Difference between revisions of "2017 AMC 12B Problems/Problem 16"
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| − | If <math>21!</math> prime factorizes into <math>p</math> prime factors with exponents <math>e_1</math> through <math>e_k</math>, then the product of the sums of each of these exponents plus <math>1</math> should be <math> | + | If <math>21!</math> prime factorizes into <math>p</math> prime factors with exponents <math>e_1</math> through <math>e_k</math>, then the product of the sums of each of these exponents plus <math>1</math> should be <math>60,000</math>. If we divide this product by the exponent of <math>2</math> plus <math>1</math>, then we should get the number of odd factors. Then, the fraction of odd divisors over total divisors is <math>\dfrac{1}{e_2+1}</math> if <math>e_2</math> is the exponent of <math>2</math>. We can find <math>e_2</math> easily using Legendre's, so our final answer is <math>\dfrac{1}{10 + 5 + 3 + 1} = \boxed{\text{(B)} \dfrac{1}{19}}.</math> |
~ icecreamrolls8 | ~ icecreamrolls8 | ||
Revision as of 20:25, 31 October 2022
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 
prime factorizes into 
prime factors with exponents 
through 
, then the product of the sums of each of these exponents plus 
should be . If we divide this product by the exponent of 
plus , then we should get the number of odd factors. Then, the fraction of odd divisors over total divisors is 
if 
is the exponent of . We can find easily using Legendre's, so our final answer is 
~ icecreamrolls8
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:
![\[21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17\times19\]](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. 
