How does this solution include the odd factors like 1 or 3 or 5 or 7 or 9 or 1*3, 1*5,etc?
Bunuel wrote:
OfficialSolution:
If one of the positive factors of\(10!\) is picked at random, what is the probability that it will be odd?
A.\(\frac{1}{8}\)
B.\(\frac{1}{9}\)
C.\(\frac{1}{10}\)
D.\(\frac{1}{11}\)
E.\(\frac{1}{12}\)
To determine the probability, first calculate the highest power of 2 in\(10!\) by summing the quotients of 10 divided by powers of 2. This is computed as\( \frac{10}{2} + \frac{10}{4} + \frac{10}{8} = 5 + 2 + 1 =8\) . Therefore,\(10!\) includes a factor of\(2^8\) .
Now, consider that EACH odd factor of 10! can be combined with powers of 2, ranging from\(2^0\) , which is 1, to\(2^8\) . This results in 9 possible combinations for each oddfactor:
\( \text{(odd factor)}*2^0, \ \text{(odd factor)}*2^1, \ \text{(odd factor)}*2^2, \ ..., \ \text{(oddfactor)}*2^8\).
Among these combinations, only the first one,\( \text{(oddfactor)*1}\) , is odd. Thus, the probability of selecting an odd factor from the factors of
...
If one of the positive factors of\(10!\) is picked at random, what is the probability that it will be odd?
A.\(\frac{1}{8}\)
B.\(\frac{1}{9}\)
C.\(\frac{1}{10}\)
D.\(\frac{1}{11}\)
E.\(\frac{1}{12}\)
To determine the probability, first calculate the highest power of 2 in\(10!\) by summing the quotients of 10 divided by powers of 2. This is computed as\( \frac{10}{2} + \frac{10}{4} + \frac{10}{8} = 5 + 2 + 1 =8\) . Therefore,\(10!\) includes a factor of\(2^8\) .
Now, consider that EACH odd factor of 10! can be combined with powers of 2, ranging from\(2^0\) , which is 1, to\(2^8\) . This results in 9 possible combinations for each oddfactor:
\( \text{(odd factor)}*2^0, \ \text{(odd factor)}*2^1, \ \text{(odd factor)}*2^2, \ ..., \ \text{(oddfactor)}*2^8\).
Among these combinations, only the first one,\( \text{(oddfactor)*1}\) , is odd. Thus, the probability of selecting an odd factor from the factors of
...
Statistics : Posted by unicornilove • on 31 Dec 2023, 09:20 • Replies 2 • Views 474
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