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A perfect square is a number that becomes an integer when sq
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14 Apr 2019, 10:11
14 Apr 2019, 10:11
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A perfect square is a number that becomes an integer when square rooting it. A, B, and C are three positive integers. The ratio of the three numbers is 1 : 2 : 3, respectively. Which one of the following expressions must be a perfect square?
(A) \(A + B + C\)
(B) \(A^2 + B^2 + C^2\)
(C) \(A^3 + B^3 + C^3\)
(D) \(3A^2 + B^2 + C^2\)
(E) \(3A^2 + 4B^2 + 4C^2\)
(A) \(A + B + C\)
(B) \(A^2 + B^2 + C^2\)
(C) \(A^3 + B^3 + C^3\)
(D) \(3A^2 + B^2 + C^2\)
(E) \(3A^2 + 4B^2 + 4C^2\)
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Re: A perfect square is a number that becomes an integer when sq
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24 Oct 2021, 07:56
24 Oct 2021, 07:56
1
raghav4202 wrote:
option C gives answer as 36 so why cant that be our answer ?
The key word here is must as in: Which one of the following expressions must be a perfect square?
You already discovered that, if A = 1, B = 2 and C = 3, then A³ + B³ + C³ = 1³ + 2³ + 3³ = 36, and 36 is a perfect square.
So, at this point, we know that A³ + B³ + C³ COULD be a perfect square
That is, we can't yet conclude that A³ + B³ + C³ MUST be a perfect square.
Notice that the values A = 2, B = 4 and C = 6 are also in the ratio 1 : 2 : 3
When we test these values, we get: A³ + B³ + C³ = 2³ + 4³ + 6³ = 288, which is NOT a perfect square.
Eliminate C.
Does that help?
General Discussion
Re: A perfect square is a number that becomes an integer when sq
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13 Oct 2019, 02:32
13 Oct 2019, 02:32
1
Since the numbers are in ratio 1:2:3 it would be wise to choose A=1, B=2, C=3 and replace them in the available choices.
When the result of one of those will be a perfect square, that will be the answer.
Answer D is the only one which returns a perfect square: 3*1 + 4 + 9 = 7 + 9 = 16 = 4*4
Hence, D.
When the result of one of those will be a perfect square, that will be the answer.
Answer D is the only one which returns a perfect square: 3*1 + 4 + 9 = 7 + 9 = 16 = 4*4
Hence, D.
