Question

Find the largest no which divides 398 and 436 and 542 and leaving the remainder 7,11 and 15

Answer 1

Hey!!


Required number ( 398 - 7 ) = 391, ( 436 - 11 ) = 425 and ( 542 - 15 ) = 527

So, the required number are 391, 425 and 527

By prime factorization method

391 = 17 × 23

425 = 5 × 5 × 17

527 = 17 × 31

Therefore, HCF[ 391, 425 and 527 ] = 17>>>Answer

____________________________

Hope it will helps you:-)

Answer 2

Given :-

398 , 436 and 542

To Find :-

The largest number

Solution :-

Let’s assume the integer is x

According to the condition given in the question

⇒ xy+7 = 398

⇒  xz+11 = 436

⇒  xk+15 = 542

⇒ xy =391

⇒  xz = 425

⇒  xk = 527

⇒  17 × 23 = 391

⇒  17 × 25 = 425

⇒ 17 × 31 = 527

So, the largest possible integer that will divide 398,436,542 & leaves reminder 7,11 and 15 respectively was 17.