What is the least number which when divided by the number 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 22 leaves no remainder ?

Solution (By Examveda Team)

$$\eqalign{ & {\text{LCM of }}\left( {3,5,6,8,10,12} \right) \cr & = 3 \times 5 \times 2 \times 4 \cr & = 120 \cr & {\text{Required number is }} \cr & \frac{{120K + 2}}{{22}} = \frac{{10K + 2}}{{22}} \cr & {\text{at }}K{\text{ = 2,}}\frac{{10K + 2}}{{22}} \cr & \Rightarrow {\text{Remainder = 0}} \cr & {\text{The given condition satisfied}} \cr & {\text{ = }}120K + 2 \cr & = 240 + 2 \cr & = 242 \cr} $$