Question

Using digits 1 to 6 (each at most once), how many 4-digit numbers can be formed that are divisible by 4?

Hint:

For permutation problems with divisibility constraints, always start by filling the places that are restricted by the rule (e.g., the last digit for divisibility by 2 or 5, the last two for divisibility by 4).

Answer 1

Correct Answer: 96

Approach: Divisibility by 4 depends only on the last two digits, so fix a valid ending first, then freely fill the front \(-\) this separates the hard constraint from the easy counting.

Step 1: A number is divisible by 4 iff its last two digits form a multiple of 4. List two-digit multiples of 4 using distinct digits from \(\{1,2,3,4,5,6\}\):
\[ 12,\ 16,\ 24,\ 32,\ 36,\ 52,\ 56,\ 64. \]
That is 8 valid endings (20, 28, 44, 48, 60 etc. are ruled out by the 0/8 or repeated digit).

Step 2: For each ending, two digits are used up, leaving 4 digits for the thousands and hundreds places. Number of ways \(= {}^{4}P_2 = 4 \times 3 = 12\).

Step 3: Multiply:
\[ \text{Total} = 8 \times 12 = 96. \]
Final Answer: 96 four-digit numbers.

Answer 2

Last-two-digits method: A number is divisible by 4 exactly when its last two digits (as a 2-digit number) are divisible by 4. Checking every ordered pair of distinct digits from \( \{1,2,3,4,5,6\} \), the pairs whose two-digit value is divisible by 4 are: \( 12, 16, 24, 32, 36, 52, 56, 64 \) — eight valid endings in all.

For each such ending, the two digits used are fixed, leaving \( 4 \) digits still unused for the thousands and hundreds places. These two places can be filled in \( 4\times3=12 \) ways.

Total count \( =8\times12 \):
\[ \boxed{96} \]