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Bunuel

If you look at all of those questions, you'll notice not one of them is official. The GMAT never tests infinite sums and products like the ones you see in these questions, because to answer questions like this, first you need to establish that these infinite expressions "converge" (they don't become infinite or undefined) and to do that, you need calculus. Calculus is not tested on the GMAT.

So these questions are all out of scope, and I don't see how the technique one uses to solve them could be relevant to any actual GMAT question, but if we need to solve, we can do as follows. Let x equal the expression:

\(\\
x = 10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\\
\)

Now notice that after the first "\(\sqrt[n]{ }\)", what we see is identical to x. So we can replace it with "x", and we have

\(x = 10*\sqrt[n]{x}\)

Now we just need to solve this equation: x/10 = x^(1/n), so x^n / 10^n = x, and x^(n-1) = 10^n, so x = 10^(n/(n-1))

With these answer choices, it's even faster to notice that we're multiplying 10 here by something greater than 1. So the answer must be greater than 10, and so must be D or E. But the larger n is, the smaller the roots all get, so the smaller the eventual answer should be. So E can't be right, because E gets bigger if you make n bigger, and only D is possible.
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Bunuel
If n is an integer greater than 1, what is the value of \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\), where the given expression extends to an infinite number of roots?


A. 10

B. \(10^{\frac{1}{n}}\)

C. \(10^{\frac{n-1}{n}}\)

D. \(10^{\frac{n}{n-1}}\)

E. \(10^{n}\)



Are You Up For the Challenge: 700 Level Questions

Asked: If n is an integer greater than 1, what is the value of \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\), where the given expression extends to an infinite number of roots?

Let x = \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\)

\(x = 10*\sqrt[n]{x}\)
\(x^n = 10^n * x\)
\(x^n - 10^n * x = 0 \)
\(x^{n-1} - 10^n = 0\) since \(x \neq 0\)
\(x^{n-1} = 10^n\)
\(x = 10^{\frac{n}{n-1}}\)

IMO D
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Bunuel
If n is an integer greater than 1, what is the value of \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\), where the given expression extends to an infinite number of roots?


A. 10

B. \(10^{\frac{1}{n}}\)

C. \(10^{\frac{n-1}{n}}\)

D. \(10^{\frac{n}{n-1}}\)

E. \(10^{n}\)


Solution


    • Given n is an integer greater than 1
      o This means, \(\frac{1}{n} < 1\)
    • Now, \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{….,}}}} = 10*10^{\frac{1}{n}}*10^{\frac{1}{n}*\frac{1}{n}}*10^{\frac{1}{n}*\frac{1}{n}*\frac{1}{n}}* ……,\)
      o \(⟹ 10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{….,}}}} = 10^{1+\frac{1}{n} + \frac{1}{n^2} +\frac{1}{n^3}…..}\)
    • We can observe that \({\frac{1}{n} + \frac{1}{n^2} +\frac{1}{n^3}…..}\) is an infinite G.P.,
      o \(⟹ 10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{….,}}}} = 10^{1+\frac{1}{n} + \frac{1}{n^2} +\frac{1}{n^3}…..}\) = \( 10^{\frac{1}{(1-\frac{1}{n})}}\) = \(10^{\frac{n}{n-1}}\)

Thus, the correct answer is Option D.
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Bunuel
If n is an integer greater than 1, what is the value of \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\), where the given expression extends to an infinite number of roots?


A. 10

B. \(10^{\frac{1}{n}}\)

C. \(10^{\frac{n-1}{n}}\)

D. \(10^{\frac{n}{n-1}}\)

E. \(10^{n}\)



Solution:

Letting x = the expression, we can create the equation:

x = 10 * n^√x

x = 10x^(1/n)

x^(1 - 1/n) = 10

x^[(n - 1)/n] = 10

x = 10^[n/(n - 1)]

Answer: D
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Asked: If n is an integer greater than 1, what is the value of \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\), where the given expression extends to an infinite number of roots?

Let x= \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\)

\(x = 10*\sqrt[n]{x}\)

\(x^n = 10^n*x\)
\(x^{n-1} = 10^n\)
\(x = 10^{\frac{n}{n-1}}\)

IMO D
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Even if you have no clue what to think of this questions you can instantly establish 2 things.

1: N is a positive integer greater than 1, meaning the nth root of a term is going to reduce it
2:10 times a root of N is going to be larger than 10 (the root will never be less than 1, even if n=100000000)

A) we already established this isn't possible as out answer must be greater than 10
B) 1/n is the equivalent to the nth root, this is one of the many terms in the expression and not the answer.
C) n-1/n is going to be less than 1, if n is less than 1 we cannot satisfy the first condition of the question tan n is greater than 1
D) if n/n-1, n can be an integer greater than 1
E) this answer is going to be dramatically larger than the real answer

D is the correct answer, and we were able to solve this while making no attempt to determine the value of this equation.
Bunuel
If n is an integer greater than 1, what is the value of \(10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{10*\sqrt[n]{...}}}}\), where the given expression extends to an infinite number of roots?


A. 10

B. \(10^{\frac{1}{n}}\)

C. \(10^{\frac{n-1}{n}}\)

D. \(10^{\frac{n}{n-1}}\)

E. \(10^{n}\)



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will this kind of question actually come in GMAT Focus Edition?
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architkap
will this kind of question actually come in GMAT Focus Edition?
It is quite unlikely that you will see a question exactly like this on the actual GMAT. However, this and similar questions can still be useful for sharpening your knowledge of exponents and roots and improving your algebraic skills, which are relevant to GMAT Quant.
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