NIMCET 2025 Paper

A 3-digit number is a multiple of 6→ divisible by 2 and divisible by 3 .→ Last digit must be even ( $2,4,6$ ).→ Sum of digits must be divisible by 3 .Step 1: Choose last digit (must be even)Possible last digits $=2,4,6 \longrightarrow 3$ choicesStep 2: After choosing the last digit, the remaining 5 digits include:Residues mod 3:- $\{1,4\} \longrightarrow$ remainder 1- $\{2,5\} \longrightarrow$ remainder 2- $\{3,6\} \longrightarrow$ remainder 0No matter which even digit you remove, the remaining residues always allow $4$ valid pairs whose divisible by 3 :- One pair from (1-group, 2-group): $2 \times 2 = 4$- (0-group, 0-group) gives no pair- (1-group, 1-group) or (2-group, 2-group) invalid for 3-digit sumTherefore:4 valid digit-pairs for the first two positions.Each pair can be arranged in $2$ ways → total $=8$ numbers per last digit.Step 3: Multiply$3 \;\text{ (last digit choices) } \;\times 8 = 24 \;\text{. }\;$