Big Powers

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 $5$.

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