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 1==
 
==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>
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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>.
  
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.
+
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==
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==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: http://www.cut-the-knot.oI lick toesrg/blue/LegendresTheorem.shtml and 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{1/19, \space \text{B}}</math>
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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>
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==Solution 3==
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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==

Latest revision as of 14:53, 4 September 2022

Problem

The number $21!=51,090,942,171,709,440,000$ has over $60,000$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

$\textbf{(A)}\ \frac{1}{21}\qquad\textbf{(B)}\ \frac{1}{19}\qquad\textbf{(C)}\ \frac{1}{18}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{11}{21}$

Solution 1

We note that the only thing that affects the parity of the factor are the powers of 2. There are $10+5+2+1 = 18$ factors of 2 in the number. Thus, there are $18$ cases in which a factor of $21!$ would be even (have a factor of $2$ in its prime factorization), and $1$ case in which a factor of $21!$ would be odd. Therefore, the answer is $\boxed{\textbf{(B)} \frac 1{19}}$.

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 $D$ of $21!$. 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. $D$ can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for $D$ to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases = $\boxed{\textbf{(B)} \frac 1{19}}$

Solution 3

We can find the prime factorization of $21!$. We do this by finding the prime factorization of each of 21, 20, ..., 2, and 1 and multiplying them together. This gives us $2^{18} \cdot 3^{9} \cdot 5^{4} \cdot 7^{3} \cdot 11 \cdot 13 \cdot 17 \cdot 19$. To find the number of odd divisors, we pretend as if the $2^{18}$ doesn't exist and count the other divisors: $10 \cdot 5 \cdot 4 \cdot 2 \cdot 2 \cdot 2 \cdot 2$. The total number of divisors are $19 \cdot 10 \cdot 5 \cdot 4 \cdot 2 \cdot 2 \cdot 2 \cdot 2$. Dividing gives $\boxed{\textbf{(B)} ~\frac{1}{19}}$.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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