1. ## Find last digit

Find the last digit of (a). 13^121, (b). 8^99, (c). 777^777

I know there's something I do to write these as smaller numbers, but am unsure on how to do this.

2. The last digit of the result is determined by the exponent and the last digit of the base. Also, for any single digit d, the last digits of d^1, d^2, d^3, ... form a periodic sequence.

18^(2011)=(-2)^2011=-2^2011=-2^3*2^2008=-2^3*(2^4)^502=-2^3*6=8*6=8=2(mod10)
I guess I need to do something like this, but from this example, I don't see where the -2 came from and the mod 10

4. from this example, I don't see where the -2 came from and the mod 10
If 0 <= d < 10 and n = d (mod 10), then d is the last digit of n. Also, 18 = -2 (mod 10).

5. Ok so 13^121
3^121

8^99
(2^3)^99

777^777
7^777

That's as far as I get with all of those

6. Originally Posted by mathematic
[SIZE=2]Ok so 13^121
3^121
So start making a list.
(All of these are mod 10)
3^1 = 3
3^2 = 9
3^3 = 7
3^4 = 1
3^5 = 3
etc.

The sequence of numbers 3, 9, 7, 1 keep repeating. There are four of them, so what is 121 mod 4? That will tell you which of 3, 9, 7, or 1.

-Dan

7. 13^121
3^121
3^1=3 mod 5
13^121
1^121 mod 2
1 mod 2

8^99
=0 mod2
3^99 mod 5
3^4 mod 5
=1 mod 5

777^777
7^777
7^2 mod 5
2^2 mod 5
=4 mod 4
=1 mod 2 since 7 is odd

8. 8^99
=0 mod2
3^99 mod 5
3^4 mod 5
=1 mod 5
You are correct that $8^{99}\equiv0\pmod{2}$. On the other hand, since $n^4\equiv1\pmod{5}$ for all integer $1\le n<5$ by Fermat's little theorem, we have $8^{99}\equiv3^{99}\equiv(3^4)^{24}\cdot3^3\equiv3^ 3\equiv2\pmod{5}$. Now you probably plan to use the Chinese remainder theorem, which says that all solutions to the system of equations $x\equiv0\pmod{2}$, $x\equiv2\pmod{5}$ are congruent modulo $2\cdot 5$. Since 2 and $8^{99}$ are solutions, $8^{99}\equiv2\pmod{10}$.

One benefit of this approach is that use can use Fermat's little theorem. In contrast, $8^n\not\equiv1\pmod{10}$ for all n > 0. However, $8^n$ modulo 10 forms a periodic sequence 8, 4, 2, 6, 8, 4, ... with period 4, as has been said above (and to which you have not responded). Since $99\equiv3\pmod{4}$, we conclude that $8^{99}$ is the 3rd element of the four-element period, i.e., 2. Formally, $8^{99}\equiv(8^4)^{24}\cdot 8^3\equiv6^{24}\cdot2\equiv6\cdot2\equiv2\pmod{10}$.