Number System - Aptitude MCQ Questions and Solutions with Explanations

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.

Let the natural number (n) be 1.
∴ n(n + 1)(n + 2)
= 1(1 + 1)(1 + 2)
= 6

9991 × 10009
= 9991(10000 + 9)
= 99910000 + 89919
= 99999919

Alternate solution
9991 × 10009
= (10000 - 9)(10000 + 9)
= (10000)² - (9)²
= 100000000 - 81
= 99999919

Let the number be = x
Then according to question
21 × x + 100 = x²
x² - 21x - 100 = 0
x² - 25x + 4x - 100 = 0
x(x - 25) + 4(x - 25) = 0
(x - 25)(x + 4) = 0
x = -4 is not possible
x = 25

x = (633)²⁴ - (277)³⁸ + (266)⁵⁴
x = 1 - 9 + 6
x = 7 - 9
x = . . . . . 7 - 9
x = 17 - 9
x = 8

S₁ = 2, 9, 16, . . . . . ., 632
S₂ = 7, 11, 15, . . . . . ., 743
S₁ = 2, 9, 16, 23, 30, 37, 44, 51 . . . . . .
S₂ = 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, . . . . . .
Number of terms in $${{\text{S}}_1} = \frac{{632 - 2}}{7} + 1 = 91$$
∵ In S₁ there is no further terms beyond 632 therefore there is no common terms.
∴ Number of terms in series S₁ = 91
If this series start from the 23 then number of terms = 91 - 3 = 88
Now, number of common terms in both series S₁ and S₂ = $$\frac{{88}}{7}$$ = 22 = n
Now, common terms 23, 51, . . . . . . 22
$$\eqalign{ & {\text{Sum}} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = 11\left[ {46 + 21 \times 28} \right] \cr & = 11\left[ {46 + 588} \right] \cr & = 11 \times 634 \cr & = 6974 \cr} $$

$$\eqalign{ & {\text{Sum of A}}{\text{.P}}{\text{.}} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & {n^{{\text{th}}}}{\text{term}} = a + \left( {n - 1} \right)d \cr & {3^{{\text{rd}}}}{\text{term}} = a + \left( {3 - 1} \right)d = 19 \cr & a + 2d = 19\,......\,\left( {\text{i}} \right) \cr & {\text{Sum of 10 term}} = \frac{{10}}{2}\left[ {2a + 9d} \right] = 390 \cr & 2a + 9d = 78\,......\,\left( {{\text{ii}}} \right) \cr & a + 2d = 19\,......\,\left( {\text{i}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \boxed{a = 3}\,\,\boxed{d = 8} \cr} $$

x = (164)¹⁶⁹ + (333)³³⁷ - (727)⁷²⁶
x = 4 + 3 - 9
x = . . . . . . 7 - 9
x = 17 - 9
x = 8

$$\eqalign{ & \frac{{{{10}^{100}}}}{{{5^{75}}}} \cr & = \frac{{{2^{100}} \times {5^{100}}}}{{{5^{75}}}} \cr & = {2^{100}} \times {5^{25}} \cr & = {2^{25}} \times {2^{75}} \times {5^{25}} \cr & = {2^{75}} \times {10^{25}} \cr} $$

Since x, y and z are distinct prime number (x < y < z)

y + z = 70 - 2 = 68
31 + 37 = 68
Hence, x = 2, y = 31 and $$\boxed{{\text{z}} = 37}{\text{ Answer}}$$