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 $=2,4,6⟶3$ choices
Step 2: After choosing the last digit, the remaining 5 digits include:
Residues mod 3:
- $\left\{1,4\right\}⟶$ remainder 1
- $\left\{2,5\right\}⟶$ remainder 2
- $\left\{3,6\right\}⟶$ remainder 0
No 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 sum
Therefore:
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