Number System - Aptitude MCQ Questions and Solutions with Explanations

Let 10531 + 4813 - 728 = x × 87
Then, (15344 - 728) = 87 × x
x = $$\frac{14616}{87}$$ = 168

$$\eqalign{ & \Leftrightarrow {n^{65}} - n \cr & = n\left( {{n^{64}} - 1} \right) \cr & = n\left( {{n^{32}} - 1} \right)\left( {{n^{32}} + 1} \right) \cr & = n\left( {{n^{16}} - 1} \right)\left( {{n^{16}} + 1} \right)\left( {{n^{32}} + 1} \right) \cr} $$
  $$ = n\left( {{n^8} - 1} \right)\left( {{n^8} + 1} \right)\left( {{n^{16}} + 1} \right)$$     $$\left( {{n^{32}} + 1} \right)$$
  $$ = n\left( {{n^4} - 1} \right)\left( {{n^4} + 1} \right)\left( {{n^8} + 1} \right)$$     $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$
  $$ = n\left( {{n^2} - 1} \right)\left( {{n^2} + 1} \right)\left( {{n^4} + 1} \right)$$     $$\left( {{n^8} + 1} \right)$$  $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$
  $$ = \left( {n - 1} \right)n\left( {n + 1} \right)\left( {{n^2} + 1} \right)\left( {{n^4} + 1} \right)$$       $$\left( {{n^8} + 1} \right)$$  $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{64}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$
Clearly, (n - 1), n and (n + 1) are three consecutive numbers and they have to be multiples of 2, 3 and 4 as n is odd.
Thus, the given number is definitely a multiple of 24

Let 414 × x × 7 = 127512
Then,
$$\eqalign{ & x = \frac{{127512}}{{414 \times 7}} \cr & \,\,\,\,\,\, = \frac{{18216}}{{414}} \cr & \,\,\,\,\,\, = \frac{{2024}}{{46}} \cr & \,\,\,\,\,\, = \frac{{1012}}{{23}} \cr & \,\,\,\,\,\, = 44 \cr} $$

In 978626, we have (6 + 6 + 7) - (2 + 8 + 9) = 0
Hence, 978626 is completely divisible by 11

Given expression :
= 884697 - (773697 +102479)
= 884697 - 876176
= 8521

Let the dividend be (x + 71) and the divisor be y.
Then, [2(x + 71) - 43] is divisible by y
⇒ (2x + 142 - 43) is divisible by y
⇒ (2x + 99) is divisible by y
∴ Divisor = 99

Shortcut method
Divisor = (2 × 71 - 43) = (142 - 43) = 99

30⁷ × 22⁵ × 34¹¹
= (2 × 3 × 5)⁷ × (2 × 11)⁵ × (2 × 7)¹¹
= 2^{(7 + 5 + 11)} × 3⁷ × 5⁷ × 11⁵ × 17¹¹
= (2²³ × 3⁷ × 5⁷ × 11⁵ × 17¹¹)
Number of prime factors
= (23 + 7 + 7 + 5 + 11)
= 53

= (217 × 217) + (183 × 183)
= (217)² + (183)²
= (200 + 17)² + (200 - 17)²
= (a + b)² + (a - b)²
= 2(a² + b²), where a = 200, b = 17
= 2 [(200)² + (17)² ]
= 2 [40000 + 289]
= (2 × 40289)
= 80578

$$\eqalign{ & = 1260 \div 14 \div 9 \cr & = \left( {1260 \times \frac{1}{{14}} \times \frac{1}{9}} \right) \cr & = 10 \cr} $$

Statement 1. Let x = 4 and y = 15. Then, each one of x and y is a composite number.
But, x + y = 19, which is not composite.
∴ Statement 1 is not true.

Statement 2. We know that 1 is neither prime nor composite.
∴ Statement 2 is not true.
Thus, neither 1 nor 2 is correct.