Find the greatest number which will divide 2112 and 2792 leaving the remainder 4 in each case ........

Quantitative Aptitude ( CCS) LCM, HCF

Find the greatest number which will divide 2112 and 2792 leaving the remainder 4 in each case.

In order to make 2112 and 2792 completely divisible by x, we need to deduct the remainder 4 from both 2112 and 2792.

2112 - 4 = 2108

2792 - 4 = 2788

Now we've to find the H.C.F of 2108 and 2788.

2108 = 4 x 17 x 31
2788 = 4 x 17 x 41

HCF is 4 x 17 = 68

Therefore, the greatest number which will divide 2112 and 2792 leaving the remainder 4 in each case is 68.

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