Question

The greatest no which will divide 2112 and 2792 leaving 4 as remainder in each case the no is

Answer 1

Let us assume the required number be x

In order to make 2112 and 2792 compeletly divisible,we need to deduct the remainder 4 from both 2112 and 2792

2112-4=2108

2792-4=2788

Now we find the HCF for 2108 and 2788

2108=4×17×31

2788=4×17×41

HCF =4×17=68

Therefore the greatest number which will divide 2112 and 2792 and leave reminder 4 is 68.

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