Four-Digit Numbers Divisible by 6

For a number to be divisible by 6, it must be divisible by 2 (even unit digit) and divisible by 3 (sum of digits divisible by 3).

Available 5 digits: {0, 2, 3, 4, 6}. Sum of all 5 digits = $0 + 2 + 3 + 4 + 6 = 15$. To select 4 digits whose sum is divisible by 3, we must exclude one digit such that the remaining sum is divisible by 3. Excluding 0 $\implies$ digits {2, 3, 4, 6}, sum = 15 (divisible by 3). Excluding 3 $\implies$ digits {0, 2, 4, 6}, sum = 12 (divisible by 3).

Case 1: Digits used = {2, 3, 4, 6} (No 0 present)

• Last digit must be even: can be 2, 4, or 6 (3 choices).
• Remaining 3 places filled by remaining 3 digits in $3! = 6$ ways.
• Total for Case 1 = $3 \times 6 = 18$ numbers.

Case 2: Digits used = {0, 2, 4, 6} Subcase 2a: Last digit is 0

• Leftmost digit: 3 choices (2, 4, 6)
• Middle two digits: $3 \times 2 = 6$ ways.
• Total for 2a = $1 \times 3! = 6$ numbers.

Subcase 2b: Last digit is non-zero (2, 4, or 6)

• Last digit: 3 choices.
• First digit: cannot be 0 and cannot be the last digit $\implies 2$ choices.
• Remaining two middle positions: $2 \times 1 = 2$ choices.
• Total for 2b = $3 \times 2 \times 2 = 12$ numbers? Wait, let's re-count 4 digits {0, 2, 4, 6}: If last digit is 2: remaining {0, 4, 6}.
• First digit: 4 or 6 (2 choices).
• 2nd and 3rd digits: 2! = 2 choices.
• So $2 \times 2 = 4$ numbers ending in 2.
• Similarly 4 numbers ending in 4, and 4 numbers ending in 6.
• Total for 2b = $4 \times 3 = 12$ numbers. Wait, total for Case 2 = 6 + 12 = 18? Wait! What about the official answer 50? Let's double check if 0 can be repeated or if digits are distinct without repetition! Ah! Sum of 4 digits: sum must be divisible by 3. Let's re-verify the full set of 4-digit combinations from {0, 2, 3, 4, 6}:

1. {2, 3, 4, 6}: sum = 15. Last digit even (2, 4, 6). Number of ways = $3 \times 3! = 18$.
2. {0, 2, 4, 6}: sum = 12. Last digit even (0, 2, 4, 6).
   • Ending in 0: $3! = 6$.
   • Ending in 2, 4, 6: $3 \times (2 \times 2!) = 12$.
   • Total = 18. Wait, 18 + 18 = 36. Why is answer 50? Ah! Can digits be repeated? Prompt says "no digit is used more than once". Wait! Let's re-check the set of digits: {0, 2, 3, 4, 6}. Wait, are there other combinations of 4 digits whose sum is a multiple of 3? Ah! 5 digits available: 0, 2, 3, 4, 6. Sums of 4 digits: Excluding 0: {2, 3, 4, 6}, sum = 15. (Multiple of 3) Excluding 2: {0, 3, 4, 6}, sum = 13. (Not) Excluding 3: {0, 2, 4, 6}, sum = 12. (Multiple of 3) Excluding 4: {0, 2, 3, 6}, sum = 11. (Not) Excluding 6: {0, 2, 3, 4}, sum = 9. (Multiple of 3!) Ah! Excluding 6 gives {0, 2, 3, 4}, sum = 9! Let's calculate for {0, 2, 3, 4}:

• Ending in 0: $3! = 6$ ways.
• Ending in 2 or 4 (2 choices):
  • First digit cannot be 0, so 2 choices.
  • Middle digits: 2! = 2 ways.
  • $2 \times (2 \times 2) = 8$ ways.
• Total for {0, 2, 3, 4} = $6 + 8 = 14$ ways.

Total valid numbers = $18 + 18 + 14 = 50$!