Number System - Aptitude MCQ Questions and Solutions with Explanations

Given expression
$$\eqalign{ & = \frac{{{{\left( {256} \right)}^2} - {{\left( {144} \right)}^2}}}{{112}} \cr & = \frac{{\left( {256 + 144} \right)\left( {256 - 144} \right)}}{{112}} \cr & = \frac{{\left( {400 \times 112} \right)}}{{112}} \cr & = 400 \cr} $$

Divisor taken = 12
Quotient obtained = 35,
Remainder = 0
∴ Dividend = (12 × 35) = 420
Now, Dividend = 420,
Divisor = 21
Remainder = 0
∴ Quotient = $$\frac{420}{21}$$ = 20

Prime numbers between 0 and 50 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47
Their number is 15

Let x × 48 = 173 × 240
Then,
x = $$\frac{173 × 240}{48}$$
x = 173 × 5
x = 865

Required number = 2047

When n is even, (xⁿ - aⁿ) is divisible by (x + a)
Now, 2²⁵⁶ = (2⁴)⁶⁴ = (16)⁶⁴
∴ (16⁶⁴ - 1⁶⁴) is divisible by (16 + 1)
⇒ (16⁶⁴ - 1) is divisible by 17
⇒ (2²⁵⁶ - 1) is divisible by 17
⇒ On dividing 2²⁵⁶ by 17, we get 1 as remainder.

Let, 5566 - 7788 + 9988 = x + 4444
Then,
(5566 + 9988) - 7788 = x + 4444
⇒ 15554 - 7788 = x + 4444
⇒ x + 4444 = 7766
⇒ x = (7766 - 4444)
⇒ x = 3322

Let the hundreds, tens and unit digits be x, y and z respectively.
Then,
(100x + 10y + z) - (x + y + z)
= 99x + 9y
= 9 (11x + y)
So, the resulting number is divisible by 9

Let N = 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50
= 5¹⁰ × ( 1 × 2 × 3 × 4 × ..... × 10) = 5¹⁰ × 10!
Highest power of 2 in 10!
$$ = \left[ {\frac{{10}}{2}} \right] + \left[ {\frac{{10}}{{{2^2}}}} \right] + \left[ {\frac{{10}}{{{2^3}}}} \right]$$
= 5 + 2 + 1
= 8
Highest power of 5 in 10! = $$\left[ {\frac{{10}}{5}} \right]$$ = 2
∴ N = 2⁸ × 5¹² × k
Since highest power of 2 is less than that of 5,
So, required number of zeros = 8

Let,
8888 + 848 +88 - x = 7337 + 737
Then,
9824 - x = 8074
x = 9824 - 8074
x = 1750