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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?

R1w
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Ludwigthestud
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    An integer is divisible by $3$ iff the sum of its digits is divisible by $3$. Permuting the digits can't change divisibility by $3$. – lulu Sep 11 '18 at 18:26
  • Also, even number means that the last digit is one of 0,2,4,6, or 8. You want the smallest possible, so larger one can be at tens or hundreds or thousands. – Andrei Sep 11 '18 at 18:32
  • It would be helpful to include the full problem statement in the body of the Question (not only in the title). This tends to avoid confusion (and gives you the OP a way to add meaningful context). – hardmath Sep 11 '18 at 18:40
  • It appears that I didn't read the problem thoroughly even when I was posting. However, this did help me. – Ludwigthestud Sep 17 '18 at 13:21

2 Answers2

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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.

Ahmad Lamaa
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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.

marty cohen
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