What are the last two digits of 3^1234
if you perform the multiplication at least 6 or 10 powers of 9 (i.e up to , you should notice that the last digit is 9 for odd powers and 1 for even powers; and the second to the last digit increases from 0, 2 ,4, 6, 8 then 0 again, from one odd power to the next odd power starting from 9^1, while it decreases from 8,6,4,2,0 then 8 again, from one even power to the next even power starting from 9^2..
hence, has 69 as the last two digits..
Yet another approach is to use mod 100. This just depends on what course is asking you the question. If it's a number theory course, try this technique:
Note that , so .
Now let's determine . You can do this by straight pen and paper division to get , so .
So now we know that , which is a much simpler problem to solve.
We know , so the problem again reduces to . Note , so we reduce further to , and since , then the problem simplifies to solving
Working out this calculation, you get .
So the final answer is 69.. again. --- this is a good problem showing how different areas of math can be used to solve a single problem.