So possible values of n are 1,2,3 & 6....
Is that fine?
different possible values of n
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gmatmachoman
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selango
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n cant be 6,if v don't assume x as 3^4gmatmachoman wrote:So possible values of n are 1,2,3 & 6....
Is that fine?
(3 ^4 )(5 ^ 6 )(7 ^ 3 ) = (5 ^ 6 )(7^6)( 3^4/7^3 )
Note that X must be 3^4/7^3 which is not possible as X is an integer.
So possible values of n are 1,2 and 3,if v don't assume X as 3^4
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gmatmachoman
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AM i correct?gmatmachoman wrote:
Soln :
3^4 * 5^6 * 7^3 = (35 ^ n )( x )
case n=6
(V7)^6 . 5^6 .3^4
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selango
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yes machoman..this is absolutely correct.gmatmachoman wrote:AM i correct?gmatmachoman wrote:
Soln :
3^4 * 5^6 * 7^3 = (35 ^ n )( x )
case n=6
(V7)^6 . 5^6 .3^4
The problem is approached in 2 ways.
Taking X as 3^4 and solving for n.[the case u followed,n=2,3,6]
Raising common exponent on both sides and representing other factors as X[the case we followed,n=1,2,3]
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kvcpk
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I think if we go with using roots, many numbers would be possible:gmatmachoman wrote:AM i correct?gmatmachoman wrote:
Soln :
3^4 * 5^6 * 7^3 = (35 ^ n )( x )
case n=6
(V7)^6 . 5^6 .3^4
(fourthroot(7))^12 * (sqrt(5))^12 * 3^4
12 is also possible.
What you say??
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GMATGuruNY
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Wow, lots of discussion about this problem!adi_800 wrote:If (3 ^4 )(5 ^ 6 )(7 ^ 3 ) = (35 ^ n )( x ), where x and n are both positive integers, how many different possible values of n are there?
A. 1
B. 2
C. 3
D. 4
E. 6
OA is C
3^4 * 5^6 * 7^3 = 5^n * 7^n * x
Since x and n must be positive integers, we can't have more factors of 7 on the right side of the equation than we have on the left side, so largest value of n=3. Since n>0, n = 1, 2, 3.
The correct answer is C.
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akhpad
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adi_800 wrote:If (3 ^4 )(5 ^ 6 )(7 ^ 3 ) = (35 ^ n )( x ), where x and n are both positive integers, how many different possible values of n are there?
A. 1
B. 2
C. 3
D. 4
E. 6
OA is C
Govi Bhai
I believe that you must be confused on this problem. This is so easy that you can click without making any calculation.
Here, key is that x and n both are positive integers.
