2011 AMC 12B Problems/Problem 15
Contents
[hide]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 |
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All AMC 12 Problems and Solutions |
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