Number System

If all the numbers from $5$ to $80$, which are divisible by $5$ are written in descending order, then which of the numbers will occupy ${11}^{th}$ place from the top ?

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

The least perfect square number which is divisible by$3,4, 5, 6$ and $8$ is_____.

Which of the following number is irrational number?

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

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

If $r$ is the remainder when each of $4749, 5601$ and$7092$ is divided by the greatest possible number $d (>1)$, then the value of $(d + r)$ will be _____.

Write the composite numbers from the following:

$21 12 18 17 11$

Write the composite numbers from the following:

$14 7 9 6 5$

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 :-

Suppose $p,q,r$ are positive rational numbers such that $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is also rational. Then

Let $N_1=2^{55}+1$ and $N_2=165$. Then

Find the negative value of $x$ satisfying the inequality $-3\frac{1}{2}\le 2x+\frac{1}{2}<6\frac{1}{2},x\in R$.
 

Identify the smallest irrational number.

Between which of the following numbers can you find an irrational number?

i) 2 and 4 ii) $ \frac{1}{4}$ and $ \frac{3}{4}$ iii) $ \surd 5$ and $ \surd 7$

Arrange the irrational numbers in ascending order: $ \nthroot{4}{9},\nthroot{3}{5},\nthroot{6}{26}$

$ \sqrt{27}$ + $ \sqrt{243}$ = ____________

Prove that $\frac{1}{\sqrt{2}}$ is irrational.