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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
65% (01:48) correct
35%
(02:04)
wrong
based on 570
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If \(s\) is a positive integer and \(s^4 = 120t\), which of the following must be integers?
(I) \(\frac{t}{2^2*3^2*5^2}\)
(II) \(\frac{t}{2*3*5}\)
(III) \(\frac{t}{2*3*5*15}\)
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III
(I) \(\frac{t}{2^2*3^2*5^2}\)
(II) \(\frac{t}{2*3*5}\)
(III) \(\frac{t}{2*3*5*15}\)
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III
31 May 2022, 00:24
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Key details from the question stem
(i) \(s^4 = 120t\)
(ii) s is an integer
Key inferences
1. \(s^4\) is an integer.
The question is missing one key information. i.e., the value of t is also an integer.
We are going to work on the premise that t is an integer. Without this piece of information, none of the three expressions will be an integer. And we do not have an answer option that states none of the above.
For instance, if t = \(\frac{2}{15}\), \(s^4\) will be 16 and hence it will satisfy all conditions. None of the 3 expressions given in I, II, and III will be an integer.
Approach: Find the minimum value of 't' and check whether the expressions will be an integer for the minimum value of 't'.
Step 1: Prime factorize 120:
120 =\(2^3 * 3 * 5\)
Therefore, 120t can be expressed as \(2^3 * 3 * 5 * t\) (t here is not a prime number)
\(s^4 = 2^3 * 3 * 5 * t\)
On the left side of the equation, we have an integer that is raised to power 4.
So, every prime factor on the right side should also have its powers as a multiple of 4.
Thus, the minimum value of t should be \(2 * 3^3 * 5^3\) to make the powers of all the prime factors a multiple of 4 - which is 4 in this instance.
Please note that this is the minimum value t can take, and not the only possibility.
If the following expressions (I), (II), and (III) are integers for the minimum value of t, it will be an integer for all other possibilities too.
Expression (I) \(\frac{t}{(2^2 * 3^2 * 5^2)}\)
Substituting the value of t as \(2 * 3^3 * 5^3\), we get
\(\implies\) \(\frac{(2 * 3^3 * 5^3)}{(2^2 * 3^2 * 5^2)}\)
\(\implies\) \(\frac{(3 * 5)}{2}\)
This number is not an integer.
So, options (A), (C), and (E) can be eliminated.
Expression (II) \(\frac{t}{(2 * 3 * 5)}\)
Substituting the value of t as \(2 * 3^3 * 5^3\), we get
\(\implies\) \(\frac{(2 * 3^3 * 5^3)}{(2 * 3 * 5)}\)
\(\implies\) \(\frac{(3^2 * 5^2)}{1}\)
We get an integer.
Expression (III) \(\frac{t}{(2 * 3 * 5 * 15)}\)
Substituting the value of t as \(2 * 3^3 * 5^3\), we get
\(\implies\) \(\frac{(2 * 3^3 * 5^3)}{(2 * 3 * 5 * 15)}\)
\(\implies\) \(\frac{(3 * 5)}{1}\)
We get an integer.
So, expressions in II and III are integers.
The correct answer is Option D
(i) \(s^4 = 120t\)
(ii) s is an integer
Key inferences
1. \(s^4\) is an integer.
The question is missing one key information. i.e., the value of t is also an integer.
We are going to work on the premise that t is an integer. Without this piece of information, none of the three expressions will be an integer. And we do not have an answer option that states none of the above.
For instance, if t = \(\frac{2}{15}\), \(s^4\) will be 16 and hence it will satisfy all conditions. None of the 3 expressions given in I, II, and III will be an integer.
Approach: Find the minimum value of 't' and check whether the expressions will be an integer for the minimum value of 't'.
Step 1: Prime factorize 120:
120 =\(2^3 * 3 * 5\)
Therefore, 120t can be expressed as \(2^3 * 3 * 5 * t\) (t here is not a prime number)
\(s^4 = 2^3 * 3 * 5 * t\)
On the left side of the equation, we have an integer that is raised to power 4.
So, every prime factor on the right side should also have its powers as a multiple of 4.
Thus, the minimum value of t should be \(2 * 3^3 * 5^3\) to make the powers of all the prime factors a multiple of 4 - which is 4 in this instance.
Please note that this is the minimum value t can take, and not the only possibility.
If the following expressions (I), (II), and (III) are integers for the minimum value of t, it will be an integer for all other possibilities too.
Expression (I) \(\frac{t}{(2^2 * 3^2 * 5^2)}\)
Substituting the value of t as \(2 * 3^3 * 5^3\), we get
\(\implies\) \(\frac{(2 * 3^3 * 5^3)}{(2^2 * 3^2 * 5^2)}\)
\(\implies\) \(\frac{(3 * 5)}{2}\)
This number is not an integer.
So, options (A), (C), and (E) can be eliminated.
Expression (II) \(\frac{t}{(2 * 3 * 5)}\)
Substituting the value of t as \(2 * 3^3 * 5^3\), we get
\(\implies\) \(\frac{(2 * 3^3 * 5^3)}{(2 * 3 * 5)}\)
\(\implies\) \(\frac{(3^2 * 5^2)}{1}\)
We get an integer.
Expression (III) \(\frac{t}{(2 * 3 * 5 * 15)}\)
Substituting the value of t as \(2 * 3^3 * 5^3\), we get
\(\implies\) \(\frac{(2 * 3^3 * 5^3)}{(2 * 3 * 5 * 15)}\)
\(\implies\) \(\frac{(3 * 5)}{1}\)
We get an integer.
So, expressions in II and III are integers.
The correct answer is Option D
General Discussion
04 May 2022, 09:07
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GMATNinja
\(s^4 = 120t\)
Or, \(t = \frac{s^4}{120}\)
Or, \(t = \frac{s^4}{2^3*3*5}\)
Now check the options for the correct values , all but (I) is possible, hence, Answer must be (D) II and III only....
