Question

Find the largest possible positive integer that will divide 398, 436 and
542 leaving remainder 7, 11, 15 respectively

Answer 1

First we subtract respectively 7 from 398,11 from 436 and 15 from 542.
So,398-7=391
436-11=425
542-15=527
Then we find their HCF.
391)425(1
391
___
034)391(11
374
____
017)34(2
34
____
00
The HCF of 391 and 425 is 17.
Now we find the HCF of 17 and 527.
17)527(31
51
___
017
17
-----
00
Hence,the largest possible positive integer that will devide 398,436 and 542 leaving remaindet 7,11 and 15 respectively.

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.