CBSE Class 10 Answered

Solution:-

Since P1 is divisor for each 1332,796,528 and remainder is P2 then

1332 = P1*a + P2……………………………………..1

796 = P1*b + P2……………………………………….2

528 = P1*c + P2……………………………………….3

                                                (Since Dividend = divisor * quotient + Remainder ;here a,b,c are respectively the quotients for these three numbers)

Take the difference (1) – (2)

1332-796 = 536 = P1*(a-b)…………………….(4)

And (2)-(3)

796-528 = 268 = P1*(b-c)…………………….(5)

So P1 could be the HCF of 536 and 268 i.e.

P1 = 268

And Hence remainder P2 could be 260 So

P1 – P2 = 268-260 = 8