Sum of natural number from 1 to 10
$$\eqalign{ & = \frac{{\left( n \right)\left( {n + 1} \right)}}{2}\,\,\,\,\,\,\,\left( {{\text{where }}n = 10} \right) \cr & = \frac{{10 \times 11}}{2} \cr & = 55 \cr} $$
Clearly, we can see the number which is added twice = 58 - 55 = 3

$$\eqalign{ & \frac{{334 \times 545 \times 7p}}{{3340}} \cr & \Rightarrow \frac{{334 \times 545 \times 7p}}{{334 \times 10}} \cr & \Rightarrow \frac{{109 \times 7p}}{2} \cr & {\text{Go through by options take 2}} \cr & \Rightarrow \frac{{109 \times 72}}{2} \cr & {\text{So, answer is 2}} \cr} $$

q > p and r > s
q > p > r > s
All number have three prime factor
One digit → s, r
s → 2         22 = 4
r → 3         32 = 9
Two digit → p, q
p → 5     52 = 25
q → 7     72 = 49
$$\eqalign{ & \frac{{p - q - 1}}{{r - s}} = \frac{{25 - 49 - 1}}{{9 - 4}} = \frac{{ - 25}}{5} = - 5 \cr & - 5 = - 4 - 1 = - s - 1 \cr} $$

We know, smallest five digit number is = 10000
LCM of 12, 18, 21 = 252

Difference = 252 - 172 = 80
Then number is = 10000 + 80 = 10080

$$\frac{1}{{24 \times 7}} = 0.0059$$

Let numbers are x, x + 5, x + 10
So, x + x + 5 + x + 10 = 225
3x + 15 = 225
3x = 210
x = 70
Largest number = 70 + 10 = 80

Let, 6 digit number after rearranging the digit divisible by 198 is = 0000891
This number is divisible by 3

Given:
674pq0 is divisible by both 3 and 11
Concept used:
If the sum of digits of a number is a multiple of 3, the number will be completely divisible by 3
If the difference of the sum of alternative digits of a number is 0 or divisible by 11, then that number is divisible by 11 completely.
Calculation:
As 674pq0 divisible by 3
So, 6 + 7 + 4 + p + q + 0
⇒ 17 + p + q also divisible by 3
Now,
Possible values of p + q = 4, 7, 10, 13, 16
From the options
Option A and Option D is satisfying
Now,
For Option A
The number is 674220
So, 6 + 4 + 2 - 7 - 2 = 3 Not divisible by 11
For Option D
The number is 674520
So, 6 + 4 + 2 - 7 - 5 = 0 divisible by 11
∴ The required answer is Option D.

99 can be written as 64 + 32 + 2 + 1 = 26 + 25 + 21 + 20
⇒ 99 in binary is 1100011
∴ 99 in binary is 1100011

$$\eqalign{ & 99\frac{{11}}{{99}} + 99\frac{{13}}{{99}} + 99\frac{{15}}{{99}} + ........ + 99\frac{{67}}{{99}} \cr & {\text{Number of term}} = \frac{{67 - 11}}{2} + 1 = 29 \cr & = 99 \times 29 + \frac{1}{{99}}\left( {11 + 13 + 15 + ........ + 67} \right) \cr & = 2871 + \frac{1}{{99}}\left[ {\frac{{29}}{2}\left( {2 \times 11 + \left( {29 - 1} \right)2} \right)} \right] \cr & = 2871 + \frac{1}{{99}}\left[ {\frac{{99}}{2}\left( {22 + 56} \right)} \right] \cr & = 2871 + \frac{1}{{99}}\left( {29 \times 39} \right) \cr & = 2871 + \frac{{1131}}{{99}} \cr & = 2871 + \frac{{377}}{{33}} \cr & = \frac{{95120}}{{33}} \cr} $$

Let the number be x, x + 1, x + 2
Sum = x + x + 1 + x + 2
= 3x + 3
= 3(x + 1)
It is always divisible by 3

196 = 72 × 22
Factors ⇒ (1, 7, 49) × (1, 2, 4)
Divisible by 4 = 4, 28, 196 = 3 factors

Divisibility rule of 3 : The sum of all digit should be divide by 3

L.C.M. of 15,18, 20, 27 = 540
x = 540K + 10

x = 540 × 4 + 10 = 2170
462 < 2170 > 472
472 = 2209
To make perfect square = 2209 - 2170 = 39

Let the three number is a - d, a, a + d
a is first term, d is common difference
a + d + a + a - d = 30     (Given)
3a = 30
a = 10
(a + d)(a)(a - d) = 910
(10 + d)(10)(10 - d) = 91 × 10
(10 + d)(10 - d) = 91
Put d = 3
$$\boxed{\left( {10 + 3} \right)\left( {10 - 3} \right) = 91}$$
So, d = 3
So, greater number is = a + b = 10 + 3 = 13

Largest 4 digit number which is exactly divisible by 93
= 9999 - Remainder
= 9999 - 48
= 9951

$$\eqalign{ & {\text{A}} = \frac{{\sqrt {0.0004} \times \root 3 \of {0.000008} }}{{\root 4 \of {16000} \times \root 3 \of {125000} \times \root 4 \of {810} }} \cr & = \frac{{0.02 \times 0.02}}{{\root 4 \of {16000 \times 810} \times \root 3 \of {125000} }} \cr & = \frac{{0.02 \times 0.02}}{{20 \times 3 \times 50}} \cr & = \frac{1}{{75 \times {{10}^5}}} \cr & {\text{B}} = \frac{{\root 3 \of {0.729} \times \root 4 \of {0.0016} }}{{\sqrt {0.16} }} \cr & = \frac{{0.9 \times 0.2}}{{0.4}} \cr & = \frac{9}{{20}} \cr & \therefore A \times B = \frac{1}{{75 \times {{10}^5}}} \times \frac{9}{{20}} \cr & = \frac{3}{{25 \times 2}} \times \frac{1}{{{{10}^6}}} \cr & = \frac{3}{5} \times \frac{1}{{{{10}^7}}} \cr & = \frac{3}{5} \times {10^{ - 7}} \cr & = 6 \times {10^{ - 8}} \cr} $$

$$\eqalign{ & {\text{Lets first number}} = a \cr & {\text{Second number}} = b \cr & a + b = 65 \cr & \sqrt {a \times b} = 26 \cr & ab = 676 \cr & {\text{Sum of reciprocal}} = \frac{1}{a} + \frac{1}{b} \cr & = \frac{{b + a}}{{ab}} \cr & = \frac{{65}}{{676}} \cr & = \frac{5}{{52}} \cr} $$

We know that after 5! we get one zero at the end of the number.
We can say that in 17! we get minimum three zeros.
The hundred's place value od 17! = 0

K = 42 × 25 × 54 × 135
K = 3 × 14 × 25 × 33 × 2 × 33 × 5
= 37 × 14 × 25 × 2 × 5
= It is divided by 37, which is equal to 3a   (given)
⇒ So, 37 = 3a
a = 7