please solve this....

Asked by  | 9th Mar, 2008, 02:54: PM

Expert Answer:

 

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  

Answered by  | 27th Jun, 2008, 11:17: AM

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