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If a^2+b^2=c^2, and a,b,c are all integers.
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31 Jul 2020, 21:40
31 Jul 2020, 21:40
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Question Stats:
66% (00:52) correct
33% (00:45) wrong based on 6 sessions
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If \(a^2+b^2=c^2\), and a,b,c are all integers. Which of the following CANNOT be the value of a+b+c?
A. 2
B. 1
C. -2
D. 4
E. 6
A. 2
B. 1
C. -2
D. 4
E. 6
Re: If a^2+b^2=c^2, and a,b,c are all integers.
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01 Aug 2020, 04:02
01 Aug 2020, 04:02
A. 2 = (3) + (4) + (-5) INCORRECT
B. 1 SKIP
C. -2 = (-3) + (-4) + (5) INCORRECT
D. 4 = (3) + (-4) + (5) INCORRECT
E. 6 = (-3) + (4) + (5) INCORRECT
So, the answer is B.
Is there any other algebraic way to solve it besides thinking of examples.
B. 1 SKIP
C. -2 = (-3) + (-4) + (5) INCORRECT
D. 4 = (3) + (-4) + (5) INCORRECT
E. 6 = (-3) + (4) + (5) INCORRECT
So, the answer is B.
Is there any other algebraic way to solve it besides thinking of examples.
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Re: If a^2+b^2=c^2, and a,b,c are all integers.
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01 Aug 2020, 05:59
01 Aug 2020, 05:59
1
lazyashell wrote:
If \(a^2+b^2=c^2\), and a,b,c are all integers. Which of the following CANNOT be the value of a+b+c?
A. 2
B. 1
C. -2
D. 4
E. 6
A. 2
B. 1
C. -2
D. 4
E. 6
Another approach is to recognize that, if we let one of the variables equal ZERO, then the sum a+b+c can be any EVEN number.
For example, 0² + 1² = 1². Here, a + b + c = 0 + 1 + 1 = 2 ELIMINATE A
Likewise, 0² + (-1)² = (-1)². Here, a + b + c = 0 + (-1) + (-1) = -2 ELIMINATE C
Likewise, 0² + 2² = 2². Here, a + b + c = 0 + 2 + 2 = 4 ELIMINATE D
Likewise, 0² + 3² = 3². Here, a + b + c = 0 + 3 + 3 = 6 ELIMINATE E
Answer: B
Cheers,
Brent
