Question

Find the least positive integer that will divide 398, 436 and 542 leaving remainders 7,11 and15 respectively

Answer 1

Let the number be X
So X divides 398,436 &542 leaving remainder 7, 11 & 15 respectively
so X must divide (398–7)=391, (436–11)=425 & (542–15)=527 completely.
So X must be the HCF or GCD of 391, 425 & 527.
So the answer 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.