Take $r$ such that $1\le r\le n$, and consider all subsets of $r$ elements of the set $\{1,2,\ldots,n\}$. Each subset has a smallest element. Let $F(n,r)$ be the arithmetic mean of these smallest elements. Prove that: \[ F(n,r)={n+1\over r+1}. \]

Let $p$ be a prime number. Prove that there is no number divisible by $p$ in the $n-th$ row of Pascal's triangle if and only if $n$ can be represented in the form $n = p^sq - 1$, where $s$ and $q$ are integers with $s \geq 0, 0 < q < p$.

The number $2017$ is prime. Let $S=\sum_{k=0}^{62}\binom{2014}{k}$. What is the remainder when $S$ is divided by $2017$? $\textbf{(A) }32\qquad \textbf{(B) }684\qquad \textbf{(C) }1024\qquad \textbf{(D) }1576\qquad \textbf{(E) }2016\qquad$

Wanda the Worm likes to eat Pascal's triangle. One day, she starts at the top of the triangle and eats $\textstyle\binom{0}{0}=1$. Each move, she travels to an adjacent positive integer and eats it, but she can never return to a spot that she has previously eaten. If Wanda can never eat numbers $a,b,c$ such that $a+b=c$, prove that it is possible for her to eat 100,000 numbers in the first 2011 rows given that she is not restricted to traveling only in the first 2011 rows. (Here, the $n+1$st row of Pascal's triangle consists of entries of the form $\textstyle\binom{n}{k}$ for integers $0\le k\le n$. Thus, the entry $\textstyle\binom{n}{k}$ is considered adjacent to the entries $\textstyle\binom{n-1}{k-1}$, $\textstyle\binom{n-1}{k}$, $\textstyle\binom{n}{k-1}$, $\textstyle\binom{n}{k+1}$, $\textstyle\binom{n+1}{k}$, $\textstyle\binom{n+1}{k+1}$.) [i]Linus Hamilton.[/i]

With about six hours left on the van ride home from vacation, Wendy looks for something to do. She starts working on a project for the math team. There are sixteen students, including Wendy, who are about to be sophomores on the math team. Elected as a math team officer, one of Wendy's jobs is to schedule groups of the sophomores to tutor geometry students after school on Tuesdays. The way things have been done in the past, the same number of sophomores tutor every week, but the same group of students never works together. Wendy notices that there are even numbers of groups she could select whether she chooses $4$ or $5$ students at a time to tutor geometry each week: \begin{align*}\dbinom{16}4&=1820,\\\dbinom{16}5&=4368.\end{align*} Playing around a bit more, Wendy realizes that unless she chooses all or none of the students on the math team to tutor each week that the number of possible combinations of the sophomore math teamers is always even. This gives her an idea for a problem for the $2008$ Jupiter Falls High School Math Meet team test: \[\text{How many of the 2009 numbers on Row 2008 of Pascal's Triangle are even?}\] Wendy works the solution out correctly. What is her answer?

For any polynomial $P(x)=a_0+a_1x+\ldots+a_kx^k$ with integer coefficients, the number of odd coefficients is denoted by $o(P)$. For $i-0,1,2,\ldots$ let $Q_i(x)=(1+x)^i$. Prove that if $i_1,i_2,\ldots,i_n$ are integers satisfying $0\le i_1<i_2<\ldots<i_n$, then: \[ o(Q_{i_{1}}+Q_{i_{2}}+\ldots+Q_{i_{n}})\ge o(Q_{i_{1}}). \]

In the diagram below $a, b, c, d, e, f, g, h, i, j$ are distinct positive integers and each (except $a, e, h$ and $j$) is the sum of the two numbers to the left and above. For example, $b = a + e, f = e + h, i = h + j$. What is the smallest possible value of $d$? j h i e f g a b c d

Does there exist a row of Pascal’s Triangle containing four distinct values $a,b,c$ and $d$ such that $b = 2a$ and $d = 2c$? Recall that Pascal’s triangle is the pattern of numbers that begins as follows [img]https://cdn.artofproblemsolving.com/attachments/2/1/050e56f0f1f1b2a9c78481f03acd65de50c45b.png[/img] where the elements of each row are the sums of pairs of adjacent elements of the prior row. For example, $10 =4+6$. Also note that the last row displayed above contains the four elements $a = 5,b = 10,d = 10,c = 5$, satisfying $b = 2a$ and $d = 2c$, but these four values are NOT distinct.

Wanda the Worm likes to eat Pascal's triangle. One day, she starts at the top of the triangle and eats $\textstyle\binom{0}{0}=1$. Each move, she travels to an adjacent positive integer and eats it, but she can never return to a spot that she has previously eaten. If Wanda can never eat numbers $a,b,c$ such that $a+b=c$, prove that it is possible for her to eat 100,000 numbers in the first 2011 rows given that she is not restricted to traveling only in the first 2011 rows. (Here, the $n+1$st row of Pascal's triangle consists of entries of the form $\textstyle\binom{n}{k}$ for integers $0\le k\le n$. Thus, the entry $\textstyle\binom{n}{k}$ is considered adjacent to the entries $\textstyle\binom{n-1}{k-1}$, $\textstyle\binom{n-1}{k}$, $\textstyle\binom{n}{k-1}$, $\textstyle\binom{n}{k+1}$, $\textstyle\binom{n+1}{k}$, $\textstyle\binom{n+1}{k+1}$.) [i]Linus Hamilton.[/i]

Given a finite sequence $ S \equal{} (a_1,a_2,\ldots,a_n)$ of $ n$ real numbers, let $ A(S)$ be the sequence \[ \left(\frac {a_1 \plus{} a_2}2,\frac {a_2 \plus{} a_3}2,\ldots,\frac {a_{n \minus{} 1} \plus{} a_n}2\right) \]of $ n \minus{} 1$ real numbers. Define $ A^1(S) \equal{} A(S)$ and, for each integer $ m$, $ 2\le m\le n \minus{} 1$, define $ A^m(S) \equal{} A(A^{m \minus{} 1}(S)).$ Suppose $ x > 0$, and let $ S \equal{} (1,x,x^2,\ldots,x^{100})$. If $ A^{100}(S) \equal{} (1/2^{50})$, then what is $ x$? $ \textbf{(A) } 1 \minus{} \frac {\sqrt {2}}2\qquad \textbf{(B) } \sqrt {2} \minus{} 1\qquad \textbf{(C) } \frac 12\qquad \textbf{(D) } 2 \minus{} \sqrt {2}\qquad \textbf{(E) } \frac {\sqrt {2}}2$

A triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in given diagram). In each square of the eleventh row, a $0$ or a $1$ is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of $0$'s and $1$'s in the bottom row is the number in the top square a multiple of $3$? [asy] defaultpen(linewidth(0.7)); path p=origin--(1,0)--(1,1)--(0,1)--cycle; int i,j; for(i=0; i<12; i=i+1) { for(j=0; j<11-i; j=j+1) { draw(shift(i/2+j,i)*p); }}[/asy]

The first of an infinite triangular spreadsheet the line contains one number, the second line contains two numbers, the third line contains three numbers, and so on. In doing so is in any $k$-th row ($k = 1, 2, 3,...$) in the first and last place the number $k$, each other the number in the table is found, however, than in the previous row the least common of the two numbers above it multiple (the adjacent figure shows the first five rows of this table). We choose any two numbers from the table that are not in their row in the first or last place. Prove that one of the selected numbers is divisible by another. [img]https://cdn.artofproblemsolving.com/attachments/3/7/107d8999d9f04777719a0f1b1df418dbe00023.png[/img]

Wanda the Worm likes to eat Pascal's triangle. One day, she starts at the top of the triangle and eats $\textstyle\binom{0}{0}=1$. Each move, she travels to an adjacent positive integer and eats it, but she can never return to a spot that she has previously eaten. If Wanda can never eat numbers $a,b,c$ such that $a+b=c$, prove that it is possible for her to eat 100,000 numbers in the first 2011 rows given that she is not restricted to traveling only in the first 2011 rows. (Here, the $n+1$st row of Pascal's triangle consists of entries of the form $\textstyle\binom{n}{k}$ for integers $0\le k\le n$. Thus, the entry $\textstyle\binom{n}{k}$ is considered adjacent to the entries $\textstyle\binom{n-1}{k-1}$, $\textstyle\binom{n-1}{k}$, $\textstyle\binom{n}{k-1}$, $\textstyle\binom{n}{k+1}$, $\textstyle\binom{n+1}{k}$, $\textstyle\binom{n+1}{k+1}$.) [i]Linus Hamilton.[/i]

Ben has a big blackboard, initially empty, and Francisco has a fair coin. Francisco flips the coin $2013$ times. On the $n^{\text{th}}$ flip (where $n=1,2,\dots,2013$), Ben does the following if the coin flips heads: (i) If the blackboard is empty, Ben writes $n$ on the blackboard. (ii) If the blackboard is not empty, let $m$ denote the largest number on the blackboard. If $m^2+2n^2$ is divisible by $3$, Ben erases $m$ from the blackboard; otherwise, he writes the number $n$. No action is taken when the coin flips tails. If probability that the blackboard is empty after all $2013$ flips is $\frac{2u+1}{2^k(2v+1)}$, where $u$, $v$, and $k$ are nonnegative integers, compute $k$. [i]Proposed by Evan Chen[/i]

In the arithmetic triangle below each number (apart from those in the first row) is the sum of the two numbers immediately above. $0 \, 1\, 2\, 3 \,4\, ... \,1991 \,1992\, 1993$ $\,\,1\, 3\, 5 \,7\, ......\,\,\,\,3983 \,3985$ $\,\,\,4 \,8 \,12\, .......... \,\,\,7968$ ······································· Prove that the bottom number is a multiple of $1993$.

For any polynomial $P(x)=a_0+a_1x+\ldots+a_kx^k$ with integer coefficients, the number of odd coefficients is denoted by $o(P)$. For $i-0,1,2,\ldots$ let $Q_i(x)=(1+x)^i$. Prove that if $i_1,i_2,\ldots,i_n$ are integers satisfying $0\le i_1<i_2<\ldots<i_n$, then: \[ o(Q_{i_{1}}+Q_{i_{2}}+\ldots+Q_{i_{n}})\ge o(Q_{i_{1}}). \]