Difference between revisions of "2020 AMC 8 Problems/Problem 17"
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/VnOecUiP-SA | https://youtu.be/VnOecUiP-SA | ||
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+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=782 | ||
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+ | ~Interstigation | ||
==See also== | ==See also== |
Latest revision as of 14:33, 18 April 2021
Contents
Problem
How many positive integer factors of have more than factors? (As an example, has factors, namely and )
Solution 1
Since , we can simply list its factors: There are of these; only (i.e. of them) don't have over factors, so the remaining factors have more than factors.
Solution 2
As in Solution 1, we prime factorize as , and we recall the standard formula that the number of positive factors of an integer is found by adding to each exponent in its prime factorization, and then multiplying these. Thus has factors. The only number which has one factor is . For a number to have exactly two factors, it must be prime, and the only prime factors of are , , and . For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of is . Thus, there are factors of which themselves have , , or factors (namely , , , , and ), so the number of factors of that have more than factors is .
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=782
~Interstigation
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.