Q.8 Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.

$$ \begin{array}{|l} \llap{2~~~~} 6, 15, 18 \\ \hline \llap{3~~~~} 3, 15, 9 \\ \hline 1, 5, 3 \end{array} $$ LCM = 2×3×5×3 = 90
Here, 90 is the least number exactly divisible by 6, 15 and 18.
To get a remainder 5, the least number will be 90 + 5 = 95.
Hence, 95 is the required number.