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# What Numbers Can We Make Now?

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### Adding All Nine

### Summing Consecutive Numbers

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Age 11 to 14

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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!

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

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!

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!

15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?