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Re: A teacher writes 3^4 on the board [#permalink]
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]
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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A teacher writes 3^4 on the board [#permalink]
Expert Reply
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 \(9-1=8\)..
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Re: A teacher writes 3^4 on the board [#permalink]
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 [#permalink]
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Re: A teacher writes 3^4 on the board [#permalink]
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