Difference between revisions of "2017 AMC 12B Problems/Problem 16"
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Solution by: vedadehhc | Solution by: vedadehhc | ||
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+ | ==Solution 2== | ||
+ | We can write <math>21!</math> as its prime factorization: | ||
+ | <cmath>21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17</cmath> | ||
+ | |||
+ | Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; <math>2^{18}</math> is going to have <math>19</math> factors: <math>2^0, 2^1, 2^2,...\text{ }2^{18}</math>, and the other exponents will behave identically. | ||
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+ | In other words, <math>21!</math> has <math>(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)</math> factors. | ||
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+ | We are looking for the probability that a randomly chosen factor of <math>21!</math> will be odd--numbers that do not contain multiples of <math>2</math> as factors. | ||
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+ | From our earlier observation, the only factors of <math>21!</math> that are even are ones with at least one multiplier of <math>2</math>, so our probability of finding an odd factor becomes the following: | ||
+ | <cmath>P(\text{odd})=\dfrac{\text{number of odd factors}}{\text{number of all factors}}=\dfrac{(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)}{(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)}=\dfrac{1}{(18+1)}=\boxed{\dfrac{1}{19}}</cmath> | ||
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+ | Solution submitted by TrueshotBarrage | ||
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:13, 16 February 2017
Contents
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 TrueshotBarrage
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 |
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