if 2007^2007 is multiplied out, what is the ones digit of the final product?

2007^2007 = 2007^3 * 2007^(5 * 401 - 1) = 2007^3 = 3 (mod 5)

∴ the ones digit is 3.

You only care about the last digit. We will start multiplying by 2007, but at each step remove everything but the last digit.
2007 ends in 7
7*2007 ends in 9
9*2007 ends in 3
3*2007 ends in 1
and so on.

Thus
2007^1 ends in 7.
2007^2 ends in 9.
2007^3 ends in 3.
2007^4 ends in 1.
2007^5 ends in 7.
2007^6 ends in 9.
2007^7 ends in 3.
2007^8 ends in 1.
and so on

Notice how it cycles through 7,9,3,1, a cycle of length 4.
If the exponent is a multiple of 4, it will end in 1.
So 2007^2004 ends in 1.
Now go 3 more through the cycle 7,9,3.
So 2007^2007 ends in 3.

The final digit is 3.

The last digit cycles 7 - 9 - 3 - 1. So 2007^(4n) ends in "1".
For example, 2007^2004 ends in "1".
2007^2005 ends in "7"
2007^2006 ends in "9"
2007^2007 ends in "3".

With similar logic, it can be shown that the last three digits are "543".

It's harder to show, but the last six digits are "423,543".

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