Found problems: 310
For any integer $n$ greater than $1$, the number of prime numbers greater than $n!+1$ and less than $n!+n$ is:
$\textbf{(A) }0\qquad
\textbf{(B) }1\qquad
\textbf{(C) }\dfrac n2\text{ for }n\text{ even,}\,\dfrac{n+1}2\text{ for }n\text{ odd}$
$\textbf{(D) }n-1\qquad
\textbf{(E) }n$
Find all positive integers $n$ such that $(n + 1)!$ is divisible by $1! + 2! + ... + n!$.
Determine the largest integer $k$ for which $12^k$ is a factor of $120! $
Which is greater: $300!$ or $100^{300}$?
Let $a$ and $b$ be positive integers such that $a! + b!$ divides $a!b!$. Prove that $3a \ge 2b + 2$.
A permutation $ \pi$ is selected randomly through all $ n$-permutations.
a) if \[ C_a(\pi)\equal{}\mbox{the number of cycles of length }a\mbox{ in }\pi\] then prove that $ E(C_a(\pi))\equal{}\frac1a$
b) Prove that if $ \{a_1,a_2,\dots,a_k\}\subset\{1,2,\dots,n\}$ the probability that $ \pi$ does not have any cycle with lengths $ a_1,\dots,a_k$ is at most $ \frac1{\sum_{i\equal{}1}^ka_i}$
Three legendary knights are fighting against a multiheaded dragon.
Whenever the first knight attacks, he cuts off half of the current number of heads plus one more. Whenever the second knight attacks, he cuts off one third of the current number of heads plus two more. Whenever the third knight attacks, he cuts off one fourth of the current number of heads plus three more. They repeatedly attack in an arbitrary order so that at each step an integer number of heads is being cut off. If all the knights cannot attack as the number of heads would become non-integer, the dragon eats them. Will the knights be able to cut off all the dragon’s heads if it has $41!$ heads?
Alexey Zaslavsky
The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$, where $T$, $M$, and $H$ denote digits that are not given. What is $T+M+H$?
$\textbf{(A) }3
\qquad\textbf{(B) }8
\qquad\textbf{(C) }12
\qquad\textbf{(D) }14
\qquad\textbf{(E) } 17 $
What is the tens digit in the sum $ 7! \plus{} 8! \plus{} 9! \plus{} \cdots \plus{} 2006!$?
$ \textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$
Put $f(n) = \frac{n^n - 1}{n - 1}$. Show that $n!^{f(n)}$ divides $(n^n)! $.
Find as many positive integers as possible for which $n!^{f(n)+1}$ does not divide $(n^n)!$ .
The product $1 \cdot 2 \cdot \ldots \cdot n$ is denoted by $n!$ and called [i]n-factorial[/i]. Prove that the product
\[ 1!2!3!\ldots 49!51! \ldots 100! \]
(the factor $50!$ is missing)
is the square of an integer number.
Prove that there exists a sequence of unbounded positive integers $a_1\leq a_2\leq a_3\leq\cdots$, such that there exists a positive integer $M$ with the following property: for any integer $n\geq M$, if $n+1$ is not prime, then any prime divisor of $n!+1$ is greater than $n+a_n$.
Prove that all positive rational numbers can be written as a fraction, which numerator and denominator are products of factorials of not necessarily different prime numbers. For example
$\frac{10}{9}=\frac{2!5!}{3!3!3!}$.
Prove that $(2m)!(2n)!$ is a multiple of $m!n!(m+n)!$ for any non-negative integers $m$ and $n$.
The number $695$ is to be written with a factorial base of numeration, that is, $695=a_1+a_2\times2!+a_3\times3!+ . . . a_n \times n!$ where $a_1, a_2, a_3 ... a_n$ are integers such that $0 \le a_k \le k$, and $n!$ means $n(n-1)(n-2)...2 \times 1$. Find $a_4$
${{ \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3}\qquad\textbf{(E)}\ 4} $
Consider a non-zero real number $a$ such that $\{a\} + \left\{\frac{1}{a}\right\}=1$, where $\{x\}$ denotes the fractional part of $x$. Prove that for any positive integer $n$, $\{a^n\} + \left\{\frac{1}{a^n}\right\}= 1$.
(a) For each integer $k\ge 3$, find a positive integer $n$ that can be represented as the sum of exactly $k$ mutually distinct positive divisors of $n$.
(b) Suppose that $n$ can be expressed as the sum of exactly $k$ mutually distinct positive divisors of $n$ for some $k\ge 3$. Let $p$ be the smallest prime divisor of $n$. Show that \[\frac1p+\frac1{p+1}+\cdots+\frac{1}{p+k-1}\ge1.\]
If $S=1!+2!+3!+ \cdots +99!$, then the units' digit in the value of $S$ is:
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 0$
We define $N !!$ to be $N(N - 2)(N -4)...5 \cdot 3 \cdot 1$ if $N$ is odd and $N(N -2)(N -4)... 6\cdot 4\cdot 2$ if $N$ is even .
For example, $8 !! = 8 \cdot 6\cdot 4\cdot 2$ , and $9 !! = 9v 7 \cdot 5\cdot 3 \cdot 1$ .
Prove that $1986 !! + 1985 !!$ i s divisible by $1987$.
(V.V . Proizvolov , Moscow)
Find all functions $f:\mathbb{Z}^+ \rightarrow \mathbb{Z}^+$ (where $\mathbb{Z}^+$ is the set of positive integers) such that $f(n!) = f(n)!$ for all positive integers $n$ and such that $m-n$ divides $f(m) - f(n)$ for all distinct positive integers $m, n$.
Kevin has a set $S$ of $2014$ points scattered on an infinitely large planar gameboard. Because he is bored, he asks Ashley to evaluate \[ x = 4f_4 + 6f_6 + 8f_8 + 10f_{10} + \cdots \] while he evaluates \[ y = 3f_3 + 5f_5+7f_7+9f_9 + \cdots, \] where $f_k$ denotes the number of convex $k$-gons whose vertices lie in $S$ but none of whose interior points lie in $S$.
However, since Kevin wishes to one-up everything that Ashley does, he secretly positions the points so that $y-x$ is as large as possible, but in order to avoid suspicion, he makes sure no three points lie on a single line. Find $\left\lvert y-x \right\rvert$.
[i]Proposed by Robin Park[/i]
The digit in unit place of $1!+2!+\ldots+99!$ is
$\textbf{(A)}~3$
$\textbf{(B)}~0$
$\textbf{(C)}~1$
$\textbf{(D)}~7$
Find the last three digits in the product $1 \cdot 3\cdot 5\cdot 7 \cdot . . . \cdot 2009 \cdot 2011$.
Find all natural numbers such that each is equal to the sum of the factorials of its digits.
Please remember to hide your solution. (by using the hide tags of course.. I don't literally mean that you should hide it :ninja: )
Find all positive integers $n$ such that $n!$ can be written as the product of two Fibonacci numbers.