12.The greatest number which willdivide 2112 and 2792 leaving4 as the

1.	12.The greatest number which willdivide 2112 and 2792 leaving4 as the remainder in each caseIJNV 1999(i) 68 (ii) 58 (iii) 78 (iv)1881S
Answer» In order to find the greatest number which will divide 2112 and 2792 leaving the remainder 4 in each case follow the below steps:- Let us assume the requirednumber is x. In order to make 2112 and 2792 completely divisible by x,deduct remainder 4 fromboth 2112 and 2792. 2112 - 4 = 2108 2792 - 4 = 2788 Now we've to find the HCF of 2108 and 2788. 2108 = 4 x 17 x 312788 = 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.

12.The greatest number which willdivide 2112 and 2792 leaving4 as the remainder in each caseIJNV 1999(i) 68 (ii) 58 (iii) 78 (iv)1881S

Answer»

In order to find the greatest number which will divide 2112 and 2792 leaving the remainder 4 in each case follow the below steps:-

Let us assume the requirednumber is x.

In order to make 2112 and 2792 completely divisible by x,deduct remainder 4 fromboth 2112 and 2792.

2112 - 4 = 2108

2792 - 4 = 2788

Now we've to find the HCF of 2108 and 2788.

2108 = 4 x 17 x 312788 = 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.