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.

Printable View

- Feb 21st 2011, 01:44 PMmathematicFind 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. - Feb 21st 2011, 02:12 PMemakarov
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.

- Feb 21st 2011, 02:23 PMmathematic
Well, the example we did in class was this:

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 - Feb 21st 2011, 02:54 PMemakarovQuote:

from this example, I don't see where the -2 came from and the mod 10

- Feb 21st 2011, 03:28 PMmathematic
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 - Feb 21st 2011, 04:02 PMtopsquark
- Feb 21st 2011, 05:18 PMmathematic
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 - Feb 22nd 2011, 01:51 AMemakarovQuote:

8^99

=0 mod2

3^99 mod 5

3^4 mod 5

=1 mod 5

One benefit of this approach is that use can use Fermat's little theorem. In contrast, $\displaystyle 8^n\not\equiv1\pmod{10}$ for all n > 0. However, $\displaystyle 8^n$ 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 $\displaystyle 99\equiv3\pmod{4}$, we conclude that $\displaystyle 8^{99}$ is the 3rd element of the four-element period, i.e., 2. Formally, $\displaystyle 8^{99}\equiv(8^4)^{24}\cdot 8^3\equiv6^{24}\cdot2\equiv6\cdot2\equiv2\pmod{10}$.