Hi Lana.
There might be a "trick" to this which makes the computation even quicker than what I will show you, but I don't see it right away.
First two facts:
 If you are looking for the units digit of a sum of whole numbers, you only need to look at the units digits of each of the addends. So we can ignore the real "bigness" of the numbers and just watch what is happening to the units digits of the addends. Thus for example the units digit of 2346 + 2377 + 2378 + 2379 is the units digit of 6 + 7 + 8 + 9.
 If a and n are whole numbers then the units digit of a^{n} is (the units digit of a)^{n}. Thus for example the units digit of 2346^{5} is the units digit of 6^{5}.
You need to convince yourself that these facts are true.
Now look at the first few quintic numbers (numbers raised to the fifth power) maybe we'll see a pattern:
1^{5} = 1
2^{5} = 32
3^{5} = 343
4^{5} = 1024
5^{5} = 3125
(notice that the units digits are 1, 2, 3, 4, 5 !! let's see if it continues)
6^{5} = 7776
7^{5} = 16807
8^{5} = 32768
9^{5} = 59049
10^{5} = 100000
It does. We could perhaps find a reason for this pattern, but it is sufficient for us to have shown it exists. Further, using the second fact above the pattern repeats. This means that your sum of the first 2006 quintic numbers has the same units digit as the sum of the first 2006 numbers.
Thus you can use the technique of Gauss to find the sum of 1 + 2 + 3 + ... + 2006 and then read off its unit digit.
Stephen and Penny
