Difference between revisions of "2017 AMC 12B Problems/Problem 16"
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Solution by: vedadehhc | Solution by: vedadehhc | ||
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We can write <math>21!</math> as its prime factorization: | 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> | <cmath>21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17\times19</cmath> | ||
Revision as of 21:03, 30 September 2024
Problem
The number 
has over 
positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
Solution 1
We can consider a factor of 
to be odd if it does not contain a 
; hence, finding the exponent of in the prime factorization of will help us find our answer. We can start off with all multiples of up to 
, which is 
. Then, we find multiples of 
, which is 
. Next, we look at multiples of 
, of which there are . Finally, we know there is only one multiple of 
in the set of positive integers up to . Now, we can add all of these to get 
. We know that, in the prime factorization of , we have 
, and the only way to have an odd number is if there is not a in that number's prime factorization. This only happens with 
, which is only one of the 19 different exponents of 2 we could have (of which having each exponent is equally likely). Hence, we have 
Solution by: armang32324
Solution 2
If prime factorizes into 
prime factors with exponents 
through 
, then the product of the sums of each of these exponents plus 
should be over . If 
is the exponent of in the prime factorization of 21!, then we can find the number of odd factors of by dividing the total by 
. Then, the number of odd divisors over total divisors is 
. We can find easily using Legendre's, so our final answer is 
~ icecreamrolls8
Solution 3
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 4
We can write as its prime factorization:
![\[21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17\times19\]](http://latex.artofproblemsolving.com/d/8/5/d850ffa06ddfcc6bead1e8c26aa7df40d48fd171.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:
![\[P(\text{odd})=\dfrac{\text{number of odd factors}}{\text{number of all factors}}=\dfrac{(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}{(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}=\dfrac{1}{(18+1)}=\boxed{\dfrac{1}{19}}\]](http://latex.artofproblemsolving.com/e/1/6/e16ecc5770af515dde2d0f8feafd5e952344f1c4.png)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
