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A teacher writes 3^4 on the board

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A teacher writes 3^4 on the board  [#permalink]

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New post Updated on: 02 Feb 2018, 05:47
1
7
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

56% (01:41) correct 44% (01:51) wrong based on 156 sessions

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A teacher writes \(3^4\) on the board and asks a student to come and add one more term of \(3^4\) to existing term, such that it now becomes \(3^4+3^4\). The teacher asks other students to repeat the step one after another till the total on board becomes a cube of a positive integer. How many students added the term to the series?

A) 2
B) 3
C) 8
D) 9
E) 10

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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Originally posted by chetan2u on 14 Jan 2016, 00:38.
Last edited by chetan2u on 02 Feb 2018, 05:47, edited 3 times in total.
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Re: A teacher writes 3^4 on the board  [#permalink]

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New post 14 Jan 2016, 01:00
Answer is C.

3^4 + 3^4 + 3^4 .............. so on till we get a number that is a perfect cube. To get a perfect cube, the final number should have a total number of prime factors that are in sets of 3. To get this, we have to add 3^4 - 9 times and then factorize.

3^4(1+1+1+1+1+1+1+1+1)
3^4 (9)
3^4 * 3^2
3^6
2 groups of 3^3 each.

Please note that out of 9 terms of 3^4, the first one was added by the teacher. The question is asking us the number of terms that students added, hence the answer is 8.

This is my understanding of the problem. Please correct me if i'm wrong.
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Re: A teacher writes 3^4 on the board  [#permalink]

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New post 14 Jan 2016, 03:01
1
3^4 = 81
81*X= perfect cube (so rephrasing the question: How many times 81 has to be written for a perfect cube )
or, 3*3*3*3*X= perfect cube
X=9
But teacher has already used up 1 ; remaining 8
Thus C
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Re: A teacher writes 3^4 on the board  [#permalink]

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New post 14 Jan 2016, 09:42
The factors of a perfect cube will be in the form of \(a^{3b}*c^{3d}*e^{3f}*\)... where a, c, e are the prime factors and b, d, f are integers. In our case, we only have one distinct prime factor, 3. So we need to add a bunch of \(3^4\) terms together to get something in the form of \(3^{3b}\)

We have to increase the exponent on \(3^4\) until it is the next multiple of 3. So the exponent needs to be 6.

\(3^6=3^{(4+2)} = 3^4*3^2 = 9*3^4\)

We need to add \(3^4\) together 9 times. The teacher already wrote one, so the students need to write out 8 terms of \(3^4\).

Answer C
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Re: A teacher writes 3^4 on the board  [#permalink]

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New post 16 Jan 2016, 07:00
chetan2u wrote:
A teacher writes \(3^4\) on the board and asks a student to come and add one more term of \(3^4\) to existing term, such that it now becomes \(3^4+3^4\). The teacher asks other students to repeat the step one after another till the total on board becomes a cube of a positive integer. How many students added the term to the series?
A) 2
B) 10
C) 8
D) 3
E) 9

OA and OE on Sun, 17th Jan 2016



3^4 =81

81 * X = Y^3
So X= Y^3/81
i.e X = Y^3/9 * 9

If Y = 9 then Y^3 /9 is an integer

So X = 9*9*9/ 9*9 = 9
Since X =9 we have to subtract 1 for 1 is set by teacher . Therefor 8 is the Answer.

IMO: C
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Re: A teacher writes 3^4 on the board  [#permalink]

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New post 16 Jan 2016, 23:29
chetan2u wrote:
A teacher writes \(3^4\) on the board and asks a student to come and add one more term of \(3^4\) to existing term, such that it now becomes \(3^4+3^4\). The teacher asks other students to repeat the step one after another till the total on board becomes a cube of a positive integer. How many students added the term to the series?
A) 2
B) 10
C) 8
D) 3
E) 9

OA and OE on Sun, 17th Jan 2016



SOLUTION:-
It has been correctly found by all of you, who have tried....
all methods are correct.

the CONCEPT is:-
since we are talking of sum of 3^4, the first cube too will be of 3 or its power..
whenever we add another 3^4 to itself, it is nothing but a multiplication table of 3^4..

so it goes like 1*3^4, 2*3^4, 3*3^4, 4*3^4... and so on..
as we see the sum is always a multiple of 3^4 and therefore the cube will be something to do with 3..
cube of 3 is already smaller to 3^4 , the next would be cube of 3^2...
so what is required is 9*3^4..
thus total 9 are required, out of which ONE is already written by the teacher..
so our answer is 9-1=8..
C
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A teacher writes 3^4 on the board  [#permalink]

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New post 02 Feb 2018, 05:42
chetan2u wrote:
A teacher writes \(3^4\) on the board and asks a student to come and add one more term of \(3^4\) to existing term, such that it now becomes \(3^4+3^4\). The teacher asks other students to repeat the step one after another till the total on board becomes a cube of a positive integer. How many students added the term to the series?

A) 2
B) 3
C) 8
D) 9
E) 10


As per the given question -
\(n*3^4 = a^3\)
The nearest n value will be \(3^2\) to make it a cube.
So, \(n= 9\) out of which first \(3^2\) is written by the teacher and remaining 8 by students.
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Re: A teacher writes 3^4 on the board &nbs [#permalink] 02 Feb 2018, 05:42
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