1964 AHSME Problems/Problem 16
Problem
Let and let be the set of integers . The number of members of such that has remainder zero when divided by is:
Solution
Note that for all polynomials , .
Proof: If , then . In the second equation, we can use the binomial expansion to expand every term, and then subtract off all terms that have a factor of or higher, since subtracting a multiple of will not change congruence . This leaves , which is , so .
So, we only need to test when has a remainder of for . The set of numbers will repeat remainders, as will all other sets. The remainders are .
This means for , is divisible by . Since is divisible, so is for , which is values of that work. Since is divisible, so is for , which is more values of that work. The values of will also generate solutions each, just like . This is a total of values of , for an answer of
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.