Number System - Aptitude MCQ Questions and Solutions with Explanations

The least 6-digit number = 100000
On dividing 100000 by 349, we get 186 as remainder.
Required number :
= 100000 - (349 - 186)
= 100000 + 163
= 100163

= 5² + 6² + .... + 10² + 20²
= (1² + 2² + 3² +..... + 10²) - (1² + 2² + 3² + 4²) + 400
= $$\frac{1}{6}$$ n(n + 1) (2n + 1) - (1 + 4 + 9 + 16) + 400, where n = 10
= $$\left( {\frac{1}{6} \times 10 \times 11 \times 21} \right)$$     - 30 + 400
= (385 - 30 + 400)
= 755

y = 2 × 1 + 1 = 3
x = 3 × y + 1 = 3 × 3 + 1 = 10
Clearly, 10 when divided by 6, leaves a remainder 4

Remainder = 48, Divisor = (5 × 48) = 240
12 × Quotient = Divisor
⇒ Quotient = $$\frac{240}{12}$$ = 20
Dividend = (240 × 20) + 48 = 4848

Sum of the digits = 12, which is divisible by 3. So, the given number is divisible by 3
(Sum of digits at odd places) - (Sum of digits at even places)
= 6 - 6 = 0
So, the given number is divisible by 11
The given number when divided by 37 gives 3003003003
So, the given number is divisible by 37
The given number when divided by 111 gives 1001001001
Clearly, it is divisible by 111 as well as by 1001
Hence, the given number is divisible by each one of 3, 11, 37, 111 and 1001

2^m > 960 when the least value of m is 10
Then, 2¹⁰ = 1024 and 1024 - 960 = 64 = 2⁶
∴ m = 10 and n = 6

Take m = 15 and n = 20
Then, each one of m and n is divisible by 5
But (m + n) is not divisible by 10
Hence, (m + n) is divisible by 10 is not true.

Let the number be x.
Let x when divided by 136 give q as quotient and 36 as remainder.
Then, x = 136q + 36
= (17 × 8q) + (17 × 2) + 2
= 17 × (8q + 2) + 2
So, the given number when divided by 17 gives 2 as remainder.

= 136 × 12 × 8
= 136 × 96
= 136 × (100 - 4)
= (136 × 100) - (136 × 4)
= 13600 - 544
= 13056

We keep on dividing 99999..... by 7 till we get 0 as remainder.

∴ Required number = 142857
Number of digit = 6