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

Question

The sum of all four-digit numbers that can be formed with the distinct non-zero digits $a$, $b$, $c$, and $d$, with each digit appearing exactly once in every number, is $153310 + n$, where $n$ is a single digit natural number. Then, the value of $(a + b + c + d)$ is ?

Answer 1

Step 1: A number is divisible by 4 only through its last two digits, and since $ 2t $ (twice the tens digit) is always even, the units digit $ u $ must itself be even. So only $ u=2,4,6 $ can ever end a multiple of 4; the odd digits $ 1,3,5 $ never work as the last digit.
Step 2: For $ u=2 $: the tens digit must be odd, giving $ t=1,3,5 $ — 3 endings ($ 12,32,52 $). For $ u=4 $: the tens digit must be even, giving $ t=2,6 $ — 2 endings ($ 24,64 $). For $ u=6 $: the tens digit must be odd, giving $ t=1,3,5 $ — 3 endings ($ 16,36,56 $).
Step 3: Total valid endings $ =3+2+3=8 $. Each fixes 2 digits, leaving 4 digits for the leading two places, fillable in $ 4\times3=12 $ ways.
\[ \boxed{8\times12=96} \]

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

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Correct Answer: 96

Solution and Explanation

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