# Dozens

*You may find the article on Divisibility Tests helpful.*

*The problem is explained below, but you may wish to scroll to the foot of the page to watch a video of the NRICH team presenting the challenge.*

Do you know a quick way to check if a number is a multiple of 2? How about 3, 4, 5..., 12..., 15..., 25...?

To start with, the interactivity below will generate two random digits.

Your task is to find the **largest possible three-digit number** which uses the computer's digits, and one of your own, to make a multiple of 2.

**Can you describe a strategy that ensures your first 'guess' is always correct?**

Clicking on the purple cog gives you a chance to change the settings.

You can vary the challenge level by changing:

- the multiple
- the number of digits in your target number
- the number of digits provided by the computer.

*To ensure you have some choice, make sure the number of digits provided by the computer is fewer than the number of digits in the target number.*

**Can you describe your strategies that ensure your first 'guess' is always correct for a variety of settings?**

Here is a video of the NRICH team presenting the challenge. You could just watch the start to check that you understand the problem, or you may like to pause the video and work on the task at various points.

**Something else to think about:**

What is the largest possible five-digit number divisible by 12 that you can make from the digits 1, 3, 4, 5 and one more digit?

*Once you've had a chance to think about this, click below to check.*

Here is a selection of follow-up problems you may now like to try:

Factors and Multiples Game

Take Three from Five

American Billions

Click here for a poster of this problem

*We are very grateful to the Heilbronn Institute for Mathematical Research for their generous support for the development of this resource.*

What is special about multiples of 2... 3... 4... 5... 6...?

You may find the article on Divisibility Tests helpful.

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:

- 84
- 48
- 52
- 32
- 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:

- 53,184
- 53,148
- 43,152
- 54,132
- 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.

### Why do this problem?

This problem offers a twist on the usual way of assessing students' knowledge of divisibility tests. Rather than asking students to check whether a number is divisible by 2, 3, 4, 5... students have to puzzle over the choices available as they are challenged to find the largest number that meets the necessary criteria. This low threshold high ceiling task has an accessible starting point, but then offers increasing levels of challenge as students can opt to work with multiples of larger and larger numbers..., and they may sometimes find that it is impossible to meet the criteria!

### Possible approach

*You may find the article on Divisibility Tests helpful. It offers clear explanations of the various divisibility rules and why they work.*

*You may be interested in watching the NRICH team discussing the use of this problem in the recording of the March 2021 NRICH secondary teacher webinar (this task begins at about 4:30), and/or the recording of the team presenting the problem to students at the Cambridge Festival. This problem also featured in an NRICH Primary webinar in April 2021.*

**Repeat this a few times and c**

**hallenge the class to develop a strategy for ALWAYS finding the highest even number on their first attempt.**- Collaboratively - students work on a particular multiple until they have entered five correct solutions in a row, on their first attempt before moving on to another multiple;
- Competitively - each student in the pair works independently to find the largest number and a point is awarded if only one of them has the correct solution. (Students who know their divisibility rules will usually draw!)

Numbers provided: 2 digits

Multiple of: 15 (sometimes impossible)

Numbers provided: 2 digits

Multiple of: 12 (sometimes impossible)

Numbers provided: 2 digits

Multiple of: 20

Numbers provided: 2 digits

Multiple of: 200 (sometimes impossible)

e.g. Target number: 3 digits

Numbers provided: 2 digits

Multiple of: 12

1 & 1

1 & 7

3 & 5

5 & 9

1 & 9 --> 912

3 & 7 --> 732

5 & 5 --> 552

7 & 9 --> 972

1 & 5 --> 516

3 & 3 --> 336

3 & 9 --> 936

5 & 7 --> 756

9 & 9 --> 996

There are ten more pairs of odd numbers, five of which sum to 1 mod 3, and so the third number can be 2, 5, or 8, and all of 12, 32, 52, 72, 92 are multiples of 4.

The last five pairs add up to a multiple of 3, so the last digit can be one of 0, 3, 6, 9. We have 16, 36, 56, 76, 96, all multiples of 4!

This argument can be used to show that it is always possible to find a four-digit multiple of 12 given two odds and an even:

If the three digits add to 2 mod 3, add a 4 and put the evens digit in the 10s column.

If the three digits add to 1 mod 3, add a 2 and put one of the odds in the 10s column, or add an 8 and the evens digit in the 10s column.

If the three digits add to 0 mod 3, add a 0 and put the even number in the 10s column, or add a 6 and put an odd digit in the 10s column.

You won't be able to find a four-digit multiple of 12 if the three digits are all odd and sum to 2 mod 3 (e.g. 3, 3, 5).

It may be tempting for students to suggest that whenever the interactivity provides two odd numbers, it will be impossible to create a three-digit multiple of 12. This is not the case; it is possible to create three-digit multiples of 12 with all other combinations of two odd numbers. Students could be asked to find which pairs of odd numbers can, and which cannot, create three-digit
multiples of 12.

**Finally**, the "something else to think about" at the end of the problem could be used as a homework task or as a final challenge.

### Key questions

How do you know if a number is a multiple of 3?

How do you know if a number is a multiple of 4?

How do you know if a number is a multiple of 6?

How do you know if a number is a multiple of 12?

How do you know you have found the biggest possible number?

### Possible support

Some students may benefit from starting with two-digit target numbers and focusing on multiples of 2, 5 and 10. This more accessible context will still require students to reason and justify. As they become more confident, they can move on to multiples of 20, 4, 3...

### Possible extension

American Billions is an engaging extension activity which uses similar ideas to the ones met in this problem.

For slightly older students, Common Divisor offers an intriguing follow-up problem.