Find the greatest number which when divides the numbers 2800, 2401 and

1.	Find the greatest number which when divides the numbers 2800, 2401 and 1470 leaves same remainder is each case.1. 1332. 1413. 1294. 149
Answer» 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

Find the greatest number which when divides the numbers 2800, 2401 and 1470 leaves same remainder is each case.1. 1332. 1413. 1294. 149

Answer» 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

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