Form 5-digit numbers without repetition using digits {0, 1, 2, 3, 4, 5}. How many such numbers are divisible by 3? (The first digit cannot be 0.)

Introduction / Context:
A number is divisible by 3 if the sum of its digits is a multiple of 3. We must count 5-digit, no-repetition numbers from the 6 digits {0,1,2,3,4,5} with a non-zero leading digit.

Given Data / Assumptions:

• Available digits: 0–5.
• Distinct digits; 5-digit numbers; first digit ≠ 0.
• Sum of all six digits is 15, divisible by 3.

Concept / Approach:
Select which single digit is omitted; the remaining 5 digits define the digit-sum. If the total sum of chosen digits is divisible by 3, every permutation yields a number divisible by 3 (subject to the no-leading-zero rule).

Step-by-Step Solution:
Let S6 = 0+1+2+3+4+5 = 15 ≡ 0 (mod 3).Sum of chosen five digits ≡ −(omitted digit) (mod 3). To get 0 (mod 3), the omitted digit must be 0 or 3.Case 1 (omit 0): digits {1,2,3,4,5} ⇒ all 5! = 120 arrangements valid.Case 2 (omit 3): digits {0,1,2,4,5} ⇒ total 5! = 120 permutations; subtract leading-zero cases (0 as first): 4! = 24 ⇒ 96 valid.Total = 120 + 96 = 216.

Verification / Alternative check:
Residue-class reasoning (mod 3) matches the omitted-digit argument; direct enumeration agrees.

Why Other Options Are Wrong:
210/217 are near but incorrect; 122/192 do not follow from the modular analysis.

Common Pitfalls:
Forgetting to exclude leading-zero permutations in the case where 0 is present.

Final Answer:
216