Thread: 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.

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.

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

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).

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

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

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

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

You are correct that . On the other hand, since for all integer by Fermat's little theorem, we have . Now you probably plan to use the Chinese remainder theorem, which says that all solutions to the system of equations , are congruent modulo . Since 2 and are solutions, .

One benefit of this approach is that use can use Fermat's little theorem. In contrast, for all n > 0. However, 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 , we conclude that is the 3rd element of the four-element period, i.e., 2. Formally, .