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M09-21

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New post 15 Sep 2014, 23:40
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A
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Question Stats:

46% (01:27) correct 54% (01:44) wrong based on 106 sessions

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Re M09-21  [#permalink]

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New post 15 Sep 2014, 23:40
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Official Solution:

If \(x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}\), then which of the following must be true?

A. \(x \gt 12\)
B. \(10 \lt x \lt 12\)
C. \(8 \lt x \lt 10\)
D. \(6 \lt x \lt 8\)
E. \(x \lt 6\)


Here is a little trick: any positive integer root from a number more than 1 will be more than 1. For example: \(\sqrt[1000]{2} \gt 1\).

Now, \(\sqrt{10} \gt 3\) (as \(3^2=9\)) and \(\sqrt[3]{9} \gt 2\) (\(2^3=8\)). Thus \(x=(\text{Number more than 3})+(\text{Number more than 2})+(\text{7 numbers more than 1})=\)
\(=(\text{number more than 5})+(\text{number more than 7})\) \(=(\text{number more than 12})\)

Answer: A
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Re: M09-21  [#permalink]

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New post 19 Jan 2016, 05:56
excellent question!! made my day!.

but can you please throw some light on how would we know for sure that all the numbers after the second number are more than 1.
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New post 19 Jan 2016, 07:28
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Re: M09-21  [#permalink]

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New post 18 Jul 2016, 03:43
Hi i still do not understand how root of 7 and 6 and 5 are more than 1 and not more than 2. Please explain.
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Re: M09-21  [#permalink]

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New post 18 Jul 2016, 23:51
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sheetaldodani wrote:
Hi i still do not understand how root of 7 and 6 and 5 are more than 1 and not more than 2. Please explain.


hi
assume 7√5 >1
=> 5>1^7 --- true

assume 7√5 >2
=> 5 >2^7 -- which is not true.
so 7√5 cannot be more than 2.
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Re: M09-21  [#permalink]

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New post 28 Jun 2017, 04:21
ShravyaAlladi wrote:
sheetaldodani wrote:
Hi i still do not understand how root of 7 and 6 and 5 are more than 1 and not more than 2. Please explain.


hi
assume 7√5 >1
=> 5>1^7 --- true

assume 7√5 >2
=> 5 >2^7 -- which is not true.
so 7√5 cannot be more than 2.

This explains my doubt about why (9)^1/3 > 2. Thanks
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New post 22 Jul 2017, 05:32
Excellent question.
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New post 03 May 2018, 04:47
\(\sqrt{10}\) is approx. 3.1
\(10\sqrt{2}\) is approx 1.0

Hence 3.1+...<8 values>+1, leads to a min value of 3.1+(9)(1.0), hence definitely larger than 12.
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New post 14 Jun 2018, 21:52
excellent question! thank you GMATCLUB.
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Re: M09-21  [#permalink]

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New post 12 Feb 2019, 10:23
Excellent question! Can really confuse you when you are in a hurry. But a calm approach ( In case you forget how the exponential function works) can help you tackle this.

Great job! :-)
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New post 20 Apr 2019, 15:35
Great question, but how does one get to (7 numbers more than 1)?
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New post 11 Aug 2019, 07:13
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Can you please elaborate more on this
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New post 27 Jan 2020, 23:00
Gmatekta wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Can you please elaborate more on this


(any number)^0 =1
so any power greater than 0 will definitely results in number greater than 1

Now, sqrt(10)>3 (since 3^2=9)
sqrt(9)>2 (since 2^3=8)
rest of them are greater than 1
now add them which gets to be greater than 12
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Re: M09-21  [#permalink]

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New post 27 Feb 2020, 03:39
HI GMATGuruNY, MentorTutoring GMATBusters

Can you please help me with this question ?
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Re: M09-21  [#permalink]

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New post 27 Feb 2020, 04:31
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Bunuel wrote:
If \(x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}\), then which of the following must be true?

A. \(x \gt 12\)
B. \(10 \lt x \lt 12\)
C. \(8 \lt x \lt 10\)
D. \(6 \lt x \lt 8\)
E. \(x \lt 6\)


\(x = \sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}\)

Consider the smallest value:
\(a = \sqrt[10]{2}\)
\(a^{10} = 2\)
Since \(1^{10} = 1\), the value of \(a\) must be GREATER THAN 1.
Implication:
The 7 smallest values are all greater than 1, summing to a value greater than 7.

Consider the two greatest values:
Since \(\sqrt{9} = 3\), \(\sqrt{10} > 3\).
Since \(\sqrt[3]{8} = 2\), \(\sqrt[3]{9} > 2\).

Thus:
x = (more than 3) + (more than 2) + (more than 7) = more than 12


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New post 27 Feb 2020, 04:42
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NandishSS wrote:
HI GMATGuruNY, MentorTutoring GMATBusters

Can you please help me with this question ?

Hello, NandishSS, and thank you for mentioning me. This question took me about 55 seconds to answer (correctly, ha ha). A little number sense goes a long way. Keep in mind, no matter what root is being asked about, if the number under the root symbol is NOT 0 or 1, either of which would lead to the same answer (e.g., √0 = 0, or √1 = 1), and, of course, you are not dealing with complex numbers such as i (which the GMAT™ does not have on it anyway), then the root must be greater than 1. How do you know? Because, quite simply, looking at, say,

\(\sqrt[9]{3}\)

you could multiply 1 * 1 * 1 * 1 * 1 * 1 * 1 * 1 * 1 and you will never get the product to equal 3. It can only equal 1. Without doing the math, you can say with certainty that 1-something times itself 9 times would work up to 3. Likewise, you could interpret each item in the chain in such a manner:

\(\sqrt{10}\) = slightly more than 3, since √9 = 3;

\(\sqrt[3]{9}\) = slightly more than 2, since the cube root of 8 is 2;

\(\sqrt[4]{8}\) = more than 1, since, again, we know the cube root, or third root, of 8 is 2;
...

There is no point in completing the rest, since the number under the root is greater than 1. Thus, the sum becomes

3-something + 2-something + (1-something) * 7

Even if we ignore the -something, the sum must equal at least

3 + 2 + 1 * 7, which gives us 12

Thus, putting the -something back into the mix, we know that the sum in the original question must be greater than 12, and we saved ourselves a whole lot of trouble trying to calculate specifics.

I hope that helps. If you have further questions, feel free to ask.

- Andrew
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M09-21   [#permalink] 27 Feb 2020, 04:42

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