Difference between revisions of "2017 AMC 12B Problems/Problem 16"
m |
(Added another solution) |
||
| Line 8: | Line 8: | ||
Solution by: vedadehhc | 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</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)</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)}{(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)}=\dfrac{1}{(18+1)}=\boxed{\dfrac{1}{19}}</cmath> | ||
| + | |||
| + | 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 22:13, 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:
![\[21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17\]](http://latex.artofproblemsolving.com/0/8/c/08c53f9a96c43c2c4dd304198340cc3bc916d114.png)
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:
![\[P(\text{odd})=\dfrac{\text{number of odd factors}}{\text{number of all factors}}=\dfrac{(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)}{(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)}=\dfrac{1}{(18+1)}=\boxed{\dfrac{1}{19}}\]](http://latex.artofproblemsolving.com/2/4/2/2420bd32cf2ffb2fea24599a8213130953b066a1.png)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
