Question

what is the units digit of 3^3887?

Answer 1

First look at some powers of 3 and observe the units digit.

Answer 2

Chart source: http://mathcentral.uregina.ca/QQ/database/QQ.09.05/alex1.html

Answer 3

3^0=1
3^1 = 3
3^2=9
3^3=27
3^4=81
3^5=243
3^6=729
3^7=2187
3^8=6561
3^9=19683

Pattern of units digit is 1,3,9,7,1,3,9,7,1,3,9,7,...
Every exponent with the same Units digit is separated from each other by 4 units in the exponent:
3^0,3^4,3^8 have units of 1
3^1,3^5,3^9 have units of 3
3^2,3^6 have units of 9
3^3,3^7 have units of 7

And so on.

Conjecture:

If the number is a perfect 4th root then units digit is a 1
If the number is a perfect cube root then units digit is a 7
If the number is a perfect square root then units digit is a 9
Otherwise, units digit is a 3

For example \(\sqrt{729}=27\), so last digit is a 9.

For our case 3^3887 is a perfect cube, so units digit must be a 7.

Answer 4

Of course, 3^3887 is a perfect cube. We need to work on the conjecture a little more. I am not certain the order we need to take them yet so I'm not certain on the units digit. Maybe you can take something from this, maybe not.

Answer 5

New Conjecture:
Take the exponent of 3^N

If N mod 4 = 0 then Units digit is a 1
If N mod 4 = 1 then Units digit is a 3
If N mod 4 = 2 then Units digit is a 9
If N mod 4 = 3 then Units digit is a 7

In our case N = 3887
So 3887 mod 4 = 3, so Units digit is a 7 (after all).

Answer 6

7 is correct.
Refer to the complete expansion attached. Calculated and formatted in
33 micro seconds by Mathemtica v9.