Question

find the largest number that divides 246 and 1030 leaving remainder 6 in each case

Answer 1

Solution :-

To find the largest number that divides 246 and 1030 leaving remainder 6 in each case, first we have to subtract 6 from both the numbers.

⇒ 246 - 6 = 240

⇒ 1030 - 6 = 1024

Now, we have to find out the Highest Common Factor of 240 and 1024.

H.C.F. of 240 and 1024 by long division method.

             _______
       240) 1024  (4
                 960
            ___________
                   64) 240  (3
                         192
                       ________
                           48) 64 (1
                                 48 
                                ______
                                 16) 48(3
                                       48
                                      ____
                                        0  
                                     _____

So, H.C.F. of 240 and 1024 is 16.

Hence, 16 is the largest number that divides 246 and 1030 leaving a remainder 6 in each case.

Let us check it.

⇒ 246 ÷ 16 

⇒ quotient = 15 and remainder = 6

⇒ 1030 ÷ 16

⇒ quotient = 64 and remainder = 6

The answer is correct

Answer 2

First subtract 6 from both the numbers & then We have to find the HCF of those numbers.

Required number = (246-6) & (1030-6)

HCF of 240 and 1024

HCF of 240 and 1024 by prime factorisation method

240 = 2 × 2 × 2 × 2 × 3 × 5 &

1024 = 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2

HCF of 240 and 1024 = 2 × 2 × 2 × 2 = 16 

Hence, 16 is the largest number which divides 246 and 1030 and leaves the remainder 6 in each case.