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 $p$.

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]

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 $3: 4: 5$?

[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 $\textbf{(A) }\dfrac{n^2-n}{2n-1}\qquad\textbf{(B) }\dfrac{n^2-n}{4n-2}\qquad\textbf{(C) }\dfrac{n^2-2n}{2n-1}\qquad\textbf{(D) }\dfrac{n^2-3n+2}{4n-2}\qquad \textbf{(E) }\text{None of these}$

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]

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 probability that any given row in Pascal’s Triangle contains a perfect square. [i] (.1 point)[/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.

In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in-box. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss delivers them in the order 1, 2, 3, 4, 5, 6, 7, 8, 9. While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing else about the morning's typing. The colleague wonder which of the nine letters remain to be typed after lunch and in what order they will be typed. Based upon the above information, how many such after-lunch typing orders are possible? (That there are no letters left to be typed is one of the possibilities.)

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$

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 $1993$.

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]

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

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,)

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]

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.