Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]
Each row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution: First, let $R(n)$ be the sum of the $n$ th row. Now, with some observation and math instinct, we can guess that $R(n) = 2^n - n$
Now we try to prove it by induction,
$R(1) = 2^n - n = 2^1 - 1 = 1$ (works for base case)
$R(k) = 2^k - k$
$R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1$
By definition from the question, the next row is always $:$
Double the sum of last row (Imagine each number from last row branches off toward left and right to the next row), plus # of new row, minus 2 (minus leftmost and rightmost's 1)
Which gives us $:$
$2(2^k - k) + (k + 1) - 2 = 2(2^k) - k - 1$
Hence, proven
Last, simply substitute $n = 2023$ , we get $R(2023) = 2^{2023} - 2023$
Last digit of $2^{2023}$ is $8$ $8-3 = \boxed{5}$

Answer: C 5
Problem: Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]
Each row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution: Let the sum of the numbers in row $2022$ be $S_{2022}$ . Let each number in row $2022$ be $x_i$ where $1 \leq i \leq 2022$
Then \begin{align*} S_{2023}&=1+(x_1+x_2+1)+(x_2+x_3+1)+...+(x_{2021}+x_{2022}+ 1)+1 \\ S_{2023}&=x_1+2(S_{2022}-x_1-x_{2022})+2023+x_{2022} \\ S_{2023} &= 2S_{2022} + 2021 \end{align*} From this we can establish: \begin{align*} S_n &= 2S_{n-1} + n-2 \\ S_{n-1} &= 2S_{n-2} + n-3 \\ S_n - S_{n-1} &= 2S_{n-1} - 2S_{n-2} + 1 \end{align*}
Let $B_{n} =  S_n - S_{n-1}$
From this we have: \begin{align*} B_n + 1 &= 2(B_{n-1} + 1) \\ B_n &= 2^{n-1} - 1 \\ S_n - S_{n-1} &= 2^{n-1} - 1 \\ S_n &= 2^{n-1} + 2^{n-2} + ... + 2^1 - (n-1) + S_1 \\ S_n &= 2^n - n \\ S_{2023} &= 2^{2023} - 2023 \end{align*}
The problem requires us to find the last digit of $S_{2023}$ . We can use modular arithmetic. \begin{align*} S_{2023} &= 2^{2023} - 2023 \\ 2^{2023} - 2023 &\equiv 8 - 3\pmod{10} \\ &\equiv \boxed{5} luckuso

Answer: C 5
Problem: Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]
Each row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution: Let the sum of the $n^{th}$ row be $S_n$
For each of the $n-2$ non-1 entries in the $n^{th}$ row, they are equal to the sum of the $2$ numbers diagonally above it in the $n-1^{th}$ row plus $1$ . Therefore all $n-3$ non-1 entries in the $n-1^{th}$ row appear twice in the sum of the $n-2$ non-1 entries in the $n^{th}$ row, the two $1$ s on each end of the $n-1^{th}$ row only appear once in the sum of the $n-2$ non-1 entries in the $n^{th}$ row. Additionally, additional $1$ s are placed at each end of the $n^{th}$ row. Hence, $S_n = 2(S_{n-1} - 1) + n-2 + 1 + 1 = 2 S_{n-1} + n - 2$
$S_4 = 1+5+5+1 = 12$ $S_5 = 1+7+11+7+1 = 27$ . By using the recursive formula, $S_5 = 2 \cdot S_4 + 5-2 = 27$
\[S_n = 2 S_{n-1} + n - 2\] \[S_n + n = 2 S_{n-1} + 2n - 2\] \[S_n + n = 2( S_{n-1} + n - 1)\] $S_n + n$ is a geometric sequence by a ratio of $2$
$\because \quad S_1 = 1$
$\therefore \quad S_n + n = 2^n$ $S_n = 2^n - n$
$S_{2023} = 2^{2023} - 2023$
The unit digit of powers of $2$ is periodic by a cycle of $4$ digits: $2$ $4$ $8$ $6$ $2023 = 3 \mod 4$ , the unit digit of $2^{2023}$ is $8$
Therefore, the unit digit of $S_{2023}$ is $8-3 = \boxed{5}$

Answer: C 5
Problem: Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]
Each row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution: Observe that: $S_n = 2{S_{n-1}} + (n-2) \\ = 2{S_{n-1}} + D_{n} \\  = 2{S_{n-1}} + D_{n-1} + 1$ where $D_1=-1$ , and $D_n=D_{n-1}+1$ for $n>1$
Make a table of values, $(S_n, D_n) \mod 2$ and $(S_n, D_n) \mod 5$ , a until the cycle completes for each. ( $\mod 2$ has period $1$ $\mod 5$ has period $20$ ).
( In both cases, there is no "head" that is not part of the cycle.)
Mod 2: $S_{2023} \equiv S_{2023 \mod 2}  \equiv   S_1  = (1\cdot 2 +0)\cdot 0 + 1 \equiv 1 \pmod 2$
Mod 5: $S_{2023} \equiv S_{2023 \mod 20}  \equiv   S_3 \equiv (1\cdot 2 +0)\cdot 2 + 1 \equiv 5 \pmod 5$
Thus $S_{2023} \mod 10 = \boxed{5}$

Answer: C 5
Problem: Let $f$ be the unique function defined on the positive integers such that \[\sum_{d\mid n}d\cdot f\left(\frac{n}{d}\right)=1\] for all positive integers $n$ . What is $f(2023)$
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$

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

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

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

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

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

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

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

Answer: B 96
Problem: How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation \[(1+2a)(2+2b)(2a+b) = 32ab?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}$

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

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

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

Answer: B 1
Problem: Let $K$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_n$ such that $n$ is a positive integer less than or equal to $10$ , each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$ , and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$ , inclusive. For example, $\{\}$ $\{5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 6, 7, 9\}$ is one such sequence, with $n = 5$ .What is the remainder when $K$ is divided by $10$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution: Consider any sequence with $n$ terms.  Every 10 number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot… and appear the first time in the $n$ th spot, which means every number has $(n+1)$ choices to show up in the sequence. Consequently, for each sequence with length $n$ , there are $(n+1)^{10}$ possible ways.
Thus, the desired value is $\sum_{i=1}^{10}(i+1)^{10}\equiv \boxed{5}$

Answer: C 5
Problem: Let $K$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_n$ such that $n$ is a positive integer less than or equal to $10$ , each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$ , and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$ , inclusive. For example, $\{\}$ $\{5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 6, 7, 9\}$ is one such sequence, with $n = 5$ .What is the remainder when $K$ is divided by $10$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution: Let $f(x,\ell)$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_\ell$ such that each $A_i$ is a subset of $\{1, 2,\dots,x\}$ , and $A_i$ is a subset of $A_{i+1}$ for $i=1$ $2\dots$ $\ell-1$ . Then $f(x,1)=2^x$ and $f(0,\ell)=1$
If $\ell\ge2$ and $x\ge1$ , we need to get a recursive formula for $f(x,\ell)$ : If $|A_1|=i$ , then $A_1$ has $\text C_x^i$ possibilities, and the subsequence $\{A_i\}_{2\le i\le\ell}$ has $f(x-i,\ell-1)$ possibilities. Hence \[f(x,\ell)=\sum_{i=0}^x\text C_x^if(x-i,\ell-1).\] By applying this formula and only considering  modulo $10$ , we get $f(1,2)=3$ $f(1,3)=4$ $f(1,4)=5$ $f(1,5)=6$ $f(1,6)=7$ $f(1,7)=8$ $f(1,8)=9$ $f(1,9)=0$ $f(1,10)=1$ $f(2,2)=9$ $f(2,3)=6$ $f(2,4)=5$ $f(2,5)=6$ $f(2,6)=9$ $f(2,7)=4$ $f(2,8)=1$ $f(2,9)=0$ $f(2,10)=1$ $f(3,2)=7$ $f(3,3)=4$ $f(3,4)=5$ $f(3,5)=6$ $f(3,6)=3$ $f(3,7)=2$ $f(3,8)=9$ $f(3,9)=0$ $f(3,10)=1$ $f(4,2)=1$ $f(4,3)=6$ $f(4,4)=5$ $f(4,5)=6$ $f(4,6)=1$ $f(4,7)=6$ $f(4,8)=1$ $f(4,9)=0$ $f(4,10)=1$ $f(5,2)=3$ $f(5,3)=4$ $f(5,4)=5$ $f(5,5)=6$ $f(5,6)=7$ $f(5,7)=8$ $f(5,8)=9$ $f(5,9)=0$ $f(5,10)=1$ $f(6,2)=9$ $f(6,3)=6$ $f(6,4)=5$ $f(6,5)=6$ $f(6,6)=9$ $f(6,7)=4$ $f(6,8)=1$ $f(6,9)=0$ $f(6,10)=1$ $f(7,2)=7$ $f(7,3)=4$ $f(7,4)=5$ $f(7,5)=6$ $f(7,6)=3$ $f(7,7)=2$ $f(7,8)=9$ $f(7,9)=0$ $f(7,10)=1$ $f(8,2)=1$ $f(8,3)=6$ $f(8,4)=5$ $f(8,5)=6$ $f(8,6)=1$ $f(8,7)=6$ $f(8,8)=1$ $f(8,9)=0$ $f(8,10)=1$ $f(9,2)=3$ $f(9,3)=4$ $f(9,4)=5$ $f(9,5)=6$ $f(9,6)=7$ $f(9,7)=8$ $f(9,8)=9$ $f(9,9)=0$ $f(9,10)=1$ $f(10,2)=9$ $f(10,3)=6$ $f(10,4)=5$ $f(10,5)=6$ $f(10,6)=9$ $f(10,7)=4$ $f(10,8)=1$ $f(10,9)=0$ $f(10,10)=1$
Lastly, we get $K\equiv\textstyle\sum\limits_{i=1}^{10}f(10,i)\equiv\boxed{5}$ .
~Quantum-Phantom

Answer: C 5
Problem: Let $K$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_n$ such that $n$ is a positive integer less than or equal to $10$ , each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$ , and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$ , inclusive. For example, $\{\}$ $\{5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 6, 7, 9\}$ is one such sequence, with $n = 5$ .What is the remainder when $K$ is divided by $10$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution: Seeing that all the answers are different modulus 5, and that 10 is divisible by 5, we cheese this problem.
Let $A_1, A_2, \cdots, A_n$ be one sequence satisfying the constraints of the problem. Let $b_1, b_2, \cdots, b_n, b_{n+1}$ be the sequence of nonnegative integers such that $A_k$ has $b_k$ elements for all $1\leq{}k\leq{}n$ , and $b_{n+1}=10$ . Note that we can generate the number of valid sequences of $A$ by first generating all sequences of $b$ such that $b_i\leq{}b_{i+1}$ for all $1\leq{}i<n$ , then choosing the elements from $A_{k+1}$ that we keep in $A_k$ , given the sequence of $b$ as the restraint for the number of elements. For each sequence $b_1, b_2, \cdots{}, b_{n+1}$ , there are $\prod_{i=2}^{n+1}\binom{b_i}{b_{i-1}}$ corresponding sequences for $A$ . Now, consider two cases - either all terms in $b$ are either 0, 5, or 10, or there is at least one term in $b$ that is neither 0, 5 nor 10. In the second case, consider the last term in $b$ that is not 10 or 5, say $b_m$ . However, that implies $b_{m+1}=10$ , and so the number of corresponding sequences of $A$ is $\binom{10}{b_m}\cdot{}$ something or $\binom{5}{b_m}\cdot$ something, which is always a multiple of $5$ . Therefore, we only need to consider sequences of $b$ where each term is $0$ $5$ or $10$ . If all terms in $b$ are 0 or 10, then for each $1\leq{}n\leq{}10$ there are $n+1$ sequences of $b$ (since there are $n+1$ places to turn from $0$ to $10$ ), for a total of $2+3+\cdots+11=65\equiv0$ (mod 5). If there exists at least one term $b_k=5$ , then we use stars and bars to count the number of sequences of $b$ , and each sequence of $b$ corresponds to $\binom{10}{5}$ sequences of $A$ . For each $n$ , we must have at least one term of $5$ . After that, there are $n-1$ stars and $2$ bars (separating $0$ to $5$ and $5$ to $10$ ), so that is $\binom{n+1}{2}$ sequences of $b$ . So the sum is $\binom{2}{2}+\binom{3}{2}+\cdots+\binom{11}{2}=\binom{12}{3}\equiv0$ (mod 5). Therefore, the answer is 0 mod 5, and it must be $\boxed{5}$

Answer: C 5
Problem: Let $K$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_n$ such that $n$ is a positive integer less than or equal to $10$ , each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$ , and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$ , inclusive. For example, $\{\}$ $\{5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 6, 7, 9\}$ is one such sequence, with $n = 5$ .What is the remainder when $K$ is divided by $10$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution: We observe that in each sequence, if element $e \in A_i$ , then $e \in A_j$ for all $j \geq i$ .
Therefore, to determine a sequence with a fixed length $n$ , we only need to determine the first set $A_i$ that each element in $\left\{ 1, 2, \cdots , 10 \right\}$ is inserted into, or an element is never inserted into any subset.
We have \begin{align*} K & = \sum_{n = 1}^{10} \left( n + 1 \right)^{10} \\ & = \sum_{m = 2}^{11} m^{10} . \end{align*}
Recalling or noticing that $x^n \equiv x^{n \mod 4} \pmod {10}$ , then, 
Modulo 10, we have \begin{align*} K & \equiv \sum_{m = 2}^{11} m^2 \\ & \equiv \sum_{m = 1}^{11} m^2 - 1^2 \\ & \equiv \frac{11 \cdot \left( 11 + 1 \right) \left( 2 \cdot 11 + 1 \right)}{6} - 1\\ & \equiv 505 \\ & \equiv \boxed{5}

Answer: C 5
Problem: There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that \[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\] whenever $\tan 2023x$ is defined. What is $a_{2023}?$
$\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023$

Solution: \begin{align*} \cos 2023 x + i \sin 2023 x &= (\cos x + i \sin x)^{2023}\\ &= \cos^{2023} x + \binom{2023}{1} \cos^{2022} x (i\sin x) + \binom{2023}{2} \cos^{2021} x (i \sin x)^{2} +\binom{2023}{3} \cos^{2023} x (i \sin x)^{3}\\ &+ \dots + \binom{2023}{2022} \cos x (i \sin x)^{2022} + (i \sin x)^{2023}\\ &= \cos^{2023} x + i \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{2} \cos^{2021} x \sin^{2} x - i\binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots\\ &- \binom{2023}{2022} \cos x \sin^{2022} x - i \sin^{2023} x\\ \end{align*}
By equating real and imaginary parts:
\[\cos 2023 x = \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x\]
\[\sin 2023 x = \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x\]
\begin{align*} \tan2023x  &= \frac{ \sin2023x }{ \cos2023x } = \frac{ \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x }{ \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x }\\ &= \frac{ \binom{2023}{1} \frac{\cos^{2022} x \sin x}{\cos^{2023} x} - \binom{2023}{3} \frac{\cos^{2020} x \sin^{3} x}{\cos^{2023} x} + \dots - \frac{\sin^{2023} x}{\cos^{2023} x} }{ \frac{\cos^{2023} x}{\cos^{2023} x} - \binom{2023}{2} \frac{\cos^{2021} x \sin^{2} x}{\cos^{2023} x} + \dots - \binom{2023}{2022} \frac{\cos x \sin^{2022} x}{\cos^{2023} x}  }\\ &= \frac{ \binom{2023}{1} \tan x - \binom{2023}{3} \tan^{3}x + \dots - \tan^{2023}x }{ 1 - \binom{2023}{2} \tan^{2}x + \dots - \binom{2023}{2022} \tan^{2022} x }\\ \end{align*}
\[a_{2023} = \boxed{1}\]

Answer: C 1