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Re: Ax(y) is an operation that adds 1 to y [#permalink]

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03 Jul 2012, 21:51

If I read the question correctly, AX(Y) = (1+Y)X Putting X=2/3, => AX(X) = (1+ 2/3)*(2/3) => AX(X) = (5/3)*(2/3) => AX(X) = 10/9 > 1

AX(AX(X) = (1+10/9)(10/9)....The first term in the bracket will be greater than 2 and next term is greater than one...so the product is greater than 2. The next operation will make the (1+Y) term greater than 3, so basically as per my understanding there is some information missing in the question or I misunderstood.

Ax(y) is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then Ax(Ax(Ax(Ax(Ax(x))))) is between

(A) 0 and ½ (B) ½ and 1 (C) 1 and 1½ (D) 1½ and 2 (E) 2 and 2½

Sorry, but I don't have the OA.

The question should read:

\(A_x(y)\) is an operation that adds 1 to y and then multiplies the result by x. If x = 2/3, then \(A_x(A_x(A_x(A_x(A_x(x)))))\) is between

(A) 0 and ½ (B) ½ and 1 (C) 1 and 1½ (D) 1½ and 2 (E) 2 and 2½

According to the stem: \(A_x(x)=(x+1)x=x^2+x\); \(A_x(A_x(x))=A_x(x^2+x)=(x^2+x+1)*x=x^3+x^2+x\); \(A_x(A_x(A_x(x)))=A_x(x^3+x^2+x)=(x^3+x^2+x+1)*x=x^4+x^3+x^2+x\); ...

We can see the pattern now, so \(A_x(A_x(A_x(A_x(A_x(x)))))=x^6+x^5+x^4+x^3+x^2+x\).

For \(x=\frac{2}{3}\) we'll get: \((\frac{2}{3})^6+(\frac{2}{3})^5+(\frac{2}{3})^4+(\frac{2}{3})^3+(\frac{2}{3})^2+(\frac{2}{3})\).

So, we have the sum of the 6 terms of the geometric progression with the first term equal to \(\frac{2}{3}\) and the common ratio also equal to \(\frac{2}{3}\).

Now, the sum of infinite geometric progression with common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term. So, if we had infinite geometric progression instead of just 6 terms then its sum would be \(Sum=\frac{\frac{2}{3}}{1-\frac{2}{3}}=2\). Which means that the sum of this sequence will never exceed 2, also as we have big enough number of terms (6) then the sum will be very close to 2, so we can safely choose answer choice D.

Answer: D.

One can also use direct formula. We have geometric progression with \(b=\frac{2}{3}\), \(r=\frac{2}{3}\) and \(n=6\);

\(S_n=\frac{b(1-r^n)}{(1-r)}\) --> \(S_{6}=\frac{\frac{2}{3}(1-(\frac{2}{3})^{6})}{(1-\frac{2}{3})}=2*(1-(\frac{2}{3})^{6})\). Since \((\frac{2}{3})^{6}\) is very small number then \(1-(\frac{2}{3})^{6}\) will be less than 1 but very close to it, hence \(2*(1-(\frac{2}{3})^{6})\) will be less than 2 but very close to it.

Re: Ax(y) is an operation that adds 1 to y [#permalink]

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15 Jan 2014, 19:53

tusharacc wrote:

If I read the question correctly, AX(Y) = (1+Y)X Putting X=2/3, => AX(X) = (1+ 2/3)*(2/3) => AX(X) = (5/3)*(2/3) => AX(X) = 10/9 > 1

AX(AX(X) = (1+10/9)(10/9)....The first term in the bracket will be greater than 2 and next term is greater than one...so the product is greater than 2. The next operation will make the (1+Y) term greater than 3, so basically as per my understanding there is some information missing in the question or I misunderstood.

If I read the question correctly, AX(Y) = (1+Y)X Putting X=2/3, => AX(X) = (1+ 2/3)*(2/3) => AX(X) = (5/3)*(2/3) => AX(X) = 10/9 > 1

AX(AX(X) = (1+10/9)(10/9)....The first term in the bracket will be greater than 2 and next term is greater than one...so the product is greater than 2. The next operation will make the (1+Y) term greater than 3, so basically as per my understanding there is some information missing in the question or I misunderstood.

Regards, Tushar

The highlighted portion is incorrect here.

Ax(Ax(X)) = (1+10/9)(2/3) Note that X is 2/3 only. Ax(X) is 10/9. Ax(10/9) = (1 + 10/9)(2/3)
_________________

Re: Ax(y) is an operation that adds 1 to y [#permalink]

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19 Feb 2015, 04:42

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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I solved this in about 1:20 by considering that eventually, this progression would stabilize, and so for any number Ax(Ax(Ax(...(Ax(y))) always approaches a number.

So I set up the equation: y = (y+1) * (2/3) 3/2y = y + 1 1/2y = 1 y = 2

So the equation will always approach the number 2.

Since we can see that the equation is approaching 2 from below, it must be close to 2, but just slightly less, hence D

Re: Ax(y) is an operation that adds 1 to y [#permalink]

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17 Sep 2016, 02:32

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: Ax(y) is an operation that adds 1 to y [#permalink]

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27 Apr 2017, 16:25

Bunuel, you never cease to amaze me with the depth of your abstract and theoretical solutions.

The simplicity of the numbers in this question and the depth of recursion in the function make it possible to calculate the value directly. This approach will not always be viable with similar questions.

Re: Ax(y) is an operation that adds 1 to y [#permalink]

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28 Apr 2017, 08:48

spence11 wrote:

Bunuel, you never cease to amaze me with the depth of your abstract and theoretical solutions.

The simplicity of the numbers in this question and the depth of recursion in the function make it possible to calculate the value directly. This approach will not always be viable with similar questions.

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