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Adding All Nine

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

Counting Factors

Is there an efficient way to work out how many factors a large number has?


Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.


Age 7 to 14
Challenge Level


Neel from Zurich International School in Switzerland, Miraya from Heckmondwike Grammar School in the UK and Skyler explained, in general, how to arrange the digits to get large numbers. Miraya wrote:

Use the larger digits in the larger units... e.g. 987 is bigger than 789!


Multiples of 2

Victoria and Michelle from Westridge in the USA and Miraya explained how you can tell if a number is a multiple of 2. Michelle wrote:

The thing that is mandatory for a multiple of 2 is that the ending of the number needs to be a multiple of 2. You can think of it this way. Multiples of 2 are 2 multiplied by another factor. It doesn't matter if the factor is odd or even. If you double any number, you are essentially adding the number to itself which always ends up even. You can try it out for yourself. Just remember, the last digit of a multiple of 2 needs to be even.

Isabel from Aiglon in Switzerland sent in some examples of finding the largest 3-digit multiple of 2:

For example use 8 and 8. 
8 is a multiple of 2 and is also the largest multiple before you get to the next unit of tens - so it should be the last number.
So we have: xx8. 
For the first unit, with the [place] value of 100, 8 could be used, however it's not the largest integer before 10. Therefore, to give the largest possible number, you could put 9 as the first value. Now we have 9x8. 
The missing value is 8, so we can substitute it in, giving us 988. 

Take another example, 3 and 7. 
Notice that both of these numbers are odd numbers, meaning the final digit won't be 3 or 7. So the highest multiple of 2 before 10 is 8 (if you go above 10, the [tens digit would change] so the solution would be incorrect), meaning the final digit must be 8. We are left with xx8. 
Now you are left with 3 and 7 to substitute into the 3 digit number.
Look at the values, clearly 7 is larger than 3, so if it's the first digit of the number it will produce the largest multiple of 2.  We are left with 738.

Giancarlo and Kayowa from Monarch Global Academy in the USA, Skyler from Westridge, Adavya from St Paul's Juniors in the UK, Sara and Kate from University of Chicago Laboratory Schools in the USA, Yash, Arnik, Arnav and Abhiram from Wilson's School in the UK, Thomas from Dulwich College Beijing in China, Daniel from DGS in the UK, Michelle, Neel and Matthew all described a general method for this.

Adavya said:

First of all, if any of the two numbers given is even, the first digit will be 9 as any three-digit number starting with 9 will be larger than any three-digit number starting with another digit. Then put the even number as the last digit, or if both other numbers are even put the smallest number as the last digit. If both digits are odd, make the last digit 8 as it is the largest even number and put the other two odd digits in descending order as the first and second digits.

Good idea, Adavya - we can also see the same method explained in Sara and Kate's slideshow.

Arnav's diagram explains the method very clearly:


Multiples of 3

Miraya and Victoria explained how to check if a number is a multiple of 3. Victoria wrote: 

Add up the digits. Then, decide if the number you got from the added up digits is a multiple of 3. For example, the number 162 is a multiple of 3 because 1+6+2=9 and 9 is a multiple of 3.

Abhiram described how to make 3-digit multiples of 3 using two numbers from the computer:

Find the total of the two numbers the computer has given, then add the largest single-digit number possible that makes sure that all the digits add up to a multiple of three.
After this, you can just rearrange the numbers to form the largest number possible.

Zaid from Wilson's School described how to do this for numbers with a four-digit target number, where the computer gives you all the digits but one:

Start by adding the three numbers and 9. If it equals a multiple of 3 then put the number in order from largest at the left to smallest at the right. If it does not equal a multiple of 3 then try again with 8 (and then 7).


Multiples of 4

Miraya explained how to check if a number is a multiple of 4:

A number is divisible by four if the number made by the last two digits can be divided by 4.

Abhiram described a method for finding three-digit multiples of 4:

Find the largest double digit number that uses at least one of the two numbers given and is a multiple of 4. Place these two numbers in the last two boxes then fill the first box in with the other number given by the computer (or with a 9).

Be careful! If the computer gives you 3 and 6, the largest double digit multiple of 4 that uses at least one of the two numbers is 96, so you end up with 396. But 936 is a larger multiple of 4.

Neel described a different method:

Let's say the numbers are 9, 7 and 3. Since all these numbers are odd I would put these in descending order like 973. The divisibility rule for 4 is that the last two digits are divisible by 4. So the only numbers divisible are 32 and 36. 36 is greater so we will take 6 and the number would be 9736.

When there is an even number, let's say [the numbers are] 9, 8 and 2. I would put 9 the way it was. But instead of putting 8 as the hundreds digit I would [use the 2 and the 8 to make a multiple of 4]. 28 is divisible by 4. I would put another 9 at the hundreds digit and then the biggest number would be 9928.

So, when the numbers are not all odd, make a multiple of 4 using the two smallest digits that can make a multiple of 4. Then fill the remaining digits to make the largest number possible.


Multiples of 5

Miraya wrote:

A number is divisible by five if the last digit is a 5 or a 0.

Finding the largest possible multiples of 5 is similar to finding the largest possible multiples of 2.


Multiples of 6

Miraya described how to tell whether a number is a multiple of 6:

A number can be divided by 6 if the last digit is even and the sum of all the digits is 3, 6, or 9 (or any other multiple of 3).

Neel explained how to find the largest four-digit multiple of 6 (with three numbers from the computer):

Let's say the numbers are 7, 2 and 9. I would add them up and I would get 18. So I would add another 9 at the thousands place and make it go in a descending order. So that would be 9972.

Let's say I have 9, 7 and 6. They add up to 22. I would put them in descending order but I would look for the largest number that adds to the nearest multiple of 3 but it's even. Here that number would be 8. So the digits would be 9, 7, 6 and 8. So since 6 is the lowest and a number that is even the largest number would be 9876.

In that example, the number added to make the multiple of 3 didn't have to be even. You only have to choose an even number if all the other numbers are odd, for example:

Let's say I have 3, 7 and 1. They add up to 11. The largest number that makes the total a multiple of 3 is 7, but you can't make an even number with 3, 7, 1 and 7. So the final number has to be 4, so the largest four-digit multiple of 6 using 3, 7 and 1 is 7314.


Multiples of 7, 8, 9, 10, 11

Miraya described how to identify multiples of these numbers:

A number can be divided by 7 if you subtract twice the last digit from the number formed by the remaining digits. E.g. 651 is divisible by 7 because 65$-$(1x2)=63 and since 63 is divisible by 7 so is 651 

A number is divisible by 8 if the number made by the last three digits will be divisible by 8.
E.g. 444,555,448 is divisible by 8 because 8x56=448 

A number is divisible by 9 if the sum of the digits add up to 9.
E.g. 179,131,590 is divisible by 9 because 1+7+9+1+3+1+5+9+0=36    3+6=9 

A number can be divided by 10 if the last digits is a 0.
E.g. 179,131,590 is divisible by 10 because the last digit is a 0.

A number can be divided by 11 if you subtract the last digit from the number formed by the remaining digits.
E.g. 396 is divisible by 11 because 39-6=33 33 is divisible by 11 so 396 is too! 


The largest 5-digit multiple of 12 using 1, 3, 4, 5 and one more digit

Nathan from Dulwich College Beijing described how to choose a multiple of 12:

For 12 we only need to find a number that has the last two digits divisible by four and the sum of digits equal to multiples of three.

Daniel found the largest possible number:

1 + 3 + 4 + 5 + X =  a number divisible by 3.

So, 13 + X =  a number divisible by 3.

So, X (our fifth digit) must be 2, 5 or 8 in order to make a number that is divisible by 3.

We need to look at the last two digits of the number. They have to be a number which is in the four times table. There are no numbers ending 1, 3 or 5 in the four times table. Looking at the numbers we have been given (1, 3, 4, 5), and the possible fifth digit we worked out above (2, 5, 8), our tens and units digits could be:

  1. 84 
  2. 48 
  3. 52 
  4. 32
  5. 12

Once we know what our tens and units numbers have to be, we can make the biggest possible number using the remaining three digits:

  1. 53,184 
  2. 53,148 
  3. 43,152 
  4. 54,132
  5. 54,312

So, the biggest number I made was 54,312.

Well done to everybody who worked out that 54,312 was the largest possible number! We also had excellent explanations for this sent in from Ziji from Mounts Bay Academy in England, Ronik from Wilson's School in England, Venus, Arin and Tan Kiu from Harrow International School in Hong Kong and Nirdvaita from Boone Meadow Elementary in the USA. Thank you to everybody who sent in their ideas.