Difference between revisions of "2017 AMC 12B Problems/Problem 16"
Rockmanex3 (talk | contribs) (→See Also) |
|||
Line 24: | Line 24: | ||
Solution submitted by [[User:TrueshotBarrage|David Kim]] | Solution submitted by [[User:TrueshotBarrage|David Kim]] | ||
+ | ==Solution 3 (Time Shortage)== | ||
+ | |||
+ | <math>\textbf{WARNING}</math>: Only use this solution if you are running out of time. | ||
+ | |||
+ | Prime factor <math>21!</math>. It is <math>2^{18}\cdot3^9\cdot5^4\cdot7^3\cdot11\cdot13\cdot17\cdot19</math>. | ||
+ | |||
+ | The number of factors is <math>19\cdot10\cdot5\cdot4\cdot2\cdot2\cdot2\cdot2</math>. | ||
+ | Looking at the choices and using some quick computation, one can figure that the only sensical answer is <math>\boxed{\textbf{(B)}\frac{1}{19}}</math>. | ||
+ | |||
+ | <math>\textbf{-RootThreeOverTwo}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}} |
Revision as of 02:35, 13 December 2018
Problem 16
The number has over positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
Solution
If a factor of is odd, that means it contains no factors of . We can find the number of factors of two in by counting the number multiples of , , , and that are less than or equal to .After some quick counting we find that this number is . If the prime factorization of has factors of , there are choices for each divisor for how many factors of should be included ( to inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of is which is .
Solution by: vedadehhc
Solution 2
We can write as its prime factorization:
Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; is going to have factors: , and the other exponents will behave identically.
In other words, has factors.
We are looking for the probability that a randomly chosen factor of will be odd--numbers that do not contain multiples of as factors.
From our earlier observation, the only factors of that are even are ones with at least one multiplier of , so our probability of finding an odd factor becomes the following:
Solution submitted by David Kim
Solution 3 (Time Shortage)
: Only use this solution if you are running out of time.
Prime factor . It is .
The number of factors is . Looking at the choices and using some quick computation, one can figure that the only sensical answer is .
See Also
2017 AMC 12B (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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.