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x and y are integers.
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06 Jul 2021, 08:11
06 Jul 2021, 08:11
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x and y are integers. \(\frac{(x^2y−1)}{2}\) has a non-zero remainder. Which of the following must be true?
I. y is odd
II. x is even
III. xy is even
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
I. y is odd
II. x is even
III. xy is even
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
Re: x and y are integers.
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06 Jul 2021, 08:23
06 Jul 2021, 08:23
\(\frac{(x^2y−1)}{2}\) has a non-zero remainder.
So the numerator will be odd.
\(x^2y-1 =\) odd
\(x^2y =\) odd \(+ 1 =\) even
\(x\) & \(y\) any one can be even. But we are asked for a choice that must be true.
so,
xy will be even.
Answer C
So the numerator will be odd.
\(x^2y-1 =\) odd
\(x^2y =\) odd \(+ 1 =\) even
\(x\) & \(y\) any one can be even. But we are asked for a choice that must be true.
so,
xy will be even.
Answer C
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Re: x and y are integers.
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06 Jul 2021, 09:07
06 Jul 2021, 09:07
Carcass wrote:
x and y are integers. \(\frac{(x^2y−1)}{2}\) has a non-zero remainder. Which of the following must be true?
I. y is odd
II. x is even
III. xy is even
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
I. y is odd
II. x is even
III. xy is even
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
Given: (x²y - 1) divided by 2 has a non-zero remainder.
If (x²y - 1) DID have a zero remainder, then (x²y - 1) would be divisible by 2 (aka EVEN)
Since (x²y - 1) DOES NOT have a zero remainder, then (x²y - 1) is NOT divisible by 2 (aka ODD)
So, x²y - 1 is ODD
This tells us that x²y is EVEN
x²y will be even IF x is even, y is even, or x and y are both even
The question asks, "Which of the following MUST be true?"
I. y is odd. y COULD be odd, but it doesn't have to be. (for example, if x = 1 and y = 2, x²y is still EVEN)
II. x is even. x COULD be even, but it doesn't have to be. (for example, if x = 2 and y = 2, x²y is still EVEN)
III. xy is even. Since at least one of the variables must be even, it MUST be the case that xy is even
Answer: C
