2008 AMC 12A Problems/Problem 15
- The following problem is from both the 2008 AMC 12A #15 and 2008 AMC 10A #24, so both problems redirect to this page.
Contents
Problem
Let . What is the units digit of ?
Solution
.
So, . Since is a multiple of four and the units digit of powers of two repeat in cycles of four, .
Therefore, . So the units digit is .
Note
Another way to get is to find the cycles of the last digit.
For , we need only be concerned with the last digit since the other digits do not affect the last digit. Since , the last digit of is .
For , note that the last digit cycles through the pattern . (You can try to see this by calculating the first powers of .)
Since is a multiple of , the last digit of is evidently
Continue as follows.
~mathboy282
Mathboy282, That is actually what solution 1 is explaining in the first sentence but I think yours is a more detailed and easier to comprehend explanation.
@graceandmymommath, I will offer my own solution below. Thanks for the comment.
Solution 2
I am going to share another approach to this problem.
A units digit for an integer implies
Let us take this step by step. First, we consider
Note that Now we calculate
Before continuing, though, we must take note of the following:
Now, we continue with the calculation.
We do the same with However, we just need to find in order to do this calculation since we have the table of
This implies that
Thus,
~mathboy282
Solution 3 (when you are limited on time)
By unit digit arithmetic, the unit digit of need to be either or Hence, we can guess one of them for a % of getting it right. This should only take 20 seconds or less. ~peelybonehead
Note: this solution is not recommended and is only advised when you have <5 minutes left.
Video Solution by OmegaLearn
https://youtu.be/-H4n-QplQew?t=36
~ pi_is_3.14
Solution 2 (Video solution)
Video: https://youtu.be/Ib-onAecb1I
Another Video
Video: https://www.youtube.com/watch?v=PzqQSOaCcnw
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2008 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.