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by [email protected] » Mon Dec 02, 2013 2:19 pm
Hi parulmahajan89,

The GMAT will NOT require you to calculate fractional "cube-roots" or fractional "quad-roots", so that's not what you should do here.

Notice how the answer choices are all "ranges"; that means that we can probably estimate the correct answer.

4^1/2 = 2 (that calculation is easy)

By comparison, 1^1/3 = 1 and 8^1/3 = 2, so...

4^1/3 is between 1 and 2

By comparison, 1^1/4 = 1 and 16^1/4 = 2, so....

4^1/4 is between 1 and 2

So, adding these values up gives us a total that is greater than 4.

Final Answer: E

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by GMATGuruNY » Mon Dec 02, 2013 9:18 pm
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Since the problem above asks only for an estimation, we can ballpark:

√4 = 2.
4^(1/3) > 1.
4^(1/4) = √2 ≈ 1.4.

So M > 4.4.

The correct answer is E.
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by Mathsbuddy » Tue Dec 03, 2013 10:16 am
This goes beyond GMat, but hey...

Let b^12 = 4
Then we get M = (b^12)^1/2 + (b^12)^1/3 + (b^12)^1/4

So M = (b^6) + (b^4) + (b^3) = b^3 * (b^3 + b^1 + 1)

In base b, this would look like M = 100 * 1011 = 101100 (Base b)

Also 4 = 1000000000000 in base b

101100 < 1000000000000

So M < 4

Who can spot the flaw...
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by [email protected] » Tue Dec 03, 2013 11:56 pm
Hi Mathsbuddy,

I'm not sure what your goal is with this last post, but you seem "know" that what you're attempting is beyond the scope of the GMAT. It also appears that you also know that you made a mistake at some point in your logic. So I have to ask WHY are spending time presenting an idea that you KNOW is not part of the GMAT AND that you know is incorrect?

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by Mathsbuddy » Wed Dec 04, 2013 9:09 am
[email protected] wrote:Hi Mathsbuddy,

I'm not sure what your goal is with this last post, but you seem "know" that what you're attempting is beyond the scope of the GMAT. It also appears that you also know that you made a mistake at some point in your logic. So I have to ask WHY are spending time presenting an idea that you KNOW is not part of the GMAT AND that you know is incorrect?

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Hi Rich,

I see your point. Basically I saw this creative 'solution' in a flash in my head, so I had to try it out; and I thought that if I get it right, then for me personally it would be a super-quick way of solving it. The problem is that the answer is wrong and I couldn't see the mistake! If there's someone out there who understand bases, then maybe they can let me know what stupid thing I've overlooked :)
I made it clear at the start that it's not GMat, so people don't waste time on it. However, if someone does get it and can fix it, either I'll have a neat reusable method for future, or I can happily discard it if it's worthless. If it can be made to work, it just offers an alternative approach. I like to share logic from all sorts of angles.

A simplified (and corrected) version would be:

Let 4 = b^12 = 1000000000000 in base b where b > 1
So M = (b^6) + (b^4) + (b^3) = 1011000 (in Base b)

I just found the mistake now:
1011000 (Base b) = 5.001614614 (base ten) > 4

So M > 4 (Correct :) )

Addendum (In base b:)
1000000 > 1
10000 > 1
1000 = 2 (base ten)
Therefore 1011000 > 4 (base ten)

I can't believe I didn't spot that before!

May that be a lesson to any base experts who expect a higher order value to be greater than a lower order value. This would only be true if the base were an integer > 1. That was my error.
The method is easily checked with a calculator. Although, it may not be ideal for this question (as the 12th root of 4 is not very practical), it could be very handy for the r = nth root of t, where r is an integer rather than a decimal.

If anyone finds this confusing, then just ignore it. (Otherwise I thought it potentially a clever quick trick).

Alternatively:

M-2 = 4^(1/3) + 4^(1/4) > 1 + 1.4 = 2.4
2.4 > 4 - 2
so M > 4

which is easier to visualise, (whether or not you can think in different bases, such as binary,etc.)
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by Mathsbuddy » Wed Dec 04, 2013 9:18 am
Further to the my last post, here is the "base method" with the bases removed!

M = 4^1/2 + 4^1/3 + 4^1/4

So M = 2 + 4^1/3 + 4^1/4

Let 4 = b^12 where b > 1, (so b^n > 1 for n > 0)
So M = 2 + (b^4) + (b^3) > 2 + 1 + 1 = 4

So M > 4
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