A certain number 'C' when divided by $N_{1}$ it leaves a remainder of 13 and when it is divided by $N_{2}$ it leaves a remainder of 1, where $N_{1}$ and $N_{2}$ are the positive integers. Then the value of $N_{1} + N_{2} i s, i f \frac{N_{1}}{N_{2}} = \frac{5}{4}$ :

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can't be determined uniquely

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Solution

The correct option is D can't be determined uniquely
C = $N_{1} Q_{1}$ + 13 and C = $N_{2} Q_{2}$ + 1
where $Q_{1} Q_{2}$ are the quotients and hence $Q_{1} a n d Q_{2}$ must be integers
So, $N_{1} Q_{1} + 13 = N_{2} Q_{2} + 1$
$N_{2} Q_{2} - N_{1} Q_{1} = 12 \frac{4}{5} N_{1} Q_{2} - N_{1} Q_{1} = 12 N_{1} (4 Q_{2} - 5 Q_{1}) = 60$
where $N_{1}$ and ( $4 Q_{2} - 5 Q_{1}$ ) both will be integers so that $N_{1}$ can be one of the values of factors of 60 i.e., 1, 2, 3, 4, 15, 20, 30 and 60. But $N_{1}$ can not be less than the remainder 13. So the possible values are 15, 20, 30 and 60. Thus we see that the corresponding values of $N_{2}$ are 12, 16, 24 and 48.
So there are more than one possible values of $N_{1} a n d N_{2}$ .
Thus there is no unique value of $N_{1} + N_{2}$ . Hence (d) is correct option.

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A certain number 'C' when divided by N1 it leaves a remainder of 13 and when it is divided by N2 it leaves a remainder of 1, where N1 and N2 are the positive integers. Then the value of N1+N2 is, if N1N2=54:

A certain number 'C' when divided by $N_{1}$ it leaves a remainder of 13 and when it is divided by $N_{2}$ it leaves a remainder of 1, where $N_{1}$ and $N_{2}$ are the positive integers. Then the value of $N_{1} + N_{2} i s, i f \frac{N_{1}}{N_{2}} = \frac{5}{4}$ :