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 a sequence $u_1,u_2\dots,$ define $\Delta^1(u_n)=u_{n+1}-u_n$ and, for all integer $k>1$, $\Delta^k(u_n)=\Delta^1(\Delta^{k-1}(u_n))$. If $u_n=n^3+n$, then $\Delta^k(u_n)=0$ for all $n$ $\textbf{(A) }\text{if }k=1\qquad$ $\textbf{(B) }\text{if }k=2,\text{ but not if }k=1\qquad$ $\textbf{(C) }\text{if }k=3,\text{ but not if }k=2\qquad$ $\textbf{(D) }\text{if }k=4,\text{ but not if }k=3\qquad$ $\textbf{(E) }\text{for no value of }k$

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 $s \geq 0, 0 < q < p$.

Consider the following array: \[ 3, 5\\3, 8, 5\\3, 11, 13, 5\\3, 14, 24, 18, 5\\3, 17, 38, 42, 23, 5\\ \ldots \] Find the 5-th number on the $ n$-th row with $ n>5$.

[asy] label("$1$",(0,0),S); label("$1$",(-1,-1),S); label("$1$",(-2,-2),S); label("$1$",(-3,-3),S); label("$1$",(-4,-4),S); label("$1$",(1,-1),S); label("$1$",(2,-2),S); label("$1$",(3,-3),S); label("$1$",(4,-4),S); label("$2$",(0,-2),S); label("$3$",(-1,-3),S); label("$3$",(1,-3),S); label("$4$",(-2,-4),S); label("$4$",(2,-4),S); label("$6$",(0,-4),S); label("etc.",(0,-5),S); //Credit to chezbgone2 for the diagram[/asy] Pascal's triangle is an array of positive integers(See figure), in which the first row is $1$, the second row is two $1$'s, each row begins and ends with $1$, and the $k^\text{th}$ number in any row when it is not $1$, is the sum of the $k^\text{th}$ and $(k-1)^\text{th}$ numbers in the immediately preceding row. The quotient of the number of numbers in the first $n$ rows which are not $1$'s and the number of $1$'s is $\textbf{(A) }\dfrac{n^2-n}{2n-1}\qquad\textbf{(B) }\dfrac{n^2-n}{4n-2}\qquad\textbf{(C) }\dfrac{n^2-2n}{2n-1}\qquad\textbf{(D) }\dfrac{n^2-3n+2}{4n-2}\qquad \textbf{(E) }\text{None of these}$

Writing down the first $4$ rows of the Pascal triangle in the usual way and then adding up the numbers in vertical columns, we obtain $7$ numbers as shown above. If we repeat this procedure with the first $1024$ rows of the Pascal triangle, how many of the $2047$ numbers thus obtained will be odd? [img]https://cdn.artofproblemsolving.com/attachments/8/a/4dc4a815d8b002c9f36a6da7ad6e1c11a848e9.png[/img]

Let $S$ be a string of $99$ characters, $66$ of which are $A$ and $33$ are $B$. We call $S$ [i]good[/i] if, for each $n$ such that $1\le n \le 99$, the sub-string made from the first $n$ characters of $S$ has an odd number of distinct permutations. How many good strings are there? Which strings are good?

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]

Consider the Pascal's triangle in the figure where the binomial coefficients are arranged in the usual manner. Select any binomial coefficient from anywhere except the right edge of the triangle and labet it $C$. To the right of $C$, in the horizontal line, there are $t$ numbers, we denote them as $a_1,a_2,\cdots,a_t$, where $a_t = 1$ is the last number of the series. Consider the line parallel to the left edge of the triangle containing $C$, there will only be $t$ numbers diagonally above $C$ in that line. We successively name them as $b_1,b_2,\cdots,b_t$, where $b_t = 1$. Show that \[b_ta_1-b_{t-1}a_2+b_{t-2}a_3-\cdots+(-1)^{t-1}b_1a_t = 1\]. For example, Suppose you choose $\binom41 = 4$ (see figure), then $t = 3$, $a_1 = 6, a_2 = 4, a_3 = 1$ and $b_1 = 3, b_2 = 2, b_3 = 1$. \[\begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1 & & \underset{b_3}{1} & & & & \\ & & & 1 & & \underset{b_2}{2} & & 1 & & & \\ & & 1 & & \underset{b_1}{3} & & 3 & & 1 & & \\ & 1 & & \boxed{4} & & \underset{a_1}{6} & & \underset{a_2}{4} & & \underset{a_3}{1} & \\ \ldots & & \ldots & & \ldots & & \ldots & & \ldots & & \ldots \\ \end{array}\]

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]

Prove that for any positive integer $n$, the number of odd integers among the binomial coefficients $\binom nh \ ( 0 \leq h \leq n)$ is a power of 2.

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

A triangle of numbers is constructed as follows. The first row consists of the numbers from $1$ to $2000$ in increasing order, and under any two consecutive numbers their sum is written. (See the example corresponding to $5$ instead of $2000$ below.) What is the number in the lowermost row? 1 2 3 4 5 3 5 7 9 8 12 16 20 28 4

In the numerical triangle $................1..............$ $...........1 ...1 ...1.........$ $......1... 2... 3 ... 2 ... 1....$ $.1...3...6...7...6...3...1$ $...............................$ each number is equal to the sum of the three nearest to it numbers from the row above it; if the number is at the beginning or at the end of a row then it is equal to the sum of its two nearest numbers or just to the nearest number above it (the lacking numbers above the given one are assumed to be zeros). Prove that each row, starting with the third one, contains an even number.

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: \[ F(n,r)={n+1\over r+1}. \]

In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below. \[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } & & & & & & & 1 & & & & & & \\\vspace{4pt} \text{Row 1: } & & & & & & 1 & & 1 & & & & & \\\vspace{4pt} \text{Row 2: } & & & & & 1 & & 2 & & 1 & & & & \\\vspace{4pt} \text{Row 3: } & & & & 1 & & 3 & & 3 & & 1 & & & \\\vspace{4pt} \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} \text{Row 5: } & & 1 & & 5 & &10& &10 & & 5 & & 1 & \\\vspace{4pt} \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 \end{array}\] In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3: 4: 5$?

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

Find the number of subsets $X$ of $\{1,2,\dots,10\}$ such that $X$ contains at least two elements and such that no two elements of $X$ differ by $1$.

Each number in the second, third, and further rows of the following triangle: [img]https://cdn.artofproblemsolving.com/attachments/1/5/589d9266749477b0f56f0f503d4f18a6e5d695.png[/img] is equal to the difference of two neighbouring numbers standing above it. Find the last number (at the bottom of the triangle). (GW Leibnitz,)

Consider the following array: \[ 3, 5\\3, 8, 5\\3, 11, 13, 5\\3, 14, 24, 18, 5\\3, 17, 38, 42, 23, 5\\ \ldots \] Find the 5-th number on the $ n$-th row with $ n>5$.

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.

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 $s \geq 0, 0 < q < p$.

Let Pascal triangle be an equilateral triangular array of number, consists of $2019$ rows and except for the numbers in the bottom row, each number is equal to the sum of two numbers immediately below it. How many ways to assign each of numbers $a_0, a_1,...,a_{2018}$ (from left to right) in the bottom row by $0$ or $1$ such that the number $S$ on the top is divisible by $1019$.

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]