Answer:

The number you want is divisible by 12, which means it is also divisible by 3 and by 4. This tells us two things right off the bat:

The sum of the number’s digits must be a multiple of 3, and

The last two digits of the number must form a number that is divisible by 4.

We know four of the five digits: 1, 3, 4, and 5, and we have one wildcard digit that could be 0 through 9. For now, we’ll refer to the wildcard digit as x.

From the first statement above, we can say that:

1 + 3 + 4 + 5 + x = a multiple of 3.

Condensing that a bit:

13 + x = a multiple of 3.

What are the possible values of x?

13 + 2 = 15 = 5(3)

13 + 5 = 18 = 6(3)

13 + 8 = 21 = 7(3)

There aren’t any other one-digit values that will work for x, so our wildcard digit must be either 2, 5, or 8.

Next, let’s look at the second statement above: the last two digits must form a number that is divisible by 4.

Using the four given digits, there aren’t any ways that we can make a number divisible by 4. Only one of the digits is even, which means that the 4 must be in the ones place; and yet we can quickly see that 14, 34, and 54 are not divisible by 4.

So the only way around this is to have the wildcard be one of the last two digits.

If the wildcard is 2, then the last two digits could be 12 or 24.

The wildcard cannot be 5, because there is no two-digit combination that will produce a number divisible by 4.

If the wildcard is 8, then the last two digits could be 48 or 84.

If the wildcard is 8, then it requires that we use 4 as one of the final two digits. But if the wild card is 2, then there’s an option that allows us to use 4 in one of the higher place values, which makes the number larger. To make the biggest number possible, we want the higher-value digits to be closer to the front of the number.

The number you’re looking for is 54,312. That’s the largest five-digit number that contains the digits 1, 3, 4, 5, and a wildcard, and is divisible by 12.