Question

A number when divided by a divisor leaves a remainder of 27. Twice the number divided by the same divisor leaves a remainder of 3. Find the divisor.

Answer 1

let q1 be quotient of first division and q2 be the quotient of second division, d be the divisor and let original number be num.
the 2 equations formed are:
1) num=d*q1+27
2)2*num=d*q2+3

multiplying equation-1 by 2 and equating the equations gives us
=>d*q2+3=2dq1+54

shifting the d term on one side and numbers on the other 
=>d*q2-2*q1=51

taking d common from the left side
=>d(q2-2q1)=51

=>d=51/q2-2*q1

so d can either be 51 or one of its factor(17 or 3).
since we have a reminder of 27 in one case d cannot be less than 27.
so divisor is 51.

Answer 2

Answer:

51

Step-by-step explanation:

Let q1 be quotient of first division and q2 be the quotient of second division, d be the divisor and let original number be num.

The 2 equations formed are:

1) num=d*q1+27

2)2*num=d*q2+3

Multiplying equation-1 by 2 and equating the equations gives us

=>d*q2+3=2dq1+54

Shifting the d term on one side and numbers on the other

=>d*q2-2*q1=51

Taking d common from the left side

=>d(q2-2q1)=51

=>d=51/q2-2*q1

So d can either be 51 or one of its factor(17 or 3)

Since we have a reminder of 27 in one case d cannot be less than 27

So divisor is 51