Number System

When $n$ is divided by $4$, the remainder is $3$. What is the remainder when $2n$ is divided by $4$?

The sum of two consecutive odd number is always divisible by:

The sum of all $3$ -digit numbers which are equal to $11$ times the sum of squares of their digits is

For natural numbers $x$ and $y$, let $\left(x,y\right)$ denote the greatest common divisor of $x$ and $y$. The pairs of natural number $x$ and $y$ with $x\le y$ which satisfy the equation $xy=x+y+\left(x,y\right)$ is

It is given that the number $43361$ can be written as a product of two distinct prime numbers $p_1, p_2.$ Further, assume that there are $42900$ numbers which are less than $43361$ and are co-prime to it. Then, $p_1+p_2$ is

If $a, b, c, d$ are four distinct numbers chosen from the set $\{1, 2, 3\dots ., 9\}$ then the minimum value of $\frac{a}{b}+\frac{c}{d}$ is

Let $S$ be the set of all ordered pairs $(x, y)$ of positive integers satisfying the condition $x^2-y^2=12345678.$ Then

Let $a_1,a_2,a_3,a_4$ be real numbers such that $a_1+a_2+a_3+a_4=0$ and $a_1^2+a_2^2+a_3^2+a_4^2=1.$ Then the smallest possible value of the expression ${\left(a_1-a_2\right)}^2+{\left(a_2-a_3\right)}^2+{\left(a_3-a_4\right)}^2+{\left(a_4-a_1\right)}^2$ lies in the interval
 

If $n$ is any natural number, then $6^n-5^n$ always ends with?

$n-1$ is divisible by $5$, if $n$ is of the form

If $n$ is any natural number, then $6^n-5^n$ always ends with?

Two prime numbers $A, B(A<B)$ are called twin primes if they differ by $2 ($e.g. $11,13,$ or $41, 43, ....)$  If $A$ and $B$ are twin primes with $B>23,$ then what is the largest number that would always divide $A+B ?$

$\mathrm{p}, \mathrm{q}$ and $\mathrm{r}$ are three prime numbers satisfying the relation $\mathrm{p} + \mathrm{q} = \mathrm{r}$ and $\mathrm{p} < \mathrm{q}$. What is the value of $\mathrm{p}$?

a and b are twin prime numbers satisfy the condition $(6\mathrm{K}\pm 1)$ and another prime number c satisfy the condition $(8\mathrm{K}+1)$. Find the least value of a prime number $\mathrm{p}$ which satisfy the condition ${\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{c}}^2+{\mathrm{p}}^2$ and $\mathrm{K}$ is a natural number.

Find the least number which is divisible by $18,24$ and $42$

The generalised form of $108=a\times 100+b\times 10+c$, then find the value of $a+b+c$.

The usual form of $100\times c+10\times b+a$ is $cba$.

A $5$-digit number $\overline{abcde}$, when multiplies by $9$, gives the $5$-digit number $\overline{edcba}$. The sum of the digits in the number is

Consider the following two statements :

I. If $n$ is a composite number, then $n$ divides $\left(n-1\right)!$

II. There are infinitely many natural numbers $n$ such that $n^3+2n^2+n$ divides $n!$

Then

Let $\left[x\right]$ be the greatest integer less than or equal to $x$, for a real number $x$. Then the following sum

$\left[\frac{2^{2020}+1}{2^{2018}+1}\right]+\left[\frac{3^{2020}+1}{3^{2018}+1}\right]+\left[\frac{4^{2020}+1}{4^{2018}+1}\right]+\left[\frac{5^{2020}+1}{5^{2018}+1}\right]+\left[\frac{6^{2020}+1}{6^{2018}+1}\right]$

is :-