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Find the number which has 8 divisors, such that the product of the divisors is 331776.

# Dozens

##### Age 7 to 14 Challenge Level:

We received a large number of well considered responses to this problem.

Ewan from Wilson's School explained his strategy for finding the largest multiple of 2:

You put the largest possible combination of three numbers, making sure the two given are there, and the last digit is even.
For instance, with 9 and 5, the largest would be 958, not 995, because 5 is odd.

Gabriel, also from Wilson's School, had strategies for identifying multiples of 3, 4 and 6:

To find out if your number is a multiple of 3, all you neeed to do is add the digits. If the digits add up to a multiple of 3, the original number is also a multiple of 3.
For example: 351
To find out if this is a multiple of 3 using normal division would be quite slow, but if you just add the didgits it takes only a few seconds:
3 + 5 + 1 = 9
9 is a multiple of 3 so 351 is a multiple of 3.

You can use this for any number: 546822
5 + 4 + 6 + 8 + 2 + 2 = 27
Note: this method only works if you want to find out if your number is a multiple of 3.
This does not work with any other number.

(It actually works for multiples of 9 as well - take a look at Divisibility Tests)

For the biggest multiple of 4, look at the last two digits. If the number formed by its last two digits is divisible by 4, the original number is a multiple of 4 as well.

For the multiples of 6, just check if the number is divisible by 2 AND 3.
If it is, then the number is also divisible by 6.

Abbas from Green Primary School had a different strategy for identifying multiples of 4:

You can divide the number by 2 and if the answer is even, that means the number that you first started with is a multiple of 4.

Patrick and Jenna from Savanna Oaks Middle School, and Thomas from Colet Court School explained how to find the largest multiples in the interactivity challenge.

Here is Thomas' detailed explanation of his strategy:

Multiple of 2:
• First check if both numbers are even. If they are, then use the highest of the 2 numbers as the tens digit and the lowest as the units digit. Use 9 as the missing number (ie the hundreds digit).
• If only one number is even, then use the odd number as the tens digit, the even number as the units and use 9 as the missing number (ie the hundreds digit).
• If both numbers are odd, then choose 8 as the units number (this will be your missing number), then use the biggest odd numbers as the hundreds digit and the smallest odd number as the tens digit.
Multiple of 3:

The sum of the three digits needs to be a multiple of 3.
Therefore we will derive the missing number by starting from 9 and checking if the sum of the two given digits plus 9 is a multiple of 9.
If this is the case, then the result will be 9 for the hundreds digit, the highest of the two given digits for the tens and the lowest of the two digits for the unit.
If 9 does not work, we try with 8 and repeat with 7 then 6 then... if 8 does not work either.
When we have found the digit, the hundreds digit is the highest of the three numbers, the tens the second highest and the unit the lowest of all the digits.

Multiple of 4:
• First we try to see if the combination of the 2 given digits is a multiple of 4. By doing so we will try first with the highest of the two digits as the tens digit and the other one as the unit. If it works then we will use 9 as the hundreds digit.
• If the combination of the highest of the two digits and the lowest does not work, we will try the other way round (lowest of the two given digits as the tens and highest as the unit). If this combination is a multiple of 4 then we will use 9 as the hundreds digit like above.
• If neither combination of the two digits is a multiple of 4, then we will use the highest of the two given digits as the hundreds digit. Then we will check if the remaining given digit is even. If this is the case we will use this number as the unit and choose as the tens digit the highest digit possible such that the last two digits are a multiple of 4.
• If the remaining given digit is not even, then we will use this number as the tens digit and choose the highest unit such that the last two digits are a multiple of 4.

Multiple of 6:

To be a multiple of 6, it needs to be a multiple of 3 and of 2.

Thomas also suggested a solution to the final challenge:

The solution to "Something else to think about" is 53184.

If the number is divisible by 12, it needs to be divisible by 4 and by 3.
To be divisible by 3 the sum of the digits need to be divisible by 3, and therefore the missing digit can only be 2, 5 or 8.
The last two digits need to be divisible by 4. Since we have 1, 3, 4 and 5, none of the combination is divisible by 4 and therefore we need to use the highest possible additional number ahead of the 4 divisible by 4, ie 8 (84 is divisible by 4).
Then for the first three digits, we rank them from the highest down to the lowest.

This problem caught many people out - we received lots of solutions suggesting that 53184 was the answer (using an 8 as the extra number).

Annabel, Harry and Liberty from Hethersett High School managed to improve on Thomas' solution to the final challenge:

The largest possible five digit number, including 1, 3, 4, 5 and another digit, which is divisible by 12 is: 54132

This is because it is divisible by 3 and 4.

Ravsimrat from Garden International School managed to improve on that:

Largest possible 5 digit number divisible by 12 is: 54312

DIVISIBILITY RULES 54312 should be divisible by 12 so...
The divisibility rule for 12 is: the number should be divisible by 3 and 4

a) Rule for 3: the sum of the digits of the number should be divisible by 3.
eg: (5 + 4 + 3 + 1 + 2) / 3 = 5 (so...it is divisible by 3)

b) Rule for 4: the last 2 digits of the number should be divisible by 4.
eg: 12 / 4 = 3 (so...it is divisible by 4)

Hence 54312 is divisible by 12.

Titus, also from Garden International School, thought about the problem in a similar way:

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.

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

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 is 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 that any number divisible by 12 has to be divisible by 4 and 3 and must be even.

Any number with 3 as a factor must have digits that add up to a multiple of 3. 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 realise 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 4).

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 4, but 54312 is obviously larger than 54132, and thus the answer 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 and a multiple of 3.

Well done to you all.