2011 AMC 12B Problems/Problem 15
Contents
Problem 15
How many positive two-digit integers are factors of ?
~ pi_is_3.14
Solution
Repeating difference of squares:
The sum of cubes formula gives us:
A quick check shows is prime. Thus, the only factors to be concerned about are , since multiplying by will make any factor too large.
Multiplying by or will give a two-digit factor; itself will also work. The next smallest factor, , gives a three-digit number. Thus, there are factors that are multiples of .
Multiplying by , , or will also give a two-digit factor, as well as itself. Higher numbers will not work, giving additional factors.
Multiply by , , or for a two-digit factor. There are no more factors to check, as all factors which include are already counted. Thus, there are an additional factors.
Multiply by or for a two-digit factor. All higher factors have been counted already, so there are more factors.
Thus, the total number of factors is
Video Solution by OmegaLearn
https://youtu.be/mgEZOXgIZXs?t=770
Video Solution by WhyMath
~savannahsolver
See also
2011 AMC 12B (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.