Difference between revisions of "2017 AMC 12B Problems/Problem 16"

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Problem 16

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

If a factor of $21!$ is odd, that means it contains no factors of $2$ . We can find the number of factors of two in $21!$ by counting the number multiples of $2$ , $4$ , $8$ , and $16$ that are less than or equal to $21$ .After some quick counting we find that this number is $10+5+2+1 = 18$ . If the prime factorization of $21!$ has $18$ factors of $2$ , there are $19$ choices for each divisor for how many factors of $2$ should be included ( $0$ to $18$ inclusive). The probability that a randomly chosen factor is odd is the same as if the number of factors of $2$ is $0$ which is $\boxed{\textbf{(B)}\frac{1}{19}}$ .

Solution by: vedadehhc

Solution 2

We can write $21!$ as its prime factorization: $21!=2^{18}\times3^9\times5^4\times7^3\times11\times13\times17\times19$

Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; $2^{18}$ is going to have $19$ factors: $2^0, 2^1, 2^2,...\text{ }2^{18}$ , and the other exponents will behave identically.

In other words, $21!$ has $(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)$ factors.

We are looking for the probability that a randomly chosen factor of $21!$ will be odd--numbers that do not contain multiples of $2$ as factors.

From our earlier observation, the only factors of $21!$ that are even are ones with at least one multiplier of $2$ , so our probability of finding an odd factor becomes the following: $P(\text{odd})=\dfrac{\text{number of odd factors}}{\text{number of all factors}}=\dfrac{(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}{(18+1)(9+1)(4+1)(3+1)(1+1)(1+1)(1+1)(1+1)}=\dfrac{1}{(18+1)}=\boxed{\dfrac{1}{19}}$

Solution submitted by David Kim

Solution 3 (Time Shortage)

$\textbf{WARNING}$ : Only use this solution if you are running out of time.

Prime factor $21!$ . It is $2^{18}\cdot3^9\cdot5^4\cdot7^3\cdot11\cdot13\cdot17\cdot19$ .

The number of factors is $19\cdot10\cdot5\cdot4\cdot2\cdot2\cdot2\cdot2$ . Looking at the choices and using some quick computation, one can figure that the only sensical answer is $\boxed{\textbf{(B)}\frac{1}{19}}$ .

$\textbf{-RootThreeOverTwo}$

@@ Line 24: / Line 24: @@
 Solution submitted by [[User:TrueshotBarrage|David Kim]]
+==Solution 3 (Time Shortage)==
+<math>\textbf{WARNING}</math>: Only use this solution if you are running out of time.
+Prime factor <math>21!</math>. It is <math>2^{18}\cdot3^9\cdot5^4\cdot7^3\cdot11\cdot13\cdot17\cdot19</math>.
+The number of factors is <math>19\cdot10\cdot5\cdot4\cdot2\cdot2\cdot2\cdot2</math>.
+Looking at the choices and using some quick computation, one can figure that the only sensical answer is <math>\boxed{\textbf{(B)}\frac{1}{19}}</math>.
+<math>\textbf{-RootThreeOverTwo}</math>
 ==See Also==
 {{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}}

2017 AMC 12B (Problems • Answer Key • Resources)
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