Difference between revisions of "2017 AMC 12B Problems/Problem 16"
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− | If a factor of <math>21!</math> is odd, that means it contains no factors of <math>2</math>. We can find the number of factors of two in <math>21!</math> by counting the number multiples of <math>2</math>, <math>4</math>, <math>8</math>, and <math>16</math> that are less than or equal to <math>21</math>.After some quick counting we find that this number is <math>10+5+2+1 = 18</math>. If the prime factorization of <math>21!</math> has <math>18</math> factors of <math>2</math>, there are <math>19</math> choices for each divisor for how many factors of <math>2</math> should be included (<math>0</math> to <math>18</math> inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of <math>2</math> is <math>0</math> which is <math>\frac{1}{19} \ | + | ==Problem== |
+ | The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd? | ||
+ | |||
+ | <math>\textbf{(A)}\ \frac{1}{21} \qquad \textbf{(B)}\ \frac{1}{19} \qquad \textbf{(C)}\ \frac{1}{18} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{11}{21}</math> | ||
+ | |||
+ | ==Solution== | ||
+ | If a factor of <math>21!</math> is odd, that means it contains no factors of <math>2</math>. We can find the number of factors of two in <math>21!</math> by counting the number multiples of <math>2</math>, <math>4</math>, <math>8</math>, and <math>16</math> that are less than or equal to <math>21</math> (Legendre's Formula). After some quick counting we find that this number is <math>10+5+2+1 = 18</math>. If the prime factorization of <math>21!</math> has <math>18</math> factors of <math>2</math>, there are <math>19</math> choices for each divisor for how many factors of <math>2</math> should be included (<math>0</math> to <math>18</math> inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of <math>2</math> is <math>0</math> which is <math>\boxed{\textbf{(B)}\frac{1}{19}}</math>. | ||
+ | |||
+ | Solution by: vedadehhc | ||
+ | |||
+ | ==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\times19</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. | ||
+ | |||
+ | In other words, <math>21!</math> has <math>(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)</math> factors. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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)(1+1)}{(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}=\dfrac{1}{(18+1)}=\boxed{\dfrac{1}{19}}</cmath> | ||
+ | |||
+ | Solution submitted by [[User:TrueshotBarrage|David Kim]] | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/ZLHNTSIcGM8 | ||
+ | |||
+ | -MistyMathMusic | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] | ||
+ | [[Category:Introductory Probability Problems]] |
Latest revision as of 13:55, 15 February 2021
Problem
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 (Legendre's Formula). 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
Video Solution
-MistyMathMusic
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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