# Big Powers

Three people chose this as a favourite problem. It is the sort of
problem that needs thinking time - but once the connection is made
it gives access to many similar ideas.

## Problem

## Getting Started

What determines whether a number is divisible by $5$ or not?

Find: $3^{1}, 3^{2}, 3^{3}, 3^{4}, 3^{5}, 3^{6}, 3^{7}, 3^{8}, 3^{9} \ldots$

What do you notice?

Find: $4^{1}, 4^{2}, 4^{3}, 4^{4}, 4^{5}, 4^{6}, 4^{7}, 4^{8}, 4^{9} \ldots$

What do you notice?

## Student Solutions

In the main, successful solvers of this problem went about it in one of two ways.

One way was to `break it down' and study the end digit for a set of powers. This is what Samantha and Zoe of Maidstone Girls' Grammar School and Angela from Hethersett High in Norfolk did.

Another way was to look at some remainders when dividing by $5$. This is what Avishek, James, Martin, Thomas, Kintel and Marcus from Simon Langton Boys' School did.

This is Angela's solution:

You have to think about how you distinguish numbers that are
divisible by five, they are numbers that end in a five or a zero.
So you know that for this number to divide by five it has to end in
a five or zero and it is on the basis of this knowledge that we can
work out this problem.

There is a pattern in the units digits for powers of
$3$:

$3$

$3\times 3=9$

$3\times 3\times 3=27$

$3\times 3\times 3\times 3=81$

$3\times 3\times 3\times 3\times 3=243$

$3\times 3\times3\times 3\times 3\times 3=729$

From this pattern we know that $3^{444}$ is going to end in a
$1$ because $444$ is divisible by $4$. $4^{333}$ works in much the
same way. The units digit for powers of $4$ are alternately $4$ and
$6$. Using these results we see that $4^{333}$ will end in a
$4$.

So now combine all the information we have so far which
is:

$3^{444}$ will end in a $1$

$4^{333}$ will end in a $4$

and $1+4=5$.

So $3^{444}+4^{333}$ will end in a $5$ and so it is divisible
by $5$.

There is an
assumption here that these patterns in the units digits continue to
hold for ALL positive powers of $3$ and $4$ and you might like to
take up the challenge of proving this.

Students who
have been introduced to some Advanced Level Mathematics might be
interested in reading the following solution sent in by
Giridhar:

$$3^{444} + 4^{333} = (3^4)^{111} + (4^3)^{111} = a^{111} +
b^{111}$$ where we define $a = 3^4$ and $b=4^3$.

Now, any expression of the form: $a^n + b^n$, has $(a+b)$ as a
factor, when $n = 1, 3, 5 \dots$

That is, we can write $$a^n + b^n = (a+b )( \dots)$$ We know
that: $$ a + b = 3^4 + 4^3 = 81 + 64 =145,$$ and since $145$ is
divisible by $5$, $3^{444} + 4^{333}$ must be divisible by $5$
too.

## Teachers' Resources

At first glance, a challenging problem; but no algebra is required to justify the solution.

Students who meet this problem for the first time may need a significant amount of support in structuring a solution so it is useful to be able to find similar tasks to which they may apply their new-found understanding.

It is important to be able to justify any pattern. How can you be sure it continues?