Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) : GMAT Problem Solving (PS)
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# Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.))))

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Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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29 Aug 2012, 06:51
1
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Question Stats:

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Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))).. [PS: nested sq. root sequence is repeated infinite times.]

A. 0
B. $$\sqrt{3}$$
C. 3
D. 2.9
E. cannot be determined.

[Reveal] Spoiler:
a=√3(√3(√3(√3(√3…….inf.))))
sq. both sides
a^2=3(√3(√3(√3(√3(√3…….inf.)))))
or
a^2=3a ; since, a=√3(√3(√3(√3(√3…….inf.))))
a^2-3a=0
a(a-3)=0
a=0/3
0 logically doesn't fit so 3 is the ans

Interesting question.
Mods, if the question needs reformatting please do so, as i was not able to properly use the sqrt symbol, given in the editor.
Added a pic, pls pardon my drawing
[Reveal] Spoiler: OA

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File comment: The RHS of question.

seq.png [ 11.13 KiB | Viewed 2619 times ]

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Last edited by conty911 on 30 Aug 2012, 04:10, edited 1 time in total.
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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29 Aug 2012, 07:15
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Expert's post
conty911 wrote:
Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))).. [PS: nested sq. root sequence is repeated infinite times.]

A. 0
B. $$\sqrt{3}$$
C. 3
D. 2.9
E. cannot be determined.

[Reveal] Spoiler:
a=√3(√3(√3(√3(√3…….inf.))))
sq. both sides
a^2=3(√3(√3(√3(√3(√3…….inf.)))))
or
a^2=3a ; since, a=√3(√3(√3(√3(√3…….inf.))))
a^2-3a=0
a(a-3)=0
a=0/3
0 logically doesn't fit so 3 is the ans

Interesting question.
Mods, if the question needs reformatting please do so, as i was not able to properly use the sqrt symbol, given in the editor.

Similar questions to practice:
tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029228
if-the-expression-x-sqrt-2-sqrt-2-sqrt-2-sqrt-2-extends-98647.html
find-the-value-of-x-75403.html

Hope it helps.
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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30 Aug 2012, 01:20
a = sqrt(3*a)

squaring both sides

a^2 = 3a

=> a = 3.

Is this correct?
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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30 Aug 2012, 01:40
Here a=0 or a=3, both are possible. Don't know which one to pick A or C. Should we take a as +ve by default?
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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30 Aug 2012, 01:42
SOURH7WK wrote:
Here a=0 or a=3, both are possible. Don't know which one to pick A or C. Should we take a as +ve by default?

Zero is neither positive nor negative number.

Also, a=0 does not satisfy the equation because the left hand side of the equation is clearly positive.
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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30 Aug 2012, 02:02
I think this will be an endless sequence of multiples of 3....

for me this should be E) cannot be determined
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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30 Aug 2012, 03:48
cracked wrote:
a = sqrt(3*a)

squaring both sides

a^2 = 3a

=> a = 3.

Is this correct?

yes it is correct , you can see the spoiler in the question and also checkout similar examples given by Bunnuel .
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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30 Aug 2012, 06:17
conty911 wrote:
cracked wrote:
a = sqrt(3*a)

squaring both sides

a^2 = 3a

=> a = 3.

Is this correct?

yes it is correct , you can see the spoiler in the question and also checkout similar examples given by Bunnuel .

Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7090

Kudos [?]: 93327 [1] , given: 10557

Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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30 Aug 2012, 06:24
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
conty911 wrote:
Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))).. [PS: nested sq. root sequence is repeated infinite times.]

A. 0
B. $$\sqrt{3}$$
C. 3
D. 2.9
E. cannot be determined.

[Reveal] Spoiler:
a=√3(√3(√3(√3(√3…….inf.))))
sq. both sides
a^2=3(√3(√3(√3(√3(√3…….inf.)))))
or
a^2=3a ; since, a=√3(√3(√3(√3(√3…….inf.))))
a^2-3a=0
a(a-3)=0
a=0/3
0 logically doesn't fit so 3 is the ans

Interesting question.
Mods, if the question needs reformatting please do so, as i was not able to properly use the sqrt symbol, given in the editor.
Added a pic, pls pardon my drawing

For those who are confused by formatting:

If $$a=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}}$$, what is the value of a?

A. 0
B. $$\sqrt{3}$$
C. 3
D. 2.9
E. cannot be determined.

$$a=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}}$$ --> $$a=\sqrt{3(\sqrt{3\sqrt{3\sqrt{3...}}})}$$. Now, as the expression under the square root extends infinitely, then expression in brackets would equal to $$a$$ itself, so we can safely replace it with $$a$$ and rewrite the given expression as $$a=\sqrt{3a}$$.

Square it: $$a^2=3a$$ --> $$a=0$$ or $$a=3$$ --> since $$a=0$$ does not satisfy given expression, then we have only one solution: $$a=3$$.

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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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02 Sep 2012, 12:54
Found another way of solving it:
We can rewrite the equation:

a=3^(1/2)*3^(1/4)*3^(1/8)*...
a=3^(1/2+1/4+1/8+1/16+...)

1/2+1/4+1/8+...=1 (ok, one has to know this equation, I think)

a=3^1=3
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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20 Aug 2014, 10:39
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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20 Aug 2014, 18:53
Bunuel wrote:
conty911 wrote:
Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))).. [PS: nested sq. root sequence is repeated infinite times.]

A. 0
B. $$\sqrt{3}$$
C. 3
D. 2.9
E. cannot be determined.

[Reveal] Spoiler:
a=√3(√3(√3(√3(√3…….inf.))))
sq. both sides
a^2=3(√3(√3(√3(√3(√3…….inf.)))))
or
a^2=3a ; since, a=√3(√3(√3(√3(√3…….inf.))))
a^2-3a=0
a(a-3)=0
a=0/3
0 logically doesn't fit so 3 is the ans

Interesting question.
Mods, if the question needs reformatting please do so, as i was not able to properly use the sqrt symbol, given in the editor.
Added a pic, pls pardon my drawing

For those who are confused by formatting:

If $$a=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}}$$, what is the value of a?

A. 0
B. $$\sqrt{3}$$
C. 3
D. 2.9
E. cannot be determined.

$$a=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}}$$ --> $$a=\sqrt{3(\sqrt{3\sqrt{3\sqrt{3...}}})}$$. Now, as the expression under the square root extends infinitely, then expression in brackets would equal to $$a$$ itself, so we can safely replace it with $$a$$ and rewrite the given expression as $$a=\sqrt{3a}$$.

Square it: $$a^2=3a$$ --> $$a=0$$ or $$a=3$$ --> since $$a=0$$ does not satisfy given expression, then we have only one solution: $$a=3$$.

Another genius explanation from Bunuel, Kudos to you

I had in mind that $$\sqrt{3} = 1.732$$ To square root it multiple times will give the value 1.00000000000xxxxxxxxxxx & so on

Now that 1 is not the option in the OA, nearest to it was 0, which I selected & went wrong

Bunuel, can you kindly explain

I opened the scientific calculator & square rooted 3 multiple times, which stuck at 1
Attachment:

calc.png [ 17.3 KiB | Viewed 1523 times ]

(I agree that answer = 3, however the result 1 from calculator is contradicting)
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Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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20 Aug 2014, 19:29
Just take 4 occurrences of square root

$$a = \sqrt{3\sqrt{3\sqrt{3\sqrt{3}}}}$$

$$a = \sqrt{3\sqrt{3\sqrt{3^1 * 3^{\frac{1}{2}}}}}$$

$$a = \sqrt{3\sqrt{3\sqrt{3^{\frac{3}{2}}}}}$$

$$a = \sqrt{3\sqrt{3 * 3^{\frac{3}{2} * \frac{1}{2}}}}$$

$$a = \sqrt{3\sqrt{3^1 * 3^{\frac{3}{4}}}}$$

$$a = \sqrt{3\sqrt{3^{\frac{7}{4}}}}$$

$$a = \sqrt{3 * 3^{\frac{7}{4} * \frac{1}{2}}}$$

$$a = \sqrt{3^{\frac{15}{8}}}$$

$$a = 3^{\frac{15}{8} * \frac{1}{2}}$$

$$a = 3^{\frac{15}{16}}$$

For n (Infinite) number of occurrences, the equation would be built up as follows

$$a = 3^{\frac{n}{n+1}}$$

We may deem here $$n\approx{n+1}$$

$$a = 3^1 = 3$$

What is find here is that, if you have a calculator in mind for value of $$\sqrt{3},$$then we tend to avoid the algebraic approach (shown above) & end up wrong
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.)))) [#permalink]

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26 Aug 2016, 00:23
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Re: Find the value of a. Given a=√3(√3(√3(√3(√3…….inf.))))   [#permalink] 26 Aug 2016, 00:23
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