Dozens

This problem caught many people out - we received lots of solutions suggesting that 53184 was the answer (using an 8 as the extra number). It is indeed a multiple of 12 but it is not the largest multiple that could be made; that is 54312 (using a 2 as the extra number).

We received correct solutions from Livvy and Connor, Lily, Maryanne, Adam and Mitchell, Nur, Harry, Thomas, Arpan, Titus, Ed C and Ed O, Charline, Ravsimrat, Luke and Jordan, Andrew, William and Ron. Well done to you all.

Ravsimrat from Garden International School and Titus from GIS thought about the problem in a similar way. Here is what Titus wrote:

Multiples of 12 are also multiples of 2, 3 and 4.
Multiples of 3 can be found by the sum of digits which can be divided by 3. So the one more digit is limited to 8, 5 or 2.
Multiples of 4 are obtained if the last two digits together can be divided by 4.
Multiples of 2 must be even numbers.
The unit place has to be 2, 4 or 8.
My answer is 54312

Luke and Jordan from Breckland Middle School also sent us their reasoning:

We worked out some facts about this mystery number to help us:

1. It must end in an even number as 12 is an even number.

2. To find out if a number is divisible by 3, you can add up all of the digits and see if it is a multiple of 3 that you recognise.

From this, we know that any number containing only 1, 3, 4 and 5 will not be divisible by 3, and therefore not divisible by 12.

Therefore, the fifth number MUST be 2, 5 or 8 for the five-digit number to be divisible by 12.

3. It cannot begin with 8. There are only 6 options beginning with 8 that are divisible by 2 and 3 (81354; 81534; 83154; 83514; 85314; and 85134) and none of these are divisible by 12.

Continued trial and error then showed 54312 to be the biggest number that is divisible by 12.

Thomas from A.Y. Jackson School sent us a detailed explanation of his thinking:

Using the numbers 1, 3, 4, 5, and another number represented by N, the largest five digit number that is divisible by 12 is 54312. (N=2)

I began by deducing any number divisible by 12 has to be divisible by 4 and 3 and must be even.

Any number with three as a factor (its digits) must add up to a multiple of three.

The sum of 1, 3, 4, 5 is 13, therefore, N could be only three numbers: 2, 5, 8.

The first attempt will certainly be to choose the largest possible value of N which is 8, and 8 as the first digit, but I realize that then the last number must be 4, but none of the remaining values (3, 1 or 5) paired with 4 (34, 14, 54) could be divisible by 4 (since if a number is divisible by 4, the last two digits must also be divisible by four)

Therefore 8 must be stuck to the rear with the number being either 53184 or 53148.

It then becomes evident that the largest value the ten-thousands digit could hold is 5.

If N=5, the same dilemma returns (3, 1, 5 could not be paired with 4).

The only solution to N then becomes 2, as then 2 could replace 4 as the last digit and thus giving the number a greater thousands place-value.

The template of the number becomes 54 _ _ 2 either 12 or 32 is divisible by four, but 54312 is obviously larger than 54132. And thus the number is 54312.

Harry from The Beacon School also sent us a full and very clear explanation of his thinking:

As the number is a multiple of 12, it must be a multiple of 3 also. Therefore its digits must add to a multiple of 3.

1+3+4+5=13, so the extra digit must be 2, 5, or 8 to make the sum into a multiple of 3.

The number must also be a multiple of 4 (because it is a multiple of 12).

For a number to be a multiple of 4, the number formed by its last two digits must be a multiple of 4.

If the extra digit is 5, then no multiples of 4 can be formed.

The explanation is as follows:

For the number to be a multiple of 4, it must be even, and the only even number in the selection of 1, 3, 4, 5, 5 is 4, therefore the number must end with 4. None of 14, 34 and 54 are multiples of 4. Therefore, if the extra digit is 5, the number cannot be a multiple of 4 and consequently cannot be a multiple of 12.

If the extra digit is 8, the only possible places for it are in the tens and the units. This is because the only possible last 2 digits (for the number to be a multiple of 4) are 48 or 84.

Choosing 84 gives a maximum of 53184 for the solution to the original problem (if the extra digit is 8).

The only other possibility is if the extra digit is 2. Now the last two digits can be 12, 24, 32, or 52. Choosing 12 is best because this leaves 5, 4 and 3 for the larger places.

The maximum using an extra digit of 2 is therefore 54312, which is larger than the 53184 produced by using 8. Therefore the solution is 54312.

This is verified as a multiple of 12 because it is both a multiple of 4 (12 is a multiple of 4) and a multiple of 3 (5+4+3+1+2=15, which is a multiple of 3).

The largest possible multiple of 12 that can be made using the digits 5, 4, 3, 1, X is 54312.

Old solutions: The solution is 54312. Joel from ACS Barker Rd School, Singapore, sent the following solution: A number which is divisible by 12 has to be divisible by 3 and 4. A number which is divisible by 3 has the sum of its digits divisible by 3. A number which is divisible by 4 has its last two digits divisible by 4. The missing number has to be one of the last two digits, since the current 4 digits have no combination of 2 digits divisible by 4. Since the sum of the current digits is 13, the missing number is either 2, 5, or 8. Since 5, 4 and 3 are the largest numbers, we assume that the first 3 digits of our final number are 5, 4 and 3 in respective order. All numbers divisible by 4 are even numbers, and 1 is not, and so the number has to be 5431_. The only digit out of 2, 5 and 8 which will make the last two digits divisible by 4, is 2, so the final number is 54312. Other correct solutions with clear explanations came from Angela of Hethersett High School, Norfolk, Nicola of Madras College, St Andrews and Soon Xing and George from The Chinese High School in Singapore.