Let x be the last number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves remainder 0, the sum of digits of x is = ?

Solution (By Examveda Team)

$$\eqalign{ & {\text{LCM of 5, 6, 7, 8 = 840}} \cr & \frac{{840n + 3}}{9} \cr & \Rightarrow \frac{{3n + 3}}{9} \cr & \Rightarrow {\text{Take }}n{\text{ = 2}} \cr & \Rightarrow 3\left( 2 \right) + 3 \cr & \Rightarrow \frac{9}{9} = {\text{Remainder = 0}} \cr & \therefore {\text{Number is }}840n + 3 \cr & \Rightarrow 840\left( 2 \right) + 3\left[ {\because n = 2} \right] \cr & \Rightarrow 1683 \cr & {\text{Sum of digits}} \cr & {\text{ = }}\left( {1 + 6 + 8 + 3} \right){\text{ = 18}} \cr} $$