Question

How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

• A. 50
• B. 72
• C. 108
• D. 54
• E. 64

Hint:

For divisibility by 6, handle divisibility by 2 and 3 separately. Using the sum of digits modulo 3 can simplify counting.

Answer 1

The Correct Option is D

Step 1: Understand Divisibility by 6:
A number is divisible by 6 if it is divisible by both 2 and 3. Divisible by 2: Last digit must be even (0, 2, 4, 6). Divisible by 3: Sum of digits must be divisible by 3.
Step 2: Total Digits and Conditions:
Available digits: 0, 2, 3, 4, 6. We need to form 4-digit numbers (first digit not 0). No repetition.
Step 3: Sum of All Digits:
Sum of all five digits = 0+2+3+4+6 = 15, which is divisible by 3. Since we are choosing 4 out of 5 digits, the number will be divisible by 3 if the sum of the chosen 4 digits is divisible by 3. This is equivalent to the digit left out being divisible by 3 (because total sum 15 is divisible by 3, so sum of 4 digits = 15 - (excluded digit) is divisible by 3 iff the excluded digit is divisible by 3). The digits divisible by 3 in the set are 0, 3, 6. So the excluded digit can be 0, 3, or 6.
Step 4: Case 1: Excluded digit is 0.
Then the 4 digits are 2, 3, 4, 6. The number must be even (last digit even). Last digit optionss: 2, 4, 6. - For each even last digit, the first digit cannot be 0 (not an issue her(e), and the remaining 3 digits can be arranged in $3!$ ways. Number of such numbers = $3 \times 3! = 3 \times 6 = 18$.
Step 5: Case 2: Excluded digit is 3.
Digits: 0, 2, 4, 6. The number must be even (last digit even). Also, first digit cannot be 0. We count by considering the last digit. Subcase 2a: Last digit is 0. Then first digit can be any of the remaining 2,4,6 (3 choices). The middle two digits can be arranged in $2!$ ways. Total = $3 \times 2 = 6$. Subcase 2b: Last digit is 2 or 4 or 6 (3 choices). Then first digit cannot be 0. For each choice of last digit, the first digit can be chosen from the remaining 3 digits excluding 0 and the last digit. That gives 2 optionss for first digit (since 3 total remain, one is 0, so non-zero optionss = 2). Then the remaining two digits (including 0) can be arranged in $2!$ ways. Total for this subcase = $3 \times 2 \times 2 = 12$. Total for Case 2 = $6 + 12 = 18$.
Step 6: Case 3: Excluded digit is 6.
Digits: 0, 2, 3, 4. The number must be even (last digit even). Last digit optionss: 0, 2, 4. Subcase 3a: Last digit is 0. Then first digit can be any of 2,3,4 (3 choices). Middle two digits: $2!$ ways. Total = $3 \times 2 = 6$. Subcase 3b: Last digit is 2. Then first digit cannot be 0. Remaining digits: 0,3,4 with last digit fixed as 2. First digit can be 3,4 (2 choices). Then middle two digits (including 0) can be arranged in $2!$ ways. Total = $2 \times 2 = 4$. Subcase 3c: Last digit is 4. Similar to subcase 3b: first digit can be 2,3 (2 choices), middle two digits: $2!$ ways. Total = $2 \times 2 = 4$. Total for Case 3 = $6 + 4 + 4 = 14$.
Step 7: Sum All Cases:
Total valid numbers = $18 + 18 + 14 = 50$.
Step 8: Final Answer:
The number of such four-digit numbers is 50.