2010 AMC 10A Problems/Problem 24
Problem
The number obtained from the last two nonzero digits of is equal to . What is ?
Solution 1
We will use the fact that for any integer ,
First, we find that the number of factors of in is equal to . Let . The we want is therefore the last two digits of , or . Since there is clearly an excess of factors of 2, we know that , so it remains to find .
We can write as where where every number in the form is replaced by .
The number can be grouped as follows:
Hence, we can reduce to
Using the fact that ,we can deduce that . Therefore .
Finally, combining with the fact that yields .
Solution 2(bash)
First, we list out all the numbers
Since we must get rid of ending s, we get rid of and the corresponding
Next, we note that ,, and , so it can be simplified to
Then, we bash using these following difference of squares tricks and multiplication
1.
2.
3.
4.
Handling of powers of and are especially critical
In the end, we get
See also
2010 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.