Number System - Aptitude MCQ Questions and Solutions with Explanations

Given x = a (b - c), y = b (c - a), z = c (a - b)
x = a (b - c)
⇒ $$\frac{x}{a}$$ = b - c ..... (i)
Similarly, y = b (c - a)
⇒ $$\frac{y}{b}$$ = c - a ..... (ii)
And similarly z = c (a - b)
⇒ $$\frac{z}{c}$$ = c - a ..... (iii)
Adding (i), (ii) and (iii) we get
∵ $$\frac{x}{a}$$ + $$\frac{y}{b}$$ + $$\frac{z}{c}$$ = b - c + c - a + a - b
⇒ $$\frac{x}{a}$$ + $$\frac{y}{b}$$ + $$\frac{z}{c}$$ = 0
$$\eqalign{ & \therefore {\left( {\frac{x}{a}} \right)^3} + {\left( {\frac{y}{b}} \right)^3} + {\left( {\frac{z}{c}} \right)^3} \cr & = 3 \times \frac{x}{a} \times \frac{y}{b} \times \frac{z}{c} \cr & = \frac{{3xyz}}{{abc}} \cr} $$
[If a + b + c = 0, a³ + b³ + c³ = 3abc]

Required number
= (2⁵ - 2)
= (32 - 2)
= 30

Given expression,
$$ = \frac{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}}{{\left( {{a^2} + {b^2}} \right)}}$$     (where a = 963 and b = 476)
$$\eqalign{ & = \frac{{2\left( {{a^2} + {b^2}} \right)}}{{\left( {{a^2} + {b^2}} \right)}} \cr & = 2 \cr} $$

Sum of the digits in 4006020 is 12, which is divisible by 3
Hence, 4006020 is divisible by 3

(xⁿ - aⁿ) is divisible by (x - a) for all values of n
∴ (9⁶ - 1⁶) is divisible by (9 - 1)
⇒ (9⁶ - 1) is divisible by 8
⇒ on dividing (9⁶ + 1) by 8, we get 2 as remainder.

Since 13b7 is divisible by 11, we have,
(7 + 3) - (b + 1) = 0
⇒ 9 - b = 0
⇒ b = 9
Putting b = 9, a + 8 = 9 we get a = 1
Hence, (a + b) = (1 + 9) = 10

Given expression,
$$ = \frac{{\left( {{a^3} + {b^3}} \right)}}{{\left( {{a^2} - ab + {b^2}} \right)}}$$     (where a = 768 and b = 232)
$$\eqalign{ & = \left( {a + b} \right) \cr & = \left( {768 + 232} \right) \cr & = 1000 \cr} $$

Let 6x43 - 46y9 = 1904
Clearly, y = 3 and x = 5
Hence * must be replaced by 5

N = 2 × 4 × 6 × 8 × ..... × 98 × 100
    = 2⁵⁰ (1 × 2 × 3 × ..... × 49 × 50)
    = 2⁵⁰ × 50!
Clearly, the highest power of 2 in N is much higher than that of 5
∴ Number of zeros in N = Highest power of 5 in N
= $$\left[ {\frac{{50}}{5}} \right] + \left[ {\frac{{50}}{{{5^2}}}} \right]$$
= 10 + 2
= 12

Number = (68 × 260) = 17680
On dividing this number by 65, we get zero as remainder