# What numbers can we make now?

Imagine we have four bags containing numbers from a sequence. What numbers can we make now?

## Problem

*What Numbers Can We Make Now? printable sheet*

*This problem follows on from **What Numbers Can We Make?*

The interactivity below creates sets of bags similar to those in What Numbers Can We Make?

Investigate what numbers you can make when you choose three numbers from the bags. Once you think you know what is special about the numbers you can check your answer.

You can also find what is special when you choose four, five or six numbers.

When you have an efficient strategy, test it by choosing 99 or 100 numbers.

*Always enter the biggest possible multiple. The "plus" number may include zero.*

**Can you explain your strategies?**

**And finally...**

If the bags contained 3s, 7s, 11s and 15s, can you describe a quick way to check whether it is possible to choose 30 numbers that will add up to 412?

There are a few related problems that you might like to try:

## Getting Started

This problem follows on from What Numbers Can We Make?, so you could start by exploring that first.

## Student Solutions

Henry, from St. Hugh's, answered our question at the end:*If the bags contained 3s, 7s, 11s and 15s, can you describe a quick way to check whether it is possible to choose 30 numbers that will add up to 412?*

He said the following:

It is impossible to make 412.

The starting number is 3 and the difference between the numbers is 4.

If I choose 30 numbers and add them all up, I will get a number that is 30x3=90 more than a multiple of 4.

But, 90Ã·4=22 remainder 2, so I will get a number that is 2 more than a multiple of 4.

But since 412 is a multiple of 4 (not 2 more than a multiple of 4), it won't work.

Well spotted! Luke, from Cottenham Village College, said this in a more algebraic way, using a tool called 'modular arithmetic':

All of our numbers are of the form 4x-1, for x=1, 2, 3 or 4. Therefore the sum of 30 of them comes to $4(x_1+x_2+\dots +x_{30})-30$, which is 2 mod 4 (i.e. 2 more than a multiple of 4). But 412 is 0 mod 4 (i.e. 0 more than a multiple of 4), so this sum cannot be equal to 412.

If you are unfamiliar with Modular Arithmetic, you might like to take a look at __this introductory article__.

Luke also gave his thoughts on the interactivity:

The numbers in the bag always form part of a linear arithmetic sequence, and so the number in the x-th bag is mx+c. Consecutive numbers are always a fixed distance m apart. This means that we can read off the value of m easily, and then find the value of c. We can then conclude that, if you choose z numbers from the bags, their sum will be of the form $m(x_1+x_2+\dots +x_z) + cz$, as all the numbers are of the form mx+c, for different values of x. This is obviously cz more than a multiple of m.

To illustrate this method with an example, take the numbers in the bag to be 1, 4, 7 and 10, as in the original example, with z=4.

Their differences are all a multiple of 3, so by analysing them mod 3 (i.e. by looking at their remainders when they are divided by 3) we find that they are all of the form 3x+1.

As z=4 and c=1, they must be cz=4 more than a multiple of 3.

Since 4 is 1 mod 3 (i.e. dividing 4 by 3 gives remainder 1), 4 numbers selected from the bags must add to give 1 more than a multiple of 3.

Nice!

## Teachers' Resources

### Why do this problem?

This problem offers students the opportunity to consider the underlying structure behind multiples and remainders, as well as leading to some very nice generalisations and justifications.

### Possible approach

**"In a while, you'll need to be able to explain to the rest of the class what happens when you add together three, four, five, six... 99, 100 numbers from your set of bags, and how you worked it out."**

### Key questions

### Possible support

Begin by asking students to explore what happens when they add numbers chosen from a set of bags containing 2s, 4s, 6s and 8s.

They could then consider what happens when they add numbers chosen from a set of bags containing 1s, 11s, 21s and 31s.

Can they explain their findings?

### Possible extension

There are a few related problems that students could work on next:

Take Three from Five

Shifting Times Tables

Charlie's Delightful Machine

A Little Light Thinking

Remainders

Where Can We Visit?

Cinema Problem