Difference between revisions of "2017 AMC 12B Problems/Problem 16"
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<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> | <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 1== | ||
+ | |||
+ | We can consider a factor of <math>21!</math> to be odd if it does not contain a <math>2</math>; hence, finding the exponent of <math>2</math> in the prime factorization of <math>21!</math> will help us find our answer. We can start off with all multiples of <math>2</math> up to <math>21</math>, which is <math>10</math>. Then, we find multiples of <math>4</math>, which is <math>5</math>. Next, we look at multiples of <math>8</math>, of which there are <math>2</math>. Finally, we know there is only one multiple of <math>16</math> in the set of positive integers up to <math>21</math>. Now, we can add all of these to get <math>10+5+2+1=18</math>. We know that, in the prime factorization of <math>21!</math>, we have <math>2^{18}</math>, and the only way to have an odd number is if there is not a <math>2</math> in that number's prime factorization. This only happens with <math>2^{0}</math>, 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 <math>\boxed{\text{(B)} \dfrac{1}{19}}.</math> | ||
+ | |||
+ | Solution by: armang32324 | ||
==Solution== | ==Solution== | ||
− | 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 over <math>60,000</math>. If | + | 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 over <math>60,000</math>. If <math>e_2</math> is the exponent of <math>2</math> in the prime factorization of 21!, then we can find the number of odd factors of <math>21!</math> by dividing the total by <math>e_2+1</math>. Then, the number of odd divisors over total divisors is <math>\dfrac{1}{e_2+1}</math>. We can find <math>e_2</math> easily using Legendre's, so our final answer is <math>\dfrac{1}{10 + 5 + 2 + 1+1} = \boxed{\text{(B)} \dfrac{1}{19}}.</math> |
~ icecreamrolls8 | ~ icecreamrolls8 |
Latest revision as of 13:55, 10 July 2023
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
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.