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# M09-21

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:40
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Difficulty:

65% (hard)

Question Stats:

43% (01:20) correct 57% (01:46) wrong based on 87 sessions

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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. $$x \lt 6$$

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16 Sep 2014, 00:40
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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 $$x=(\text{Number more than 3})+(\text{Number more than 2})+(\text{7 numbers more than 1})=$$
$$=(\text{number more than 5})+(\text{number more than 7})$$ $$=(\text{number more than 12})$$

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19 Jan 2016, 06:56
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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19 Jan 2016, 08:28
1
gangeyyo wrote:
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.

Say $$\sqrt[1000]{2} \leq 1$$. Then $$2 \leq 1^{1000}$$, which is not true. Thus $$\sqrt[1000]{2}$$ must be more than 1.
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18 Jul 2016, 04:43
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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19 Jul 2016, 00:51
2
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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28 Jun 2017, 05:21
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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22 Jul 2017, 06:32
Excellent question.
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03 May 2018, 05:47
$$\sqrt{10}$$ 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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14 Jun 2018, 22:52
excellent question! thank you GMATCLUB.
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12 Feb 2019, 11:23
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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20 Apr 2019, 16:35
Great question, but how does one get to (7 numbers more than 1)?
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11 Aug 2019, 08:13
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Can you please elaborate more on this
Re M09-21   [#permalink] 11 Aug 2019, 08:13
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