### N000ughty Thoughts

Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in 10 000! and 100 000! or even 1 000 000!

### Mod 3

Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.

### Novemberish

a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.

# Big Powers

##### Stage: 3 and 4 Challenge Level:

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.