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]

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.)

Find the probability that any given row in Pascal’s Triangle contains a perfect square. [i] (.1 point)[/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 $b_1 = 3, b_2 = 2, b_3 = 1$. \[\begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1 & & \underset{b_3}{1} & & & & \\ & & & 1 & & \underset{b_2}{2} & & 1 & & & \\ & & 1 & & \underset{b_1}{3} & & 3 & & 1 & & \\ & 1 & & \boxed{4} & & \underset{a_1}{6} & & \underset{a_2}{4} & & \underset{a_3}{1} & \\ \ldots & & \ldots & & \ldots & & \ldots & & \ldots & & \ldots \\ \end{array}\]

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

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 $1$.

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?

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?

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$.

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

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

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

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 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

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 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]

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

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 $S$ 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?

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 $3: 4: 5$?

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]

Compute the prime factorization of $1007021035035021007001$. (You should write your answer in the form $p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k}$ where $p_1,\ldots,p_k$ are distinct prime numbers and $e_1,\ldots,e_k$ are positive integers.)

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]