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# M09-21

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:40
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65% (hard)

Question Stats:

46% (01:27) correct 54% (01:44) wrong based on 106 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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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 $$x=(\text{Number more than 3})+(\text{Number more than 2})+(\text{7 numbers more than 1})=$$
$$=(\text{number more than 5})+(\text{number more than 7})$$ $$=(\text{number more than 12})$$

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19 Jan 2016, 05:56

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, 07:28
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gangeyyo wrote:

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, 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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18 Jul 2016, 23:51
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, 04:21
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, 05:32
Excellent question.
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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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14 Jun 2018, 21:52
excellent question! thank you GMATCLUB.
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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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20 Apr 2019, 15:35
Great question, but how does one get to (7 numbers more than 1)?
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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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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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27 Feb 2020, 03:39
HI GMATGuruNY, MentorTutoring GMATBusters

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27 Feb 2020, 04:31
1
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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27 Feb 2020, 04:42
2
NandishSS wrote:
HI GMATGuruNY, MentorTutoring GMATBusters

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

# M09-21

Moderators: chetan2u, Bunuel