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: \[ o(Q_{i_{1}}+Q_{i_{2}}+\ldots+Q_{i_{n}})\ge o(Q_{i_{1}}). \]

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$.

Feeling excited over her successful explorations into Pascal's Triangle, Wendy formulates a second problem to use during a future Jupiter Falls High School Math Meet: \[\text{How many of the first 2010 rows of Pascal's Triangle (Rows 0 through 2009)} \ \text{have exactly 256 odd entries?}\] What is the solution to Wendy's second problem?

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.

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,)

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

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.

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 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 $1019$.

Let $p$ be a prime. A [i]complete residue class modulo $p$[/i] is a set containing at least one element equivalent to $k \pmod{p}$ for all $k$. (a) Show that there exists an $n$ such that the $n$th row of Pascal's triangle forms a complete residue class modulo $p$. (b) Show that there exists an $n \le p^2$ such that the $n$th row of Pascal's triangle forms a complete residue class modulo $p$.

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 given triangular number table is as follows: [img]https://cdn.artofproblemsolving.com/attachments/a/0/123b7511850047f3cc51494f107703f2757085.png[/img] Among them, the numbers in the first row are $1, 2, 3, ..., 98, 99, 100$. Starting from the second row, each number is equal to the sum of the left and right numbers in the row above it. Find the value of $M$.

Find all positive integers $n$ such that there are $\infty$ many lines of Pascal's triangle that have entries coprime to $n$ only. In other words: such that there are $\infty$ many $k$ with the property that the numbers $\binom{k}{0},\binom{k}{1},\binom{k}{2},...,\binom{k}{k}$ are all coprime to $n$.

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.

If $(3x-1)^7 = a_7x^7 + a_6x^6 + \cdots + a_0$, then $a_7 + a_6 + \cdots + a_0$ equals \[ \text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 64 \qquad \text{(D)}\ -64 \qquad \text{(E)}\ 128 \]

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]