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M26-07

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M26-07  [#permalink]

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New post 16 Sep 2014, 01:24
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58% (01:57) correct 42% (02:36) wrong based on 135 sessions

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If \(x\) is a positive integer then the value of \(\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}\) is closest to which of the following?

A. \(2^{11x}\)
B. \(11^{11x}\)
C. \(22^{11x}\)
D. \(2^{22x}*11^{11x}\)
E. \(2^{22x}*11^{22x}\)

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Re M26-07  [#permalink]

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New post 16 Sep 2014, 01:24
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Official Solution:

If \(x\) is a positive integer then the value of \(\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}\) is closest to which of the following?

A. \(2^{11x}\)
B. \(11^{11x}\)
C. \(22^{11x}\)
D. \(2^{22x}*11^{11x}\)
E. \(2^{22x}*11^{22x}\)


Note that we need approximate value of the given expression. Now, \(22^{22x}\) is much larger number than \(22^{2x}\). Hence \(22^{22x}-22^{2x}\) will be very close to \(22^{22x}\) itself, basically \(22^{2x}\) is negligible in this case. The same way \(11^{11x}-11^x\) will be very close to \(11^{11x}\) itself.

Thus \(\frac{22^{22x}-22^x}{11^{11x}-11^x} \approx \frac{22^{22x}}{11^{11x}}=\frac{2^{22x}*11^{22x}}{11^{11x}}=2^{22x}*11^{11x}\).

You can check this algebraically as well: \(\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}=\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}\). Again, -1, both in denominator and numerator is negligible value and we'll get the same expression as above: \(\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)} \approx \frac{22^{2x}*22^{20x}}{11^x*11^{10x}}=\frac{22^{22x}}{11^{11x}}=2^{22x}*11^{11x}\).


Answer: D
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Re M26-07  [#permalink]

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New post 28 Oct 2015, 08:06
I think this is a high-quality question and I agree with explanation. Doesn't the x go away in 11^11x?? I'm getting 2^22x*11^11
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Re: M26-07  [#permalink]

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New post 29 Oct 2015, 01:12
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danjbon wrote:
I think this is a high-quality question and I agree with explanation. Doesn't the x go away in 11^11x?? I'm getting 2^22x*11^11


Check below:

\(\frac{22^{22x}}{11^{11x}}=\frac{(2*11)^{22x}}{11^{11x}}=\frac{2^{22x}*11^{22x}}{11^{11x}}=2^{22x}*11^{22x-11x}=2^{22x}*11^{11x}\)

Does it make sense?
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Re: M26-07  [#permalink]

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New post 10 Mar 2018, 14:06
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Great question

It very easy to waste a lot of time trying to solve this question algebraically (because it looks a little like you need to use difference of squares) if you miss the hint "closet to" and don't realise you're being asked to approximate, thus can take out 22^2X and 11^2X as factors and discard the +1s they produce when factoring out.

However, even if you don't notice that, you should realise that once you break down 22 to prime factors (22^22=2^22*11^22) you have powers of 2 up on the numerator and nothing to cancel that or decrease it on the denominator. In contrast, you do have powers of 11 on the denominator - so look for an answer choice where the exponent of the 2s is unchanged but the exponent on the 11s is being decreased - there's only one option which has that, which is D. (not to mention 11^22/11^11 is 11^11 so D is in the right 'range' for what you'd expect the exponent on 11 to be, as well)
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Re: M26-07  [#permalink]

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New post 10 May 2018, 12:17
This is a very nice question. I spent lots of time trying to figure out the solution. The explanation give is very nice.
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Re: M26-07  [#permalink]

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New post 25 Jul 2018, 07:16
+1 for option D. One needs to use approximation here.
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Re: M26-07 &nbs [#permalink] 25 Jul 2018, 07:16
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