Found problems: 1340
Prove the following inequality.
\[\frac{e-1}{n+1}\leqq\int^e_1(\log x)^n dx\leqq\frac{(n+1)e+1}{(n+1)(n+2)}\ (n=1,2,\cdot\cdot\cdot) \]
1994 Kyoto University entrance exam/Science
Let $d_i(k)$ the number of divisors of $k$ greater than $i$.
Let $f(n)=\sum_{i=1}^{\lfloor \frac{n^2}{2} \rfloor}d_i(n^2-i)-2\sum_{i=1}^{\lfloor \frac{n}{2} \rfloor}d_i(n-i)$.
Find all $n \in N$ such that $f(n)$ is a perfect square.
At a competition with $N$ players, the number of players given elite status is equal to \[2^{1+\lfloor\log_2{(N-1)}\rfloor} - N. \] Suppose that $19$ players are given elite status. What is the sum of the two smallest possible values of $N$?
$ \textbf{(A)}\ 38\qquad
\textbf{(B)}\ 90 \qquad
\textbf{(C)}\ 154 \qquad
\textbf{(D)}\ 406 \qquad
\textbf{(E)}\ 1024$
Find the number of integer solutions of $\left[\frac{x}{100} \left[\frac{x}{100}\right]\right]= 5$
A natural number $n$ is given. Determine all $(n - 1)$-tuples of nonnegative integers $a_1, a_2, ..., a_{n - 1}$ such that
$$\lfloor \frac{m}{2^n - 1}\rfloor + \lfloor \frac{2m + a_1}{2^n - 1}\rfloor + \lfloor \frac{2^2m + a_2}{2^n - 1}\rfloor + \lfloor \frac{2^3m + a_3}{2^n - 1}\rfloor + ... + \lfloor \frac{2^{n - 1}m + a_{n - 1}}{2^n - 1}\rfloor = m$$
holds for all $m \in \mathbb{Z}$.
If $a$ is any number, $\lfloor a \rfloor$ is $a$ rounded down to the nearest integer. For example, $\lfloor \pi \rfloor =$ $3$.
Show that the sequence
$\lfloor \frac{2^{1}}{17} \rfloor$, $\lfloor \frac{2^{2}}{17} \rfloor$, $\lfloor \frac{2^{3}}{17} \rfloor$, $\dots$
contains infinitely many odd numbers.
Find the total number of primes $p<100$ such that $\lfloor (2+\sqrt{5})^p \rfloor-2^{p+1}$ is divisible by $p$. Here $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.
For each positive real number $\alpha$, define
$$\lfloor \alpha \mathbb{N}\rfloor :=\{\lfloor \alpha m \rfloor\; |\; m\in \mathbb{N}\}.$$
Let $n$ be a positive integer. A set $S\subseteq \{1,2,\ldots,n\}$ has the property that: for each real $\beta >0$,
$$ \text{if}\; S\subseteq \lfloor \beta \mathbb{N} \rfloor, \text{then}\; \{1,2,\ldots,n\} \subseteq \lfloor \beta\mathbb{N}\rfloor.$$
Determine, with proof, the smallest positive size of $S$.
Let $k$ be a fixed positive integer. Consider the sequence definited by \[a_{0}=1 \;\; , a_{n+1}=a_{n}+\left\lfloor\root k \of{a_{n}}\right\rfloor \;\; , n=0,1,\cdots\] where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. For each $k$ find the set $A_{k}$ containing all integer values of the sequence $(\sqrt[k]{a_{n}})_{n\geq 0}$.
What is the largest integer less than to $\sqrt[3]{(2011)^3 + 3 \times (2011)^2 + 4 \times 2011+ 5}$ ?
(A) $2010$, (B) $2011$, (C) $2012$, (D) $2013$, (E) None of the above.
Prove that the function $f : \mathbb{N}\longrightarrow \mathbb{Z}$ defined by $f(n) = n^{2007}-n!$, is injective.
In the beginning, there is a pair of positive integers $(m,n)$ written on the board. Alice and Bob are playing a turn-based game with the following move. At each turn, a player erases one of the numbers written on the board, and writes a different positive number not less than the half of the erased one. If a player cannot write a new number at some turn, he/she loses the game. For how many starting pairs $(m,n)$ from the pairs $(7,79)$, $(17,71)$, $(10,101)$, $(21,251)$, $(50,405)$, can Alice guarantee to win when she makes the first move?
$
\textbf{(A)}\ 4
\qquad\textbf{(B)}\ 3
\qquad\textbf{(C)}\ 2
\qquad\textbf{(D)}\ 1
\qquad\textbf{(E)}\ \text{None of above}
$
Let $m, k$, and $c$ be positive integers with $k > c$, and let $\lambda$ be a positive, non-integer real root of the equation $\lambda^{m+1} - k \lambda^m - c = 0$. Let $f : Z^+ \to Z$ be defined by $f(n) = \lfloor \lambda n \rfloor$ for all $n \in Z^+$. Show that $f^{m+1}(n) \equiv cn - 1$ (mod $k$) for all $n \in Z^+$. (Here, $Z^+$ denotes the set of positive integers, $ \lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$, and $f^{m+1}(n) = f(f(... f(n)...))$ where $f$ appears $m + 1$ times.)
Prove that $[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4n+1}]$ for all $n \in N$.
Recall that $[x]$ denotes the largest integer not exceeding $x$ for real $x$. For integers $a, b$ in the interval $1 \leq a<b \leq 100$, find the number of ordered pairs $(a, b)$ satisfying the following equation.
$$[a+\frac{b}{a}]=[b+\frac{a}{b}]$$
We say that a natural number $n$ is interesting if it can be written in the form
\[
n = \left\lfloor \frac{1}{a} \right\rfloor + \left\lfloor \frac{1}{b} \right\rfloor + \left\lfloor \frac{1}{c} \right\rfloor,
\] where $a,b,c$ are positive real numbers such that $a + b + c = 1.$
Determine all interesting numbers. ( $\lfloor x \rfloor$ denotes the greatest integer not greater than $x$.)
Determine the number of positive divisors of $2008^8$ that are less than $2008^4$.
A natural number $n$ is given. Determine all $(n - 1)$-tuples of nonnegative integers $a_1, a_2, ..., a_{n - 1}$ such that
$$\lfloor \frac{m}{2^n - 1}\rfloor + \lfloor \frac{2m + a_1}{2^n - 1}\rfloor + \lfloor \frac{2^2m + a_2}{2^n - 1}\rfloor + \lfloor \frac{2^3m + a_3}{2^n - 1}\rfloor + ... + \lfloor \frac{2^{n - 1}m + a_{n - 1}}{2^n - 1}\rfloor = m$$
holds for all $m \in \mathbb{Z}$.
If $N=\lfloor \frac{2}{5} \rfloor + \lfloor \frac{2^2}{5} \rfloor +\dots \lfloor \frac{2^{2009}}{5} \rfloor$, what is the remainder when $2^{2010}$ is divided by $N$?
$ \textbf{(A)}\ 5034
\qquad\textbf{(B)}\ 5032
\qquad\textbf{(C)}\ 5031
\qquad\textbf{(D)}\ 5028
\qquad\textbf{(E)}\ 5024
$
For every real number $x$, let $\lfloor x\rfloor$ denote the greatest integer not exceeding $x$, and let \[f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).\] The set of all numbers $x$ such that $1\leq x<2014$ and $f(x)\leq 1$ is a union of disjoint intervals. What is the sum of the lengths of those intervals?
$\textbf{(A) }1\qquad
\textbf{(B) }\dfrac{\log 2015}{\log 2014}\qquad
\textbf{(C) }\dfrac{\log 2014}{\log 2013}\qquad
\textbf{(D) }\dfrac{2014}{2013}\qquad
\textbf{(E) }2014^{\frac1{2014}}\qquad$
Let $a_1,a_2,a_3,\ldots,a_n$ be a sequence of non-negative integers, where $n$ is a positive integer. Let
\[ A_n={a_1+a_2+\cdots+a_n\over n}\ . \]
Prove that
\[ a_1!a_2!\ldots a_n!\ge\left(\lfloor A_n\rfloor !\right)^n \]
where $\lfloor A_n\rfloor$ is the greatest integer less than or equal to $A_n$, and $a!=1\times 2\times\cdots\times a$ for $a\ge 1$(and $0!=1$). When does equality hold?
The numbers from 1 to $ 2013^2 $ are written row by row into a table consisting of $ 2013 \times 2013 $ cells. Afterwards, all columns and all rows containing at least one of the perfect squares $ 1, 4, 9, \cdots, 2013^2 $ are simultaneously deleted.
How many cells remain?
A person flips $2010$ coins at a time. He gains one penny every time he flips a prime number of heads, but must stop once he flips a non-prime number. If his expected amount of money gained in dollars is $\frac{a}{b}$, where $a$ and $b$ are relatively prime, compute $\lceil\log_{2}(100a+b)\rceil$.
[i]Proposed by Lewis Chen[/i]
Find all positive integers $n$ such that $$\lfloor \sqrt{n} \rfloor +
\left\lfloor \frac{n}{\lfloor \sqrt{n} \rfloor} \right \rfloor> 2\sqrt{n}.$$
If $k$ is a real number, then $\lfloor k \rfloor$ means the floor of $k$, this is the greatest integer less than or equal to $k$.
Find the largest positive integer that is equal to the cube of the sum of its digits.