Found problems: 14
Consider a sequence of 100 positive integers. Each member of the sequence, starting from the second one, is derived by either multiplying the previous number by 2 or dividing it by 16. Is it possible for the sum of these 100 numbers to be equal to $2^{2023}$?
[i]Proposed by Nika Glunchadze, Georgia[/i]
Let $n>1$ be a natural number and $x_k{}$ be the residue of $n^2$ modulo $\lfloor n^2/k\rfloor+1$ for all natural $k{}$. Compute the sum \[\bigg\lfloor\frac{x_2}{1}\bigg\rfloor+\bigg\lfloor\frac{x_3}{2}\bigg\rfloor+\cdots+\left\lfloor\frac{x_n}{n-1}\right\rfloor.\]
Let $ a, b, c$ be integers satisfying $ 0 < a < c \minus{} 1$ and $ 1 < b < c$. For each $ k$, $ 0\leq k \leq a$, Let $ r_k,0 \leq r_k < c$
be the remainder of $ kb$ when divided by $ c$. Prove that the two sets $ \{r_0, r_1, r_2, \cdots , r_a\}$ and $ \{0, 1, 2, \cdots , a\}$ are different.
We say that an integer $a$ is a quadratic, cubic, or quintic residue modulo $n$ if there exists an integer $x$ such that $x^2\equiv a \pmod n$, $x^3 \equiv a \pmod n$, or $x^5 \equiv a \pmod n$, respectively. Further, an integer $a$ is a primitive residue modulo $n$ if it is exactly one of these three types of residues modulo $n$.
How many integers $1 \le a \le 2015$ are primitive residues modulo $2015$?
[i] Proposed by Michael Ren [/i]
For an odd number n, we write $n!! = n\cdot (n-2)...3 \cdot 1$.
How many different residues modulo $1000$ do you get from $n!!$ for $n= 1, 3, 5, …$?
Find all functions $f:\mathbb{Z} \to \mathbb{Z}$ such that the numbers
\[n, f(n),f(f(n)),\dots,f^{m-1}(n)\]
are distinct modulo $m$ for all integers $n,m$ with $m>1$.
(Here $f^k$ is defined by $f^0(n)=n$ and $f^{k+1}(n)=f(f^{k}(n))$ for $k \ge 0$.)
For an odd positive integer $n>1$ define $S_n$ to be the set of the residues of the powers of two, modulo $n{}$. Do there exist distinct $n{}$ and $m{}$ whose corresponding sets $S_n$ and $S_m$ coincide?
[i]Proposed by D. Kuznetsov[/i]
For positive integers $n, m$ define $f(n,m)$ as follows. Write a list of $ 2001$ numbers $a_i$, where $a_1 = m$, and $a_{k+1}$ is the residue of $a_k^2$ $mod \, n$ (for $k = 1, 2,..., 2000$). Then put $f(n,m) = a_1-a_2 + a_3 -a_4 + a_5- ... + a_{2001}$. For which $n \ge 5$ can we find m such that $2 \le m \le n/2$ and $f(m,n) > 0$?
Find all functions $f:\mathbb{N} \rightarrow \mathbb{N}$ such that for every prime number $p$ and natural number $x$,
$$\{ x,f(x),\cdots f^{p-1}(x) \} $$
is a complete residue system modulo $p$. With $f^{k+1}(x)=f(f^k(x))$ for every natural number $k$ and $f^1(x)=f(x)$.
[i]Proposed by IndoMathXdZ[/i]
Let $a_1,\ldots , a_{p-2}{}$ be nonzero residues modulo an odd prime $p{}$. For every $d\mid p - 1$ there are at least $\lfloor(p - 2)/d\rfloor$ indices $i{}$ for which $p{}$ does not divide $a_i^d-1$. Prove that the product of some of $a_1,\ldots , a_{p-2}$ gives the remainder two modulo $p{}$.
A set of $k$ integers is said to be a [i]complete residue system modulo[/i] $k$ if no two of its elements are congruent modulo $k$. Find all positive integers $m$ so that there are infinitely many positive integers $n$ wherein $\{ 1^n,2^n, \dots , m^n \}$ is a complete residue system modulo $m$.
Do there exist 16 three digit numbers, using only three different digits in all, so that the all numbers give different residues when divided by 16?
[i]Bulgaria[/i]
We say that a prime $p$ is $n$-$\textit{rephinado}$ if $n | p - 1$ and all $1, 2, \ldots , \lfloor \sqrt[\delta]{p}\rfloor$ are $n$-th residuals modulo $p$, where $\delta = \varphi+1$. Are there infinitely many $n$ for which there are infinitely many $n$-$\textit{rephinado}$ primes?
Notes: $\varphi =\frac{1+\sqrt{5}}{2}$. We say that an integer $a$ is a $n$-th residue modulo $p$ if there is an integer $x$ such that $$x^n \equiv a \text{ (mod } p\text{)}.$$
Find all functions $f:\mathbb{N} \rightarrow \mathbb{N}$ such that for every prime number $p$ and natural number $x$,
$$\{ x,f(x),\cdots f^{p-1}(x) \} $$
is a complete residue system modulo $p$. With $f^{k+1}(x)=f(f^k(x))$ for every natural number $k$ and $f^1(x)=f(x)$.
[i]Proposed by IndoMathXdZ[/i]