Question

What is the last digit of 3^2002?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

Answer 1

\[3^0=1\\
3^1=3\\
3^2=9\\
3^3=27\to 7\\
3^4=81\to 1\] and then it repeats

Answer 2

EXAMPLE
Suppose that you are going to construct a table with 358 rows and record last digit of the the 358 powers of 7. Start the table, either using either your calculator or doing the arithmetic by hand.

Power of 7 Last digit
71 7
72 9
73 3
74 1
75 7
76 9
77 3
78 1
79 7
Notice the pattern of the last digits. They are 7,9,3,1,7,9,3,1,7,9,... The last digit repeats in a pattern that is 4 digits long, 7,9,3,1.

If you complete the table for 358 rows how many times will this pattern repeat? 358 divided by 4 is 89 with a remainder of 2 so the pattern will repeat 89 times and then there are two more rows. These rows then have 7 and 9 in the second column so the last digit of 7358 is 9.

Answer 3

or just divide \(2002\) and see that the integer remainder is 2, and \(3^2=9\)

Answer 4

Oh..
@satellite73 What do you divide 2002 by

Answer 5

pattern is 3, 9, 7, 1 i.e. every fourth one has the remainder of 1, so i divided by four

Answer 6

\(3^4\to 1\)
\[3^8\to 1\] etc

Answer 7

so \(3^{2000}\to 1\)
\[3^{20001}\to 3\\
3^{2002}\to 9\]

Answer 8

Oh...I see it now.
Thanks! c:

Answer 9

yw
once you get a remainder of 1, it starts over, you do not need to keep going

Answer 10

I know its answered already but I just wanted to share another method you can use, which is the Euler's Totient Function for this problem

\[a^{\phi(n)}= 1(\mod n) \]
so we know \[\phi(10) = 4\]
and (3,10) = 1

so \[3^{2002} = (3^{4})^{500}(3^2) = (1)^{500}(3^2) = 9 (\mod 10)\]
so you now have 9(mod10) = 9
Thus the last digit of \[3^{2002}\] is 9