Question

Find the greatest number which when divides the numbers 2800, 2401 and 1470 leaves same remainder is each case.
1. 133
2. 141
3. 129
4. 149

Answer 1

Correct Answer - Option 1 : 133

GIVEN:

Three numbers 2800, 2401 and 1470.

CALCULATION:

Let same remainder = ‘x’

Required maximum number = HCF of (2800 – x), (2401 – x) and (1470 – x)

We know that if two numbers be divisible by a certain number, then their difference is also divisible by that number.

HCF of (2800 – x), (2401 – x) and (1470 – x) = HCF of [(2800 – x) – (2401 – x)], [(2800 – x) – (1470 – x)] and [(2401 – x) – (1470 – x)] = HCF of 399, 1330 and 931 = 133

Hence, required greatest number = 133