Difference between revisions of "2017 AMC 10B Problems/Problem 20"

Latest revision as of 13:53, 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 $\boxed{\textbf{(B)} ~\frac{1}{19}}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 ==Problem==
-Placeholder
+The number <math>21!=51,090,942,171,709,440,000</math> has over <math>60,000</math> positive integer divisors. One of them is chosen at random. What is the probability that it is odd?
-==Solution==
+<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>
-We note that the only thing that affects the parity of the factor are the powers of 2. Since there are <math>10+5+2+1 = 18</math> factors of 2, and since only haveing a 0 power of 2 makes the factor odd, then the answer is <math>\boxed{\textbf{(B) } \frac 1{19}}</math>.
+==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>.
+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 <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==
 {{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}}

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