Difference between revisions of "2017 AMC 10B Problems/Problem 20"
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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== | + | ==Solution 1== |
− | We note that the only thing that affects the parity of the factor are the powers of 2. There are <math>10+5+2+1 = 18</math> factors of 2 in the number. Thus, there are <math>18</math> cases in which a factor of <math>21!</math> would be even (have a factor of <math>2</math> in its prime factorization), and <math>1</math> case in which a factor of <math>21!</math> would be odd. Therefore, the answer is <math>\boxed{\textbf{(B)} \frac 1{19}}</math> | + | We note that the only thing that affects the parity of the factor are the powers of 2. There are <math>10+5+2+1 = 18</math> factors of 2 in the number. Thus, there are <math>18</math> cases in which a factor of <math>21!</math> would be even (have a factor of <math>2</math> in its prime factorization), and <math>1</math> case in which a factor of <math>21!</math> would be odd. Therefore, the answer is <math>\boxed{\textbf{(B)} \frac 1{19}}</math>. |
− | ==Solution 2 | + | Note from Williamgolly: To see why symmetry occurs here, we group the factors of 21! into 2 groups, one with powers of 2 and the others odd factors. For each power of 2, the factors combine a certain number of 2's from the first group and numbers from the odd group. That is why symmetry occurs here. |
− | Consider how to construct any divisor <math>D</math> of <math>21!</math>. First by Legendre's theorem for the divisors of a factorial, we have that there are a total of 18 factors of 2 in the number. <math>D</math> can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for <math>D</math> to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases= <math>\boxed{1/19, \ | + | |
+ | ==Solution 2 (Constructive Counting)== | ||
+ | Consider how to construct any divisor <math>D</math> of <math>21!</math>. First by Legendre's theorem for the divisors of a factorial (see here: [[Legendre's Formula]]), we have that there are a total of 18 factors of 2 in the number. <math>D</math> can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for <math>D</math> to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases = <math>\boxed{\textbf{(B)} \frac 1{19}}</math> | ||
+ | ==Solution 3== | ||
+ | We can find the prime factorization of <math>21!</math>. We do this by finding the prime factorization of each of 21, 20, ..., 2, and 1 and multiplying them together. This gives us <math>2^{18} \cdot 3^{9} \cdot 5^{4} \cdot 7^{3} \cdot 11 \cdot 13 \cdot 17 \cdot 19</math>. To find the number of odd divisors, we pretend as if the <math>2^{18}</math> doesn't exist and count the other divisors: <math>10 \cdot 5 \cdot 4 \cdot 2 \cdot 2 \cdot 2 \cdot 2</math>. The total number of divisors are <math>19 \cdot 10 \cdot 5 \cdot 4 \cdot 2 \cdot 2 \cdot 2 \cdot 2</math>. Dividing gives <math>\boxed{\textbf{(B)} ~\frac{1}{19}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}} | ||
+ | {{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:53, 4 September 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 1
We note that the only thing that affects the parity of the factor are the powers of 2. There are factors of 2 in the number. Thus, there are
cases in which a factor of
would be even (have a factor of
in its prime factorization), and
case in which a factor of
would be odd. Therefore, the answer is
.
Note from Williamgolly: To see why symmetry occurs here, we group the factors of 21! into 2 groups, one with powers of 2 and the others odd factors. For each power of 2, the factors combine a certain number of 2's from the first group and numbers from the odd group. That is why symmetry occurs here.
Solution 2 (Constructive Counting)
Consider how to construct any divisor of
. First by Legendre's theorem for the divisors of a factorial (see here: Legendre's Formula), we have that there are a total of 18 factors of 2 in the number.
can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for
to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases =
Solution 3
We can find the prime factorization of . We do this by finding the prime factorization of each of 21, 20, ..., 2, and 1 and multiplying them together. This gives us
. To find the number of odd divisors, we pretend as if the
doesn't exist and count the other divisors:
. The total number of divisors are
. Dividing gives
.
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 10 Problems and Solutions |
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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