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06 Sep 2021, 08:18
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (02:45) correct
65%
(02:49)
wrong
based on 71
sessions
History
Date
Time
Result
Not Attempted Yet
If x is a positive integer greater than 3, is x equal to 7?
1) \(x^4 − 3^4\) is divisible by 29
2) \(x^4 + 3x^3\) is divisible by 7
1) \(x^4 − 3^4\) is divisible by 29
2) \(x^4 + 3x^3\) is divisible by 7
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
06 Sep 2021, 08:52
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If x is a positive integer greater than 3, is x equal to 7?
1) \(x^4 − 3^4=(x^2-3^2)(x^2+3^2)=(x-3)(x+3)(x^2+9)\) is divisible by 29
When x=7, \((7+3)(7-3)(7^2+9)=10*4*58\). Since, it is divisible by 29, so x can be 7.
But, when x is 31, then too \((32+3)(32-3)(32^2+9)=35*29*(32^2+9)\) is divisible by 29.
Insufficient
2) \(x^4 + 3x^3=x^3(x+3)\) is divisible by 7
So x can be 4, 7 and so on.
Insufficient
Combined
x could be 7, as seen in the examples above.
But (x+3) is common in two, and we can have x+3=29*7, multiple of both 7 and 29, and thus satisfying both statements.
Thus, x+3=203…….x=200
Insufficient
E
1) \(x^4 − 3^4=(x^2-3^2)(x^2+3^2)=(x-3)(x+3)(x^2+9)\) is divisible by 29
When x=7, \((7+3)(7-3)(7^2+9)=10*4*58\). Since, it is divisible by 29, so x can be 7.
But, when x is 31, then too \((32+3)(32-3)(32^2+9)=35*29*(32^2+9)\) is divisible by 29.
Insufficient
2) \(x^4 + 3x^3=x^3(x+3)\) is divisible by 7
So x can be 4, 7 and so on.
Insufficient
Combined
x could be 7, as seen in the examples above.
But (x+3) is common in two, and we can have x+3=29*7, multiple of both 7 and 29, and thus satisfying both statements.
Thus, x+3=203…….x=200
Insufficient
E
29 Sep 2021, 08:01
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chetan2u
chetan2u,
13*7*109= 9919, whereas 29*342=9918. Am i missing something here? I think the answer should be A. Please share your thoughts.
