Problem Solving

A number is successively divided by5,6and 8 and leaving remainders 3,4,7 respectively.What will be the remainders if the order of divisors be reversed?

rattan123 wrote:A number is successively divided by5,6and 8 and leaving remainders 3,4,7 respectively.What will be the remainders if the order of divisors be reversed?

If a number is divided by 5 the remainder is 3, then if the number is divided by 6 the remainder is 4, and if the number is divided by 8 the remainder is 7... is that what you are saying?

first the number is divided by 5 then the quotient is divided by 6...then the quotient by 8...

Then how do you get 3 different remainders as stated above.

lets take an eg....156...divide it by 5 then the remainder is 1 and quotient is 31...then we divide it by 4 then the remainder is 3 and quotient is 7...like this.....the values are different in the question...

rattan123 wrote:lets take an eg....156...divide it by 5 then the remainder is 1 and quotient is 31...then we divide it by 4 then the remainder is 3 and quotient is 7...like this.....the values are different in the question...

This is not a gmat question.

are the remainders 1,5,4???

rattan123 wrote:A number is successively divided by5,6and 8 and leaving remainders 3,4,7 respectively.What will be the remainders if the order of divisors be reversed?

say in 3rd & last cycle, the quotient from II cycle is divided by 8 and leaves K as quotient and 7 as remainder, then the Dividend(quotient of II) will be 8K+7.

Similarly,from 2nd cycle,as quotient is 8K+7. It means Dividend of II(quotient of I)will be 6*(8k+7)+4 = 48K+46.

Similarly, as quotient in I cycle is 48K+46. It means Dividend of I cycle will be 5*(48K+46)+3= 240k+233.

So finally the nos. is 240k+233. Lets put a convenient value of k....say k=1.

then the rqd.no. will be N=240*1+233 = 473.

Now Reverse division...

1. 473/8 => quotient = 59; remainder = 1.

2. 59/6 => quotient = 9; remainder = 5.

3. 9/5 => quotient = 1; remainder = 4.

Remainder will be 1, 5, 4.