Using digits {1, 3, 4, 6, 7}, create the largest five-digit even number that is divisible by 3. To clarify, no digits can repeat either. Can someone outline their thought process to solve? Here's mine:

I used a process of elimination to determine what numbers I could use in the ten-thousands, thousands, and hundreds place. This gave me $7 6 4$. Is this thought process safe so far?

Where I think I began to mess up(if not earlier). I was left with $3$ and $1$. Choosing between $7 6 4 1 3$ and $7 6 4 3 1$, I made the unsafe assumption that division by $3$ would not be possible with a five-digit number ending with a $1$. Little did I know that $76431 /3  = 25477$. Aside from getting to this point and manually checking what's possible, can anyone explain a better approach?
The number we seek to obtain must follow 2 conditions:
 a-The number must be divisible by 3
 b-The number must be even

You should not worry about condition (a) since a number is divisible by 3 if and only if the sum of the digits is divisible by 3, which is already obtained (1 + 3 + 4 + 6 + 7 = 21 = 3 * 7). So this condition will always be satisfied, whatever permutation you decide on.
Condition (b) implies that your number must end with an even digit(Either 4 or 6). To obtain the largest number possible, you should place the smallest of these integers in the least significant positions, namely the ones.
Also, intuitively, you must place the largest digits in the most significant places.
So your answer would be 76314.
The sum of the digits, 21,
is divisible by 3,
so any number formed
from these digits is
divisible by 3.

So just take the digits
in descending order,
ensuring that the last digit is even
(which I carelessly overlooked):
76314.

As a check,
76431/3
=25477.