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Bunuel
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If n is a positive integer, and (45*14*7^n - 15*7^(n - 1)*54)^(1/2) is 17 Mar 2020, 05:38
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39% (02:37) correct 61% (02:52) wrong based on 244 sessions

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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


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M36-52

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Bunuel
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If n is a positive integer, and (45*14*7^n - 15*7^(n - 1)*54)^(1/2) is 08 Nov 2021, 05:38
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Bunuel
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


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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
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If n is a positive integer, and (45*14*7^n - 15*7^(n - 1)*54)^(1/2) is 17 Mar 2020, 05:42
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Bunuel
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


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Similar question: https://gmatclub.com/forum/if-n-is-a-po ... 18655.html
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If n is a positive integer, and (45*14*7^n - 15*7^(n - 1)*54)^(1/2) is Updated on: 17 Mar 2020, 06:04
Originally posted by CareerGeek on 17 Mar 2020, 05:50.
Last edited by CareerGeek on 17 Mar 2020, 06:04, edited 1 time in total.
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Bunuel
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
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**Edited
\(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}\)
= \(\sqrt{2*3^2*5*7^{n + 1} - 2*3^4*5*7^{(n - 1)}}\)
= \(\sqrt{2*3^2*5*7^2*7^{n - 1} - 2*3^4*5*7^{(n - 1)}}\)
= \(\sqrt{2*3^2*5*7^{n - 1}(7^2 - 3^2)}\)
= \(\sqrt{2*3^2*5*7^{n - 1}(40)}\)
= \(\sqrt{2*3^2*5*7^{n - 1}(2^3*5)}\)
= \(\sqrt{2^4*3^2*5^2*7^{n - 1}}\)
= \(2^2*3*5\sqrt{7^{n - 1}}\)
= \(60\sqrt{7^{n - 1}}\) is NOT an integer

The above is NOT an integer for all even values of n greater than 0, \(n = {2, 4, 6, 8, 10, . . . . . . }\)

(1) n is a prime number
--> Possible value of \(n = 2\) only Sufficient

(2) n < 4
--> Possible value of \(n = 2\) only Sufficient

Option D
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If n is a positive integer, and (45*14*7^n - 15*7^(n - 1)*54)^(1/2) is 17 Mar 2020, 05:58
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√(45x14x7^n - 15x7^(n-1) x 54) = √(15x3x2x7^(n+1) - 15x3x2x9x7^(n-1)) = √(15x6x7^(n-1) x (7^2 - 9))
=√(90x7^(n-1) x 40) = √(3600x7^(n-1)) = 60√(7^(n-1)
The question is basically asking what is the value of n, given that √(7^(n-1) is not an integer.

Statement 1: n is a prime number.
Sufficient. Since we know that the only even prime number is 2. All other values of n, which is a prime number apart from 2, will cause √(7^(n-1) to be an integer.

Statement 2: n<4
Sufficient. Since we know from the question stem that n is a positive integer, so possible values of n are: 1, 2, 3, and only n=2 will satisfy the condition that √(7^(n-1) is not an integer.

The answer is therefore D.
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If n is a positive integer, and (45*14*7^n - 15*7^(n - 1)*54)^(1/2) is 02 May 2020, 21:04
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Is there a faster way to solve it? I could have done 2, 3 and 5, but I that would take more than 2 minutes to do.
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If n is a positive integer, and (45*14*7^n - 15*7^(n - 1)*54)^(1/2) is 09 May 2020, 15:10
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There has to be a faster way right? I spent two minutes on the prime factorization alone.
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If n is a positive integer, and (45*14*7^n - 15*7^(n - 1)*54)^(1/2) is 08 Oct 2023, 00:15
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