M09-21

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

Now, \(\sqrt{10} \gt 3\) (as \(3^2=9\)) and \(\sqrt[3]{9} \gt 2\) (as \(2^3=8\)).

Therefore:

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

\(=(\text{number more than 3})+(\text{number more than 2})+(\text{7 numbers more than 1})=\)

\(=(\text{number more than 5})+(\text{number more than 7})=\)

\(=(\text{number more than 12})\)


Answer: A
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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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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.

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.

Excellent question.

excellent question! thank you GMATCLUB.

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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HI GMATGuruNY, MentorTutoring GMATBusters

Can you please help me with this question ?
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\(x = \sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}\)

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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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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I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation. Can we assume that root 9 , root 10 etc. are positive? such as can we say that root 9 is +3 and not -3?

Even roots cannot give negative result.


\(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;
Similarly \(\sqrt{\frac{1}{16}} = \frac{1}{4}\), NOT +1/4 or -1/4.


Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I think this is a high-quality question and I agree with explanation.
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