Question

Let $n$ be the number of different $5$-digit numbers, divisible by $4$, using the digits 1, 2, 3, 4, 5, 6 without repetition. Find $n$.

• A. 144
• B. 168
• C. 192
• D. None of these

Hint:

When counting numbers divisible by 4, focus on the last two digits, ensure they form a multiple of 4, and then arrange the remaining digits.

Answer 1

The Correct Option is B

A number is divisible by 4 if its last two digits form a number divisible by 4. From digits $\{1,2,3,4,5,6\}$, possible 2-digit endings divisible by 4 are: \[ 12, 16, 24, 32, 36, 52, 56, 64 \] Total = $8$ endings. For each ending, the remaining $3$ places are filled with any of the remaining $4$ digits in $4 \times 3 \times 2 = 24$ ways. Thus: \[ n = 8 \times 24 = 192 \] Wait — check: We are forming 5-digit numbers, so after fixing last two digits, first digit cannot be zero (not relevant since 0 not in digits set), so no restriction. But we have $6$ digits, pick $2$ for the ending (from the 8 valid pairs), remaining $4$ digits to choose $3$ for first three positions: number of arrangements $P(4,3) = 24$. However, one case: in $8$ valid endings, do all use distinct digits? Yes, since digits are all different in set $\{1,2,3,4,5,6\}$ and no repetition allowed — each valid pair automatically uses distinct digits. So: \[ n = 8 \times 24 = 192 \] The correct option is (3) in list, so original answer key says 192. \[ \boxed{192} \]