - 99) }{     (\frac{1}{2})     (\frac{1}{2} - 1)     (\frac{1}{2} - 2)     \cdots     (\frac{1}{2} - 99) } \end{align*}
Now, note that $-\frac{1}{2}-1=\frac{1}{2}-2$ $-\frac{1}{2}-2=\frac{1}{2}-3$ , etc.; in essence, $-\frac{1}{2}-n=\frac{1}{2}-(n+1)$ . We can then simplify the numerator and cancel like terms.
\begin{align*} \frac{     (-\frac{1}{2})     (-\frac{1}{2} - 1)     (-\frac{1}{2} - 2)     \cdots     (-\frac{1}{2} - 99) }{     (\frac{1}{2})     (\frac{1}{2} - 1)     (\frac{1}{2} - 2)     \cdots     (\frac{1}{2} - 99) } &= \frac{     \cancel{(\frac{1}{2} - 1)}     \cancel{(\frac{1}{2} - 2)}     \cancel{(\frac{1}{2} - 3)}     \cdots     (\frac{1}{2} - 100) }{     (\frac{1}{2})     \cancel{(\frac{1}{2} - 1)}     \cancel{(\frac{1}{2} - 2)}     \cdots     \cancel{(\frac{1}{2} - 99)} } \\ &= \frac{\frac{1}{2}-100}{\frac{1}{2}} \\ &= \frac{-\frac{199}{2}}{\frac{1}{2}} \\ &= \boxed{199.}


Answer: 819.0

For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$

Solution:
Note that the $\Delta$ s are reminiscent of differentiation; from the condition $\Delta(\Delta{A}) = 1$ , we are led to consider the differential equation \[\frac{d^2 A}{dn^2} = 1\] This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; \[a_{n} = \frac{1}{2}(n-19)(n-92)\] as we must have roots at $n = 19$ and $n = 92$
Thus, $a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}$

Solution:
Let $\Delta^1 A=\Delta A$ , and $\Delta^n A=\Delta(\Delta^{(n-1)}A)$
Note that in every sequence of $a_i$ $a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...$
Then $a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...$
Since $\Delta a_1 =a_2 -a_1$ $a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}$
$a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153$
$a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095$
Solving, $a_1=\boxed{819}$

Solution:
The sequence $\Delta(\Delta A)$ is the second finite difference sequence, and the first $k-1$ terms of this sequence can be computed in terms of the original sequence as shown below.
$\begin{array}{rcl} a_3+a_1-2a_2&=&1\\ a_4+a_2-2a_3&=&1\\ &\vdots\\ a_k + a_{k-2} - 2a_{k-1} &= &1\\ a_{k+1} + a_{k-1} - 2a_k &=& 1.\\ \end{array}$
Adding the above $k-1$ equations we find that
\[(a_{k+1} - a_k) = k-1 + (a_2-a_1).\tag{1}\]
We can sum equation $(1)$ from $k=1$ to $18$ , finding \[18(a_1-a_2) - a_1 = 153.\tag{2}\]
We can also sum equation $(1)$ from $k=1$ to $91$ , finding \[91(a_1-a_2) - a_1 = 4095.\tag{3}\] Finally, $18\cdot (3) - 91\cdot(2)$ gives $a_1=\boxed{819}$

Solution:
Since all terms of $\Delta(\Delta A)$ are 1, we know that $\Delta A$ looks like $(k,k+1,k+2,...)$ for some $k$ . This means $A$ looks like $(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)$ . More specifically, $A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}$ . Plugging in $a_{19}=a_{92}=0$ , we have the following linear system: \[a_1+91k=-4095\] \[a_1+18k=-153\] From this, we can easily find that $k=-54$ and $a_1=\boxed{819}$ .
Solution by Zeroman

Solution: Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that $f(19) = f(92) = 0$ so the quadratic is $f(x) = a(x-19)(x-92)$ for some constant $a.$ Now we use the conditions that the finite difference is $1$ to find $a.$ We know $f(19) = 0$ and $f(20) = -72a$ and $f(18) = 74a.$ Therefore applying finite differences once yields the sequence $-74a,-72a$ and then applying finite differences one more time yields $2a$ so $a =\frac{1}{2}.$ Therefore $f(1) = 9 \cdot 91 = \boxed{819}.$

Solution: Let $a_1=a,a_2=b.$ From the conditions, we have \[a_{n-1}+a_{n+1}=2a_n+1,\] for all $n>1.$ From this, we find that \begin{align*} a_3&=2b+1-a \\ a_4&=3b+3-2a\\ a_5&=4b+6-3a, \end{align*} or, in general, \[a_n=(n-1)b+\frac{(n-2)(n-1)}{2}-(n-2)a.\] Note: we can easily prove this by induction. Now, substituting $n=19,92,$ we find that \begin{align*} 0=&18b+\frac{17\cdot18}{2}-17a\\ 0=&91b+\frac{90\cdot91}{2}-90a\\ b=&\frac{17a-\frac{17\cdot18}{2}}{18}=\frac{90a-\frac{90\cdot91}{2}}{91}. \end{align*} Now, cross multiplying, we find that \begin{align*} 91\left(17a-\frac{17\cdot18}{2}\right)&=18\left(90a-\frac{90\cdot91}{2}\right)\\ 1547a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\ 73a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\ a=\frac{18\cdot91}{2}=9\cdot91=\boxed{819}


Answer: 97.0

For any set $S$ , let $|S|$ denote the number of elements in $S$ , and let $n(S)$ be the number of subsets of $S$ , including the empty set and the set $S$ itself. If $A$ $B$ , and $C$ are sets for which $n(A)+n(B)+n(C)=n(A\cup B\cup C)$ and $|A|=|B|=100$ , then what is the minimum possible value of $|A\cap B\cap C|$
$(A) 96 \ (B) 97 \ (C) 98 \ (D) 99 \ (E) 100$

Solution:
As $|A|=|B|=100$ $n(A)=n(B)=2^{100}$
As $n(A)+n(B)+n(C)=n(A \cup B \cup C)$ $2^{|A|}+2^{|B|}+2^{|C|}=2^{|A \cup B \cup C|}$ $2^{100}+2^{100}+2^{|C|}=2^{|A \cup B \cup C|}$
$2^{101}+2^{|C|}=2^{|A \cup B \cup C|}$ as $|C|$ and $|A \cup B \cup C|$ are integers, $|C|=101$ and $|A \cup B \cup C| = 102$
By the Principle of Inclusion-Exclusion $|A \cup B| = |A| + |B| - |A \cap B| = 200 - |A \cap B|$
$|A|=|B| \le |A \cup B| \le |A \cup B \cup C|$ $100 \le |A \cup B| \le 102$ $98 \le |A \cap B| \le 100$
By the Principle of Inclusion-Exclusion $|A \cup C| = |A| + |C| - |A \cap C| = 201 - |A \cap C|$
$|C| \le |A \cup C| \le |A \cup B \cup C|$ $101 \le |A \cup C| \le 102$ $99 \le |A \cap C| \le 100$
By the Principle of Inclusion-Exclusion $|B \cup C| = |B| + |C| - |B \cap C| = 201 - |B \cap C|$
$|C| \le |B \cup C| \le |A \cup B \cup C|$ $101 \le |B \cup C| \le 102$ $99 \le |B \cap C| \le 100$
By the Principle of Inclusion-Exclusion $|A \cap B \cap C|=|A \cup B \cup C|- |A| - |B| - |C| + |A \cap B| + |A \cap C|+|B \cap C| = 102-100-100-101+ |A \cap B| + |A \cap C|$ $+|B \cap C|=|A \cap B| + |A \cap C|+|B \cap C| -199$
\[98 + 99 + 99 - 199 \le |A \cap B \cap C| \le 100+100+100-199\]
\[\boxed{97} \le |A \cap B \cap C| \le 101\]


Answer: 72.0

For certain ordered pairs $(a,b)\,$ of real numbers , the system of equations
has at least one solution, and each solution is an ordered pair $(x,y)\,$ of integers.  How many such ordered pairs $(a,b)\,$ are there?

Solution:
The equation $x^2+y^2=50$ is that of a circle of radius $\sqrt{50}$ , centered at the origin. By testing integers until the left side becomes too big, we see that the lattice points on this circle are $(\pm1,\pm7)$ $(\pm5,\pm5)$ , and $(\pm7,\pm1)$ where the signs are all independent of each other, for a total of $3\cdot 2\cdot 2=12$ lattice points. They are indicated by the blue dots below.

Since $(x,y)=(0,0)$ yields $a\cdot 0+b\cdot 0=0 \neq 1$ , we know that $ax+by=1$ is the equation of a line that does not pass through the origin. So, we are looking for the number of lines which pass through at least one of the $12$ lattice points on the circle, but do not pass through the origin or through any non-lattice point on the circle. An example is the green line above. It is straightforward to show that a line passes through the origin precisely when there exist two opposite points $(p,q)$ and $(-p,-q)$ through which it passes. And example is the red line above.
There are $\binom{12}{2}=66$ ways to pick two distinct lattice points, and subsequently $66$ distinct lines which pass through two distinct lattice points on the circle. Then we subtract the lines which pass through the origin by noting that the lattice points on the circle can be grouped into opposite pairs $(p,q)$ and $(-p,-q)$ , for a total of $\frac{12}{2}=6$ lines. Finally, we add the $12$ unique tangent lines to the circle at each of the lattice points.
Therefore, our final count of distinct lines which pass through one or two of the lattice points on the circle, but do not pass through the origin, is \[66-6+12=\boxed{72}.\]


Answer: 125.0

For certain pairs $(m,n)$ of positive integers with $m\geq n$ there are exactly $50$ distinct positive integers $k$ such that $|\log m - \log k| < \log n$ . Find the sum of all possible values of the product $mn$

Solution:
We have $\log m - \log k = \log \left( \frac mk \right)$ , hence we can rewrite the inequality as follows: \[- \log n < \log \left( \frac mk \right) < \log n\] We can now get rid of the logarithms, obtaining: \[\frac 1n < \frac mk < n\] And this can be rewritten in terms of $k$ as \[\frac mn < k < mn\]
From $k<mn$ it follows that the $50$ solutions for $k$ must be the integers $mn-1, mn-2, \dots, mn-50$ .
This will happen if and only if the lower bound on $k$ is in a suitable range -- we must have $mn-51 \leq \frac mn < mn-50$
Obviously there is no solution for $n=1$ . For $n>1$ the left inequality can be rewritten as $m\leq\dfrac{51n}{n^2-1}$ , and the right one as $m > \dfrac{50n}{n^2-1}$
Remember that we must have $m\geq n$ . However, for $n\geq 8$ we have $\dfrac{51n}{n^2-1} < n$ , and hence $m<n$ , which is a contradiction.
This only leaves us with the cases $n\in\{2,3,4,5,6,7\}$
Therefore the answer is $34\cdot 2 + 19\cdot 3 = 68 + 57 = \boxed{125}$


Answer: 7007.0

For certain real numbers $a$ $b$ , and $c$ , the polynomial \[g(x) = x^3 + ax^2 + x + 10\] has three distinct roots, and each root of $g(x)$ is also a root of the polynomial \[f(x) = x^4 + x^3 + bx^2 + 100x + c.\] What is $f(1)$
$\textbf{(A)}\ -9009 \qquad\textbf{(B)}\ -8008 \qquad\textbf{(C)}\ -7007 \qquad\textbf{(D)}\ -6006 \qquad\textbf{(E)}\ -5005$

Solution:
$f(x)$ must have four roots, three of which are roots of $g(x)$ . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that
\[f(x)=g(x)(x+p)\]
where $-p\in\mathbb{R}$ is the fourth root of $f(x)$ .
(Using $(x+p) = (x-r))$ instead of $(x-r)$ makes the following computations less messy.)
Substituting $g(x)$ and expanding, we find that
\begin{align*}f(x)&=(x^3+ax^2+x+10)(x+p)\\ &=x^4+(a+p)x^3+(1+ap)x^2+(10+p)x+10r.\end{align*}
Comparing coefficients with $f(x)$ , we see that
\begin{align*} a+p&=1\\ 1+ap=b\\ 10+p&=100\\ 10p&=c.\\ \end{align*}
Let's solve for $a,b,c,$ and $p$ . Since $10+p=100$ $p=90$
Since $a+p=1$ $a=-89$
(Solution 1.1 branches from here and takes a shortcut.)
$c=(10)(90)=900$
Then, since $b=1+ap$ $b=-8009$ . Thus,
\[f(x)=x^4+x^3-8009x^2+100x+900.\]
(Solution 1.2 branches from here and takes another shortcut)
Taking $f(1)$ , we find that
\begin{align*} f(1)&=1^4+1^3-8009(1)^2+100(1)+900\\ &=1+1-8009+100+900\\ &=\boxed{7007}

Solution:
A faster ending to Solution 1 is as follows.
\begin{align*} f(1)&=(1+p)(1^3+a\cdot1^2+1+10)\\ &=(91)(-77)\\ &= (7)(13)(11)(-7) = (1001)(-7) \\ &=\boxed{7007}

Solution:
Also a faster ending to Solution 1 is as follows.
To find $f(1)$ we just need to find the sum of the coefficients which is $1 + 1 - 8009 + 100 + 900= \boxed{7007}.$

Solution:
We notice that the constant term of $f(x)=c$ and the constant term in $g(x)=10$ . Because $f(x)$ can be factored as $g(x) \cdot (x- r)$ (where $r$ is the unshared root of $f(x)$ , we see that using the constant term, $-10 \cdot r = c$ and therefore $r = -\frac{c}{10}$ .
Now we once again write $f(x)$ out in factored form:
\[f(x) = g(x)\cdot (x-r) = (x^3+ax^2+x+10)(x+\frac{c}{10})\]
We can expand the expression on the right-hand side to get:
\[f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c\]
Now we have $f(x) = x^4+(a+\frac{c}{10})x^3+(1+\frac{ac}{10})x^2+(10+\frac{c}{10})x+c=x^4+x^3+bx^2+100x+c$
Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations: \[10+\frac{c}{10}=100 \Rightarrow c=900\] \[a+\frac{c}{10} = 1, c=900 \Rightarrow a + 90 =1 \Rightarrow a= -89\]
and finally,
\[1+\frac{ac}{10} = b = 1+\frac{-89 \cdot 900}{10} = b = -8009\]
We know that $f(1)$ is the sum of its coefficients, hence $1+1+b+100+c$ . We substitute the values we obtained for $b$ and $c$ into this expression to get $f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{7007}$

Solution:
Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$ . Let $r_4$ be the additional root of $f(x)$ . Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$ , we obtain the following:
\begin{align*} r_1+r_2+r_3&=-a \\  r_1+r_2+r_3+r_4&=-1 \end{align*}
Thus $r_4=a-1$
Now applying Vieta's formulas on the constant term of $g(x)$ , the linear term of $g(x)$ , and the linear term of $f(x)$ , we obtain:
\begin{align*} r_1r_2r_3  & = -10\\ r_1r_2+r_2r_3+r_3r_1 &= 1\\  r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2  & = -100\\ \end{align*}
Substituting for $r_1r_2r_3$ in the bottom equation and factoring the remainder of the expression, we obtain:
\[-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100\]
It follows that $r_4=-90$ . But $r_4=a-1$ so $a=-89$
Now we can factor $f(x)$ in terms of $g(x)$ as
\[f(x)=(x-r_4)g(x)=(x+90)g(x)\]
Then $f(1)=91g(1)$ and
\[g(1)=1^3-89\cdot 1^2+1+10=-77\]
Hence $f(1)=91\cdot(-77)=\boxed{7007}$

Solution: Let the roots of $g(x)$ be $r_1$ $r_2$ , and $r_3$ . Let the roots of $f(x)$ be $r_1$ $r_2$ $r_3$ , and $r_4$ . From Vieta's, we have: \begin{align*} r_1+r_2+r_3=-a \\ r_1+r_2+r_3+r_4=-1 \\ r_4=a-1 \end{align*} The fourth root is $a-1$ . Since $r_1$ $r_2$ , and $r_3$ are common roots, we have: \begin{align*} f(x)=g(x)(x-(a-1)) \\ f(1)=g(1)(1-(a-1)) \\ f(1)=(a+12)(2-a) \\ f(1)=-(a+12)(a-2) \\ \end{align*} Let $a-2=k$ \begin{align*} f(1)=-k(k+14) \end{align*} Note that $-7007=-1001\cdot(7)=-(7\cdot(11)\cdot(13))\cdot(7)=-91\cdot(77)$ This gives us a pretty good guess of $\boxed{7007}$

Solution: First off, let's get rid of the $x^4$ term by finding $h(x)=f(x)-xg(x)$ . This polynomial consists of the difference of two polynomials with $3$ common factors, so it must also have these factors. The polynomial is $h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c$ , and must be equal to $(1-a)g(x)$ . Equating the coefficients, we get $3$ equations. We will tackle the situation one equation at a time, starting the $x$ terms. Looking at the coefficients, we get $\dfrac{90}{1-a} = 1$ \[\therefore 90=1-a.\] The solution to the previous is obviously $a=-89$ . We can now find $b$ and $c$ $\dfrac{b-1}{1-a} = a$ \[\therefore b-1=a(1-a)=-89*90=-8010\] and $b=-8009$ . Finally $\dfrac{c}{1-a} = 10$ \[\therefore c=10(1-a)=10*90=900\] Solving the original problem, $f(1)=1 + 1 + b + 100 + c = 102+b+c=102+900-8009=\boxed{7007}$

Solution: Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of $\frac{f(x)}{g(x)}$ eventually brings us the final step $(1-a)x^3 + (b-1)x^2 + 90x + c$ minus $(1-a)x^3 - (a-a^2)x^2 + (1-a)x + 10(1-a)$ after we multiply $f(x)$ by $(1-a)$ . Now we equate coefficients of same-degree $x$ terms. This gives us $10(1-a) = c, b-1 = a - a^2, 1-a = 90 \Rightarrow a = -89, c = 900, b = -8009$ . We are interested in finding $f(1)$ , which equals $1^4 + 1^3 -8009\cdot1^2 + 100\cdot1 + 900 = \boxed{7007}$ . ~skyscraper

Solution:
We first note that $f(x) = g(x) \cdot q(x) + r(x)$ where $q$ is the quotient function and $r$ is the remainder function.
Clearly, $r(x) = 0$ because every single root in $g$ is also in $f$ , thus implying $g$ divides $f$ .
So, we wish to find $f(1) = g(1) \cdot q(1)$
Such an expression for $g(1)$ is pretty clean here as we can obtain $g(1) = a + 12$ , so we rewrite $f(1) = (a + 12) \cdot q(1)$ .
Well, now we need to know how $q$ is expressed in order to obtain $q(1)$ . This motivates us to long divide to obtain the quotient function. After simple long division $q(x) = x + (1 - a)$ . In addition, what is left over, namely $r(x)$ , has a constant piece of $a + 89$ (you'll see in a few sentences why we only care about particularly the constant piece).
Now we can write: $f(1) = (a + 12) \cdot (2 - a)$
Now, as we have already established $r(x) = 0$ for ALL $x$ that means $r(0)$ or the constant piece is $0$ , so $89 + a = 0$ , in which we obtain $a = -89$ . We now plug this back into our equation for $f(1)$ to get $(-89 + 12)(2 - (89)) = -77 \cdot 91 = \boxed{7007}$ . ~triggod


Answer: 7007.0

Solution:
Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$ . Let $r_4$ be the additional root of $f(x)$ . Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$ , we obtain the following:
\begin{align*} r_1+r_2+r_3&=-a \\  r_1+r_2+r_3+r_4&=-1 \end{align*}
Thus $r_4=a-1$
Now applying Vieta's formulas on the constant term of $g(x)$ , the linear term of $g(x)$ , and the linear term of $f(x)$ , we obtain:
\begin{align*} r_1r_2r_3  & = -10\\ r_1r_2+r_2r_3+r_3r_1 &= 1\\  r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2  & = -100\\ \end{align*}
Substituting for $r_1r_2r_3$ in the bottom equation and factoring the remainder of the expression, we obtain:
\[-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100\]
It follows that $r_4=-90$ . But $r_4=a-1$ so $a=-89$
Now we can factor $f(x)$ in terms of $g(x)$ as
\[f(x)=(x-r_4)g(x)=(x+90)g(x)\]
Then $f(1)=91g(1)$ and
\[g(1)=1^3-89\cdot 1^2+1+10=-77\]
Hence $f(1)=91\cdot(-77)=\boxed{7007}$

Solution:
Since all of the roots of $g(x)$ are distinct and are roots of $f(x)$ , and the degree of $f$ is one more than the degree of $g$ , we have that
\[f(x) = C(x-k)g(x)\]
for some number $k$ . By comparing $x^4$ coefficients, we see that $C=1$ . Thus,
\[x^4+x^3+bx^2+100x+c=(x-k)(x^3+ax^2+x+10)\]
Expanding and equating coefficients we get that
\[a-k=1,1-ak=b,10-k=100,-10k=c\]
The third equation yields $k=-90$ , and the first equation yields $a=-89$ . So we have that
$f(1)=(1+90)g(1)=91(1-89+1+10)=(91)(-77)=\boxed{7007}$

Solution:
$f(x)$ must have four roots, three of which are roots of $g(x)$ . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that
\[f(x)=g(x)(x-r)\]
where $r\in\mathbb{C}$ is the fourth root of $f(x)$ . Substituting $g(x)$ and expanding, we find that
\begin{align*}f(x)&=(x^3+ax^2+x+10)(x-r)\\ &=x^4+(a-r)x^3+(1-ar)x^2+(10-r)x-10r.\end{align*}
Comparing coefficients with $f(x)$ , we see that
\begin{align*} a-r&=1\\ 1-ar&=b\\ 10-r&=100\\ -10r&=c.\\ \end{align*}
(Solution 1.1 picks up here.)
Let's solve for $a,b,c,$ and $r$ . Since $10-r=100$ $r=-90$ , so $c=(-10)(-90)=900$ . Since $a-r=1$ $a=-89$ , and $b=1-ar=-8009$ . Thus, we know that
\[f(x)=x^4+x^3-8009x^2+100x+900.\]
Taking $f(1)$ , we find that
\begin{align*} f(1)&=1^4+1^3-8009(1)^2+100(1)+900\\ &=1+1-8009+100+900\\ &=\boxed{7007}

Solution:
A faster ending to Solution 1 is as follows.
We shall solve for only $a$ and $r$ . Since $10-r=100$ $r=-90$ , and since $a-r=1$ $a=-89$ . Then, \begin{align*} f(1)&=(1-r)(1^3+a\cdot1^2+1+10)\\ &=(91)(-77)\\ &=\boxed{7007}

Solution:
We notice that the constant term of $f(x)=c$ and the constant term in $g(x)=10$ . Because $f(x)$ can be factored as $g(x) \cdot (x- r)$ (where $r$ is the unshared root of $f(x)$ , we see that using the constant term, $-10 \cdot r = c$ and therefore $r = -\frac{c}{10}$ .
Now we once again write $f(x)$ out in factored form:
\[f(x) = g(x)\cdot (x-r) = (x^3+ax^2+x+10)\left(x+\frac{c}{10}\right)\]
We can expand the expression on the right-hand side to get:
\[f(x) = x^4+\left(a+\frac{c}{10}\right)x^3+\left(1+\frac{ac}{10}\right)x^2+\left(10+\frac{c}{10}\right)x+c\]
Now we have $f(x) = x^4+\left(a+\frac{c}{10}\right)x^3+\left(1+\frac{ac}{10}\right)x^2+\left(10+\frac{c}{10} \right)x+c=x^4+x^3+bx^2+100x+c$
Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations: \[10+\frac{c}{10}=100 \Rightarrow c=900\] \[a+\frac{c}{10} = 1, c=900 \Rightarrow a + 90 =1 \Rightarrow a= -89\]
and finally,
\[1+\frac{ac}{10} = b = 1+\frac{-89 \cdot 900}{10} = b = -8009\]
We know that $f(1)$ is the sum of its coefficients, hence $1+1+b+100+c$ . We substitute the values we obtained for $b$ and $c$ into this expression to get $f(1) = 1 + 1 + (-8009) + 100 + 900 = \boxed{7007}$

Solution: Let the roots of $g(x)$ be $r_1$ $r_2$ , and $r_3$ . Let the roots of $f(x)$ be $r_1$ $r_2$ $r_3$ , and $r_4$ . From Vieta's, we have: \begin{align*} r_1+r_2+r_3=-a \\ r_1+r_2+r_3+r_4=-1 \\ r_4=a-1 \end{align*} The fourth root is $a-1$ . Since $r_1$ $r_2$ , and $r_3$ are common roots, we have: \begin{align*} f(x)=g(x)(x-(a-1)) \\ f(1)=g(1)(1-(a-1)) \\ f(1)=(a+12)(2-a) \\ f(1)=-(a+12)(a-2) \\ \end{align*} Let $a-2=k$ \begin{align*} f(1)=-k(k+14) \end{align*} Note that $-7007=-1001\cdot(7)=-(7\cdot(11)\cdot(13))\cdot(7)=-91\cdot(77)$ This gives us a pretty good guess of $\boxed{7007}$

Solution: Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of $\frac{f(x)}{g(x)}$ eventually brings us the final step $(1-a)x^3 + (b-1)x^2 + 90x + c$ minus $(1-a)x^3 - (a-a^2)x^2 + (1-a)x + 10(1-a)$ after we multiply $f(x)$ by $(1-a)$ . Now we equate coefficients of same-degree $x$ terms. This gives us $10(1-a) = c, b-1 = a - a^2, 1-a = 90 \Rightarrow a = -89, c = 900, b = -8009$ . We are interested in finding $f(1)$ , which equals $1^4 + 1^3 -8009\cdot1^2 + 100\cdot1 + 900 = \boxed{7007}$ . ~skyscraper


Answer: 7007.0

Solution: First off, let's get rid of the $x^4$ term by finding $h(x)=f(x)-xg(x)$ . This polynomial consists of the difference of two polynomials with $3$ common factors, so it must also have these factors. The polynomial is $h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c$ , and must be equal to $(1-a)g(x)$ . Equating the coefficients, we get $3$ equations. We will tackle the situation one equation at a time, starting the $x$ terms. Looking at the coefficients, we get $\dfrac{90}{1-a} = 1$ \[\therefore 90=1-a.\] The solution to the previous is obviously $a=-89$ . We can now find $b$ and $c$ $\dfrac{b-1}{1-a} = a$ \[\therefore b-1=a(1-a)=-89\cdot90=-8010\] and $b=-8009$ . Finally $\dfrac{c}{1-a} = 10$ \[\therefore c=10(1-a)=10\cdot90=900\] Solving the original problem, $f(1)=1 + 1 + b + 100 + c = 102+b+c=102+900-8009=\boxed{7007}$


Answer: 51.0

For certain real values of $a, b, c,$ and $d_{},$ the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots.  The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$ Find $b.$

Solution: Since the coefficients of the polynomial are real, it follows that the non-real roots must come in complex conjugate pairs. Let the first two roots be $m,n$ . Since $m+n$ is not real, $m,n$ are not conjugates, so the other pair of roots must be the conjugates of $m,n$ . Let $m'$ be the conjugate of $m$ , and $n'$ be the conjugate of $n$ . Then, \[m\cdot n = 13 + i,m' + n' = 3 + 4i\Longrightarrow m'\cdot n' = 13 - i,m + n = 3 - 4i.\] By Vieta's formulas , we have that $b = mm' + nn' + mn' + nm' + mn + m'n' = (m + n)(m' + n') + mn + m'n' = \boxed{051}$

Solution:
Let's assume that the 2 roots multiplied together are p+qi, and r+si, and the two roots added together are the conjugates of the previous roots. Using Vieta, we get \[b = (p+qi)(r+si) + (p+qi)(r-si) + (p-qi)(r+si) + (p-qi)(r-si) + (p+qi)(p-qi) + (r+si)(r-si) =\]
\[(p+qi+p-qi)(r+si+r-si) + (p+qi)(p-qi) + (r+si)(r-si) =\]
\[(2p)(2r) + p^2 + q^2 + r^2 + s^2 = 4pr + p^2 + q^2 + r^2 + s^2 = (p+r)^2 + 2pr + q^2 + s^2\]
We are now stuck. We can't simplify further. But, we look back to the problem and see that the two roots that are multiplied together give a product of $13+i$ , and the two roots that are added give $3+4i$ . This gets three equations necessary for solving the problem. \[p+r = 3\] \[pr-qs = 13\] \[-q-s = 4\] So, alright. Let's use the first equation to get that $(p+r)^2 = 9$ , and substitute that in.  Now, the equation becomes:
\[b = 9 + 2pr + q^2 + s^2\]
We wish that we can turn the 2pr into 2qs. Fortunately, we can do that. By using the second equation, we can manipulate it to be $pr = 13+qs$ , and substitute that in.
\[b = 9 + 2(13+qs) + q^2 + s^2 = 9 + 26 + 2qs + q^2 + s^2 = 35 + (q+s)^2\]
We can square both sides of the third equation, and get $(q+s)^2 = 16$ We substitute that in and we get
\[b = 35+16 = \boxed{051}\]


Answer: 502.0

For distinct positive integers $a$ $b < 2012$ , define $f(a,b)$ to be the number of integers $k$ with $1 \le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012.  Let $S$ be the minimum value of $f(a,b)$ , where $a$ and $b$ range over all pairs of distinct positive integers less than 2012.  Determine $S$

Solution:
Let $ak \equiv r_{a} \pmod{2012}$ and $bk \equiv r_{b} \pmod{2012}$ . Notice that this means $a(2012 - k) \equiv 2012 - r_{a} \pmod{2012}$ and $b(2012 - k) \equiv 2012 - r_{b} \pmod{2012}$ . Thus, for every value of $k$ where $r_{a} > r_{b}$ , there is a value of $k$ where $r_{b} > r_{a}$ . Therefore, we merely have to calculate $\frac{1}{2}$ times the number of values of $k$ for which $r_{a} \neq r_{b}$ and $r_{a} \neq 0$
However, the answer is NOT $\frac{1}{2}(2012) = 1006$ ! This is because we must count the cases where the value of $k$ makes $r_{a} = r_{b}$ or where $r_{a} = 0$
So, let's start counting.
If $k$ is even, we have either $a \equiv 0 \pmod{1006}$ or $a - b \equiv 0 \pmod{1006}$ . So, $a = 1006$ or $a = b + 1006$ . We have $1005$ even values of $k$ (which is all the possible even values of $k$ , since the two above requirements don't put any bounds on $k$ at all).
If $k$ is odd, if $k = 503$ or $k = 503 \cdot 3$ , then $a \equiv 0 \pmod{4}$ or $a \equiv b \pmod{4}$ . Otherwise, $ak \equiv 0 \pmod{2012}$ or $ak \equiv bk \pmod{2012}$ , which is impossible to satisfy, given the domain $a, b < 2012$ . So, we have $2$ values of $k$
In total, we have $2 + 1005 = 1007$ values of $k$ which makes $r_{a} = r_{b}$ or $r_{a} = 0$ , so there are $2011 - 1007 = 1004$ values of $k$ for which $r_{a} \neq r_{b}$ and $r_{a} \neq 0$ . Thus, by our reasoning above, our solution is $\frac{1}{2} \cdot 1004 = \boxed{502}$