Question

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

Answer 1

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 required number is x.

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 HCF 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.

Answer 2

Answer:

The required number is 68.

Step-by-step explanation:

Each time when the number divides 2112 and 2792 it leaves remainder 4

So, the number exactly divides (2112 - 4) and (2792 - 4) ⇒ the number divides 2108 and 2788

So, the required number which when divides 2112 and 2792 and leaves remainder 4 will be HCF (2108,2788)

Now, to find HCF (2108,2788) : Find the prime factors of 2108 and 2788

2108 = 2 × 2 × 17 × 31

2788 = 2 × 2 × 17 × 41

So, we can see the common factor in both the numbers are 2, 2, 17

⇒ 2 × 2 × 17 = 68

So, the greatest number which when divides 2112 and 2792 leaves remainder 4 each time is 68.