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16 Sep 2014, 01:52
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C
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28% (02:40) correct
72%
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An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1?
A. \(\frac{17}{300}\)
B. \(\frac{1}{15}\)
C. \(\frac{2}{25}\)
D. \(\frac{1}{10}\)
E. \(\frac{3}{25}\)
A. \(\frac{17}{300}\)
B. \(\frac{1}{15}\)
C. \(\frac{2}{25}\)
D. \(\frac{1}{10}\)
E. \(\frac{3}{25}\)
16 Sep 2014, 01:52
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Official Solution:
An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1?
A. \(\frac{17}{300}\)
B. \(\frac{1}{15}\)
C. \(\frac{2}{25}\)
D. \(\frac{1}{10}\)
E. \(\frac{3}{25}\)
First, make sure that you grasp the question. To find the desired probability, we need to count the integers between 1 and 300 that fit the given constraint. (We will then divide this count by 300, to determine the final answer.)
The constraint is worded in a confusing way, so we should attempt to reword it. If necessary, put in sample numbers to make the conditions make sense. We need an "integer raised to an exponent that is an integer greater than 1." In other words, we need an integer raised to the 2nd, 3rd, 4th, etc. power. In even other words, we need perfect squares, perfect cubes, etc. that are between 1 and 300, inclusive.
Now we need to count these perfect squares, etc. in an efficient way. Let's start with the squares. What's the biggest perfect square less than 300? Test numbers if necessary. \(16^2 = 256\) and \(17^2 = 289\), but \(18^2 = 324\). Thus, we have 12 through 172, for 17 perfect squares.
Now, let's count the cubes. Leave out \(1^3\), since we've already counted 1 as \(1^2\). \(5^3 = 125\) and \(6^3 = 216\), but \(7^3 = 343\). Thus, we seem to have 5 more integer powers (\(2^3\) through \(6^3\), inclusive) - but careful! \(4^3 = 64\), which we've already counted as a square (\(8^2\)), so we only have 4 more integer powers. This gives us a cumulative total of 21.
What about the fourth powers - \(2^4\), \(3^4\), etc.? We've already counted any of these that matter, because they are also perfect squares: \(2^4 = 42\), \(3^4 = 9^2\), etc. We can leave out any higher even powers for the same reason (\(2^6 = 8^2\), \(2^8 = 16^2\), etc.).
However, we must consider additional odd powers, continuing to leave out 1 raised to any power. \(2^5 = 32\), \(3^5 = 243\), but \(4^5 =\) something greater than 300, as we can see by considering that \(4^5 = 4^4 \times 4 = 16^2 \times 4 = 256 \times 4\). So we have two more integer powers (\(2^5\) and \(3^5\)), for a cumulative total of 23.
Seventh powers: \(2^7 = 128\), but \(3^7 =\) something much greater than 300 (since \(3^7 = 3^5 \times 3^2 = 243 \times 9\)). Be sure to stop calculating when you see that the result is outside the bounds of the problem.
We have 1 more power, for a cumulative total of 24.
Ninth powers: \(2^9 = 128 \times 4 =\) greater than 300. So we can stop here.
Finally, we compute \(\frac{24}{300} = \frac{2}{25}\).
Answer: C
An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1?
A. \(\frac{17}{300}\)
B. \(\frac{1}{15}\)
C. \(\frac{2}{25}\)
D. \(\frac{1}{10}\)
E. \(\frac{3}{25}\)
First, make sure that you grasp the question. To find the desired probability, we need to count the integers between 1 and 300 that fit the given constraint. (We will then divide this count by 300, to determine the final answer.)
The constraint is worded in a confusing way, so we should attempt to reword it. If necessary, put in sample numbers to make the conditions make sense. We need an "integer raised to an exponent that is an integer greater than 1." In other words, we need an integer raised to the 2nd, 3rd, 4th, etc. power. In even other words, we need perfect squares, perfect cubes, etc. that are between 1 and 300, inclusive.
Now we need to count these perfect squares, etc. in an efficient way. Let's start with the squares. What's the biggest perfect square less than 300? Test numbers if necessary. \(16^2 = 256\) and \(17^2 = 289\), but \(18^2 = 324\). Thus, we have 12 through 172, for 17 perfect squares.
Now, let's count the cubes. Leave out \(1^3\), since we've already counted 1 as \(1^2\). \(5^3 = 125\) and \(6^3 = 216\), but \(7^3 = 343\). Thus, we seem to have 5 more integer powers (\(2^3\) through \(6^3\), inclusive) - but careful! \(4^3 = 64\), which we've already counted as a square (\(8^2\)), so we only have 4 more integer powers. This gives us a cumulative total of 21.
What about the fourth powers - \(2^4\), \(3^4\), etc.? We've already counted any of these that matter, because they are also perfect squares: \(2^4 = 42\), \(3^4 = 9^2\), etc. We can leave out any higher even powers for the same reason (\(2^6 = 8^2\), \(2^8 = 16^2\), etc.).
However, we must consider additional odd powers, continuing to leave out 1 raised to any power. \(2^5 = 32\), \(3^5 = 243\), but \(4^5 =\) something greater than 300, as we can see by considering that \(4^5 = 4^4 \times 4 = 16^2 \times 4 = 256 \times 4\). So we have two more integer powers (\(2^5\) and \(3^5\)), for a cumulative total of 23.
Seventh powers: \(2^7 = 128\), but \(3^7 =\) something much greater than 300 (since \(3^7 = 3^5 \times 3^2 = 243 \times 9\)). Be sure to stop calculating when you see that the result is outside the bounds of the problem.
We have 1 more power, for a cumulative total of 24.
Ninth powers: \(2^9 = 128 \times 4 =\) greater than 300. So we can stop here.
Finally, we compute \(\frac{24}{300} = \frac{2}{25}\).
Answer: C
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21 Jul 2015, 08:14
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any other simpler/faster method?
