# Find last digit

• Feb 21st 2011, 01:44 PM
mathematic
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.
• Feb 21st 2011, 02:12 PM
emakarov
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 PM
mathematic
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 PM
emakarov
Quote:

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).
• Feb 21st 2011, 03:28 PM
mathematic
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 PM
topsquark
Quote:

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
• Feb 21st 2011, 05:18 PM
mathematic
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 AM
emakarov
Quote:

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, $8^{99}\equiv(8^4)^{24}\cdot 8^3\equiv6^{24}\cdot2\equiv6\cdot2\equiv2\pmod{10}$.