What are the last two digits of 3^2007?

Answer 1

3^1=03
3^2=09
3^3=27
3^4=81
3^5=.43
3^6=.29
3^7=.87
3^8=.01 and then the cycle begins again
So divide 2007 into 8 and take the remainder 2007 = 8*250+ 7 .So the two last digits are those of 3^7 =87

Answer 2

Below are the powers of 3 up to 3^21

x _ 3^x
0 _ 1
1 _ 3
2 _ 9
3 _ 27
4 _ 81
5 _ 243
6 _ 729
7 _ 2187
8 _ 6561
9 _ 19683
10 _ 59049
11 _ 177147
12 _ 531441
13 _ 1594323
14 _ 4782969
15 _ 14348907
16 _ 43046721
17 _ 129140163
18 _ 387420489
19 _ 1162261467
20 _ 3486784401
21 _ 10460353203

At x = 20, the last two digits will repeat (03, 09, 27, ...). So every 20 powers will have the same last two numbers. Since 2007 is equivalent to 7 mod 20 (meaning that if you divide 20 into 2007, you get a remainder of 7), then it will have the same last two digits of 3^7, so 87.

Answer 3

If you are pre-college, then you just have to see some
repeating patterns of 3^i for various i values, but by the
result below, one has to go all the way to 3^40, and that
is just too much.

If you are in college, then we know by the Euler's theorem,
3^phi(100) = 1 (mod 100) since gcd(3, 100) = 1.
Since phi(100) = phi(4)phi(25) = (2^2 - 2)(5^2 - 5) = 2 * 20 = 40, we know that 3^40 = 1 (mod 100).
Write 2007 = 50 * 40 + 7. Then
3^2007 = 3^(50 * 40 + 7) = 3^(50 * 40) * 3^7
= (3^40)^50 * 3^7 = 1^50 * 3^7 = 3^7 = 2187 = 87 (mod 100).

Answer 4

To answer this question notice that the powers of 3 cycle in their last digit as 3. 9. 7, 1
... nevermind, someone else beat me to it!

Answer 5

87

Answer 6

i'm not a math whiz, but i'll say ^17

Answer 7

5,7