Question

Find the largest number that will divide 398,436,542 leaving remainder 7,11,15 in each case

Answer 1

On dividing 398 by the required number, there is a remainder of 7. This means that 398 7 = 391 is exactly divisible by the required number. Similarly, 436 -11 = 425 and 542 15 = 527 are exactly divisible by the required number.
The HCF of two positive integers is the largest positive integer that divides both the integers.
So, the required number will be the HCF of 391, 425 and 527. And that can be found by using Euclids division algorithm.
425 = 391 x 1 + 34
391 = 34 x 11 + 17
34 = 17 x 2 + 0
Thus, HCF = 17
Hence, the required number is 17.

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.