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08 Nov 2021, 05:37
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If \(n\) is a positive integer, and \(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}\) is NOT an integer, what is the value of \(n\)?
(1) \(n\) is a prime number
(2) \(n < 4\)
(1) \(n\) is a prime number
(2) \(n < 4\)
08 Nov 2021, 05:37
Kudos
Bookmarks
Official Solution:
If \(n\) is a positive integer, and \(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}\) is NOT an integer, what is the value of \(n\)?
Simplify given expression:
\(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}=\)
\(=\sqrt{2*3^2*5*7^2*7^{(n-1)} - 2*3^4*5*7^{(n - 1)}}=\)
\(=\sqrt{2*3^2*5*7^{(n-1)}(7^2 - 3^2)}=\)
\(=\sqrt{2*3^2*5*7^{(n-1)}(2^3*5)}=\)
\(=\sqrt{2^4*3^2*5^2*7^{(n-1)}}=\)
\(=2^2*3*5*\sqrt{7^{(n-1)}}\)
So, we are given that \(2^2*3*5\sqrt{7^{(n-1)}}\) is NOT an integer.
Notice that since \(n\) is a positive integer, then \(7^{(n-1)}\) is also a positive integer.
Now, the square root of any positive integer is either an integer or an irrational number. Meaning that, \(\sqrt{positive \ integer}\) cannot be a fraction, for example it cannot equal to \(\frac{1}{2}, \ \frac{3}{7}, \ \frac{19}{2}, \ \frac{1}{60}\) ... It MUST be an integer (1, 2, 3, ...) or an irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{7}\), ...). Thus, since \(7^{(n-1)}\) is a positive integer then \(\sqrt{7^{(n-1)}}\) is either an integer itself or an irrational number.
Therefore, for \(2^2*3*5*\sqrt{7^{(n-1)}}\) NOT to be an integer, \(\sqrt{7^{(n-1)}}\) must NOT be an integer. \(\sqrt{7^{(n-1)}}\) is NOT an integer for (positive) EVEN values of \(n\).
So, from above, \(n\) is a positive even integer: 2, 4, 6, 8, ...
(1) \(n\) is a prime number
There is only one even prime, namely 2. So, \(n = 2\). Sufficient.
(2) \(n < 4\)
There is only one positive even number less than 4, namely 2. So, \(n = 2\). Sufficient.
Answer: D
If \(n\) is a positive integer, and \(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}\) is NOT an integer, what is the value of \(n\)?
Simplify given expression:
\(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}=\)
\(=\sqrt{2*3^2*5*7^2*7^{(n-1)} - 2*3^4*5*7^{(n - 1)}}=\)
\(=\sqrt{2*3^2*5*7^{(n-1)}(7^2 - 3^2)}=\)
\(=\sqrt{2*3^2*5*7^{(n-1)}(2^3*5)}=\)
\(=\sqrt{2^4*3^2*5^2*7^{(n-1)}}=\)
\(=2^2*3*5*\sqrt{7^{(n-1)}}\)
So, we are given that \(2^2*3*5\sqrt{7^{(n-1)}}\) is NOT an integer.
Notice that since \(n\) is a positive integer, then \(7^{(n-1)}\) is also a positive integer.
Now, the square root of any positive integer is either an integer or an irrational number. Meaning that, \(\sqrt{positive \ integer}\) cannot be a fraction, for example it cannot equal to \(\frac{1}{2}, \ \frac{3}{7}, \ \frac{19}{2}, \ \frac{1}{60}\) ... It MUST be an integer (1, 2, 3, ...) or an irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{7}\), ...). Thus, since \(7^{(n-1)}\) is a positive integer then \(\sqrt{7^{(n-1)}}\) is either an integer itself or an irrational number.
Therefore, for \(2^2*3*5*\sqrt{7^{(n-1)}}\) NOT to be an integer, \(\sqrt{7^{(n-1)}}\) must NOT be an integer. \(\sqrt{7^{(n-1)}}\) is NOT an integer for (positive) EVEN values of \(n\).
So, from above, \(n\) is a positive even integer: 2, 4, 6, 8, ...
(1) \(n\) is a prime number
There is only one even prime, namely 2. So, \(n = 2\). Sufficient.
(2) \(n < 4\)
There is only one positive even number less than 4, namely 2. So, \(n = 2\). Sufficient.
Answer: D
18 Aug 2023, 04:12
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In the first when we start to simplify, wouldn't "7^2∗7^(n−1)" turn into 7^(n+1)? does this change how we solve the problem?
Many thanks.
Many thanks.
