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17 May 2016, 03:44
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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70% (01:29) correct
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A perfect cube is an integer whose cube root is an integer. For example, 27, 64 and 125 are perfect cubes. If p and q are perfect cubes, which of the following will not necessarily be a perfect cube?
A. 8p
B. pq
C. pq + 27
D. -p
E. (p - q)^6
A. 8p
B. pq
C. pq + 27
D. -p
E. (p - q)^6
16 Jun 2016, 12:28
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Bunuel
Test some super easy values.
How about p = 1 and q = 1 (both are perfect cubes since 1³ = 1)
A. 8(1) = 8 = perfect cube
B. (1)(1) = 1 = perfect cube
C. (1)(1) + 27 = 28 = NOT a perfect cube
STOP TESTING VALUES. We've found an answer choice that is not necessarily a perfect cube
Answer:
General Discussion
17 May 2016, 10:25
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A. 8p
= 2^3 * p
Perfect cube
B. pq will be a perfect cube since its a product of 2 perfect cubes
C. pq + 27
Not necessarily a perfect cube
if p = 8 and q= 27
pq+ 27 = 8*27 + 27 = 27(8+1)
=3^3 * 3^2
=3^5
Not a perfect cube
D. -p
Perfect cube
E. (p - q)^6
Difference of p and q will be an integer and when its raised to 6th power , it will be a perfect cube
Answer C
= 2^3 * p
Perfect cube
B. pq will be a perfect cube since its a product of 2 perfect cubes
C. pq + 27
Not necessarily a perfect cube
if p = 8 and q= 27
pq+ 27 = 8*27 + 27 = 27(8+1)
=3^3 * 3^2
=3^5
Not a perfect cube
D. -p
Perfect cube
E. (p - q)^6
Difference of p and q will be an integer and when its raised to 6th power , it will be a perfect cube
Answer C
