# Tens

When is $7^n + 3^n$ a multiple of 10? Can you prove the result by two different methods?

## Problem

*This problem follows on from Power Mad!*

Work out $9^n + 1^n$ for a few odd values of $n$.

What do you notice?

Can you prove it?

Now try the same with the following:

$7^n + 3^n$, where $n$ is odd.

$8^n - 2^n$, where $n$ is even.

$6^n - 4^n$, where $n$ is even.

Can you find any more results like these?

## Getting Started

You could write $9^n$ as $(10-1)^n$ and consider the terms of the binomial expansion...

Or you could use Modular Arithmetic - there's a useful introduction here.

Or you could use Modular Arithmetic - there's a useful introduction here.

## Student Solutions

We received lots of great solutions for this problem, from Niharika at Rugby School, Lee from Garden International School, Joseph from Bingley Grammar, Nathan from Mossbourne Community Academy, James from Samuel Whitbread Academy, and Seemanta from St Placid's High School.

Most of you noticed that all the sums are multiples of ten; here is a solution sent in by Amir from Wallington Country Grammar School:

In order for a number to be a multiple of 10, it must end in a zero. We need to find out if $3^n + 7^n$ will end in a 0 in order to answer this question.

To do this we can construct a table to see what the 'units' digit is when we put either $3^n$ or $7^n$.

It is evident that only the odd values of N give a multiple of 10, however we can see a pattern; the unit of the number will cycle after you add 4. For example, $3^1$ ends in the same digit as $3^5$ and $3^9$. As a result the final digit of the sum $3^n + 7^n$ will always be 1 of 4 possibilities. It is either '3 + 7', '9 + 9', '7 + 3' or '1 + 1', of which two are multiples of 10 and happen to be when n is either 1 or 3. Therefore we can see that adding any multiple of 4 to 1 or 3 will always yield an odd number - hence odd number values of n will always give multiples of 10.

We can extend this by saying that when n is in the form '4k + 2', the final digit will be an 8 (9+9=18) and when n is in the form '4k' (i.e. is a multiple of 4), it will end in 2 - either way it will be a multiple of 2. Therefore the sum is always even.

Note: k is any integer greater or equal to 0.

Daniel from Chetham's School of Music provided proofs of the same result using induction and the Binomial theorem. Here is his solution: .pdf

Finally, Michael noticed that this result can be generalised in another way:

By factorising generalised expressions we can prove all four of the given expressions are divisible by ten. For example,

$$x^{2m+1}+y^{2m+1}=(x+y)(x^{2m}-x^{2m-1}y+x^{2m-2}y^2-\dots -xy^{2m-1}+y^{2m})$$

Meaning that for any two numbers that add up to 10 (or any multiple of 10), $x^n+y^n$ is divisible by 10 for odd n. Thus this proves the first two given expressions divide by ten. Secondly, observe that

$$x^{2m}-y^{2m}=(x+y)(x^{2m-1}-x^{2m-2}y+\dots -y^{2m-1})$$

Meaning that for any two numbers that add up to some multiple of 10, $x^n-y^n$ is divisible by 10 for even n. This factorisation therefore proves the last two expressions divide by ten.

Well done everyone!

Most of you noticed that all the sums are multiples of ten; here is a solution sent in by Amir from Wallington Country Grammar School:

In order for a number to be a multiple of 10, it must end in a zero. We need to find out if $3^n + 7^n$ will end in a 0 in order to answer this question.

To do this we can construct a table to see what the 'units' digit is when we put either $3^n$ or $7^n$.

n | $3^n$ unit | $7^n$ unit | Sum end in 0? |

1 | 3 | 7 | Y |

2 | 9 | 9 | N |

3 | 7 | 3 | Y |

4 | 1 | 1 | N |

5 | 3 | 7 | Y |

6 | 9 | 9 | N |

7 | 7 | 3 | Y |

It is evident that only the odd values of N give a multiple of 10, however we can see a pattern; the unit of the number will cycle after you add 4. For example, $3^1$ ends in the same digit as $3^5$ and $3^9$. As a result the final digit of the sum $3^n + 7^n$ will always be 1 of 4 possibilities. It is either '3 + 7', '9 + 9', '7 + 3' or '1 + 1', of which two are multiples of 10 and happen to be when n is either 1 or 3. Therefore we can see that adding any multiple of 4 to 1 or 3 will always yield an odd number - hence odd number values of n will always give multiples of 10.

We can extend this by saying that when n is in the form '4k + 2', the final digit will be an 8 (9+9=18) and when n is in the form '4k' (i.e. is a multiple of 4), it will end in 2 - either way it will be a multiple of 2. Therefore the sum is always even.

Note: k is any integer greater or equal to 0.

Daniel from Chetham's School of Music provided proofs of the same result using induction and the Binomial theorem. Here is his solution: .pdf

Finally, Michael noticed that this result can be generalised in another way:

By factorising generalised expressions we can prove all four of the given expressions are divisible by ten. For example,

$$x^{2m+1}+y^{2m+1}=(x+y)(x^{2m}-x^{2m-1}y+x^{2m-2}y^2-\dots -xy^{2m-1}+y^{2m})$$

Meaning that for any two numbers that add up to 10 (or any multiple of 10), $x^n+y^n$ is divisible by 10 for odd n. Thus this proves the first two given expressions divide by ten. Secondly, observe that

$$x^{2m}-y^{2m}=(x+y)(x^{2m-1}-x^{2m-2}y+\dots -y^{2m-1})$$

Meaning that for any two numbers that add up to some multiple of 10, $x^n-y^n$ is divisible by 10 for even n. This factorisation therefore proves the last two expressions divide by ten.

Well done everyone!

## Teachers' Resources

### Why do this problem?

As well as offering practice in working with indices, the proofs in this problem can be done using several different techniques, using the binomial theorem or modular arithmetic.### Possible approach

See Power Mad! for a related problem that you may wish to set students instead of or as well as this one.

Start by inviting students to explore $9^n+1^n$ for odd values of $n$. When they observe that the answers all seem to be multiples of $10$, give students a short while to come up with their own convincing arguments. They are likely to use an argument based on the last digits of odd powers of $9$, and this could be a good opportunity to introduce the language and notation of modular arithmetic, and the idea of working mod $10$. This article could be given to students as background reading.

Alternatively, if you wish to focus on using the binomial theorem, invite students to express the questions in terms that can be expanded: for example, rewriting $7^n + 3^n$ as $7^n + (10-7)^n$. Students can write out the expression for odd and even values of $n$ and construct a convincing justification why the expression is always a multiple of $10$ when $n$ is odd.

### Key questions

What happens to the units digit when you find consecutive powers of a number?Can we use the fact that the numbers occurring in each part add up to 10?