A number when divided by a divisor leaves 27 as remainder Twice the number when divided by the same number leaves remainder 3.Find the divisor. Please explain.Thanks
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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)
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)
Wonderer09:
Actually it is 51 because the dividend is always greater than the remainder. Thanks
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