Number System - Aptitude MCQ Questions and Solutions with Explanations

(a) In 183654729, 1836547 is not divisible by 7
(b) In 381654729, 38 is divisible by 2,
381 is divisible by 3,
3816 is divisible by 4,
38165 is divisible by 5,
381654 is divisible by 6,
3816547 is divisible by 7
38165472 is divisible by 8 and 381654729 is divisible by 9
(c) In 983654721, 983 is not divisible by 3
(d) In 981654723, 9816547 is not divisible by 7

Let the required prime number be p.
Let p when divided by 6 give n as quotient and r as remainder. Then p = 6n + r, where 0 $$ \leqslant $$ r < 6
Now, r = 0, r = 2, r = 3 and r = 4 do not give p as prime.
∴ r $$ \ne $$ 0, r $$ \ne $$ 2, r $$ \ne $$ 3, and r $$ \ne $$ 4
Hence, r = 1 or r = 5

On dividing 8112 by 99, we get 93 as remainder.
So, the required number to be subtracted is 93.

Let the two numbers be are a and b
∴ a + b = 14.....(i)
a - b = 10.....(ii)
By adding equation (i) and (ii) we get
2a = 24
∴ a = 12 and b = 2
∴ Product of these two numbers
= 12 × 2 = 24

Three consecutive multiples of 3 are 3m, 3(m + 1) and 3(m + 2)
Their product = 3m × 3(m + 1) × 3(m + 2)
                      = 27 × m × (m + 1) × (m + 2)
Putting m = 1, this product is (27 × 1 × 2 × 3) = 162
So, this product is always divisible by 162

$${p^3} - {q^3} = \left( {p - q} \right)$$   $$\left\{ {{{\left( {p - q} \right)}^2} - xpq} \right\}$$
$$ \Rightarrow \left( {p - q} \right)\left( {{p^2} + {q^2} + pq} \right)$$     $$ = \left( {p - q} \right)$$   $$\left\{ {{{\left( {p - q} \right)}^2} - xpq} \right\}$$
$$\left\{ {\because {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)} \right\}$$
By cancelling same terms of both sides
$$ \Rightarrow {p^2} + {q^2} + pq = {p^2} + {q^2}$$     $$ - 2pq - xpq$$   $$\left\{ {{{\left( {a - b} \right)}^2} = {a^2} + {b^2} - 2ab} \right\}$$
$$\eqalign{ & \Rightarrow 3pq = - xpq \cr & \Rightarrow x = - 3 \cr} $$

Let n = 1 + x = 1 + m (m + 1) (m + 2) (m + 3), where m is a positive integer.
Then, clearly two of m, (m + 1), (m + 2), (m + 3) are even and so their product is even.
Thus, x is even and hence n = 1 + x is odd
Also, n = 1 + m (m + 3) (m + 1) (m + 2) = 1 + (m² + 3m) (m² + 3m + 2)
⇒ n = 1 + y (y + 2), where m² + 3m = y
⇒ n = 1 + y² +2y = (1 + y)² , which is a perfect square.
Hence, I and III are true.

225 = 9 × 25, where 9 and 25 are co-primes.
So, a number is divisible by 225 if it is divisible by both 9 and 25
Given number is divisible by 25, only if 7q is divisible by 25, i,e, if q = 5
Sum of digits of given number
= (8 + 7 + 6 + p + 3 + 7 + 5)
= 36 + p, which must be divisible by 9
This is possible if p = 0
Hence, p = 0,   q = 5

2³¹
= 2 × 2³⁰
= 2 × (2² )¹⁵
= 2 × 4¹⁵
When n is odd, (xⁿ + aⁿ) is divisible by (x + a)
∴ (4¹⁵ + 1¹⁵) is divisible by (4 + 1)
⇒ (4¹⁵ + 1) is divisible by 5
⇒ (2³⁰ + 1) is divisible by 5
⇒ On dividing 2³⁰ by 5, we get (5 - 1) i.e., 4 as remainder.
∴ Remainder obtained on dividing 2³¹ by 5
= Remainder obtained in dividing (2 × 4) i.e., 8 by 5 = 3

$$\eqalign{ & \left( {800 \div 64} \right) \times \left( {1296 \div 36} \right) \cr & = \frac{{800}}{{64}} \times \frac{{1296}}{{36}} \cr & = 450 \cr} $$