1965 AHSME Problems/Problem 23
Problem
If we write for all such that , the smallest value we can use for is:
Solution
There are two cases: (i) and (ii) . In case (i), , so . In case (ii), , so . Because in case (i) can take a larger value, we use it to determine the upper bound for under the given restrictions. Now, we can see that , which is answer choice .
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.