1957 AHSME Problems/Problem 42
Revision as of 09:10, 27 July 2024 by Thepowerful456 (talk | contribs) (see also box, formatting change)
Contents
Problem 42
If , where and is an integer, then the total number of possible distinct values for is:
Solution
We first use the fact that . Note that and , so and are periodic with periods at most 4. Therefore, it suffices to check for .
For , we have .
For , we have .
For , we have .
For , we have .
Hence, the answer is .
Solution 2
Notice that the powers of cycle in cycles of 4. So let's see if is periodic.
For : we have .
For : we have .
For : we have .
For : we have .
For : we have again. Well, it can be seen that cycles in periods of 4. Select .
~hastapasta
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 41 |
Followed by Problem 43 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.