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If x and y are integers and x + y is even, which of the following must
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12 Nov 2021, 06:48
12 Nov 2021, 06:48
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If x and y are integers and x + y is even, which of the following must be odd?
(A) \(x\)
(B) \(3xy\)
(C) \(x+2y-1\)
(D) \(xy+x+1\)
(E) \(x^2+2xy+y^2\)
(A) \(x\)
(B) \(3xy\)
(C) \(x+2y-1\)
(D) \(xy+x+1\)
(E) \(x^2+2xy+y^2\)
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Re: If x and y are integers and x + y is even, which of the following must
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12 Nov 2021, 08:08
12 Nov 2021, 08:08
1
Carcass wrote:
If x and y are integers and x + y is even, which of the following must be odd?
(A) \(x\)
(B) \(3xy\)
(C) \(x+2y-1\)
(D) \(xy+x+1\)
(E) \(x^2+2xy+y^2\)
(A) \(x\)
(B) \(3xy\)
(C) \(x+2y-1\)
(D) \(xy+x+1\)
(E) \(x^2+2xy+y^2\)
We can solve many Integer Properties questions by testing NICE values. Let's do that.
If x + y is even, then it COULD be the case that x = 0 and y = 0
Now plug these values into each answer choice to see which ones yield an ODD integer
(A) \(0\) EVEN. ELIMINATE
(B) \(3(0)(0) = 0\) EVEN. ELIMINATE
(C) \(0+2(0)-1 = -1\) ODD, Keep for now.
(D) \((0)(0)+0+1= 1\) ODD, Keep for now.
(E) \(0^2+2(0)(0)+0^2= 0\) EVEN. ELIMINATE
We're already down to C and D.
Let's test another pair of values.
If x + y is even, then it COULD be the case that x = 1 and y = 1
Now plug these values into the remaining answer choices to see which ones yield an ODD integer
(C) \(1+2(1)-1=2\) EVEN. ELIMINATE
(D) \((1)(1)+1+1=3\) ODD
By the process of elimination, the answer is D
General Discussion
Re: If x and y are integers and x + y is even, which of the following must
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12 Nov 2021, 21:43
12 Nov 2021, 21:43
1
Use these simple rules -
1] odd + odd = even
2] odd + even = odd
3] even + odd = odd
4] even + even = even
5] odd x odd = odd
6] odd x even = even x odd = even
7] even x even = even
Its given that \(x+y\) is even, which mean either both \(x\) and \(y\) are even or both are odd
Option A - \(x\) can be even or odd, so eliminated
Option B - , from Rule 5 and 7 we can tell that \(3xy\) can be even or odd, so eliminated
Option C - \(x + 2y - 1\)
If both \(x\) and \(y\) are even, then [even] + 2[even] - 1 = ([even] + 2[even]) - 1 = even - 1 = odd
If both \(x\) and \(y\) are odd, then [odd] + 2[odd] - 1 = [odd] + [even] - 1 = ([odd] + [even]) - 1 = odd - 1 = even
so eliminated
Option D - \(xy + x + 1\)
If both \(x\) and \(y\) are even, [even x even] + [even] + 1 = [even] + [even] + 1 = ([even] + [even]) + 1 = [even] + 1 = odd
If both \(x\) and \(y\) are odd, [odd x odd] + [odd] + 1 = [odd] + [odd] + 1 = ([odd] + [odd]) + 1 = [even] + 1 = odd
D is possible
Apply same concept for E, though since its a single correct you dont really need to go further.
Hence, Answer is D
1] odd + odd = even
2] odd + even = odd
3] even + odd = odd
4] even + even = even
5] odd x odd = odd
6] odd x even = even x odd = even
7] even x even = even
Its given that \(x+y\) is even, which mean either both \(x\) and \(y\) are even or both are odd
Option A - \(x\) can be even or odd, so eliminated
Option B - , from Rule 5 and 7 we can tell that \(3xy\) can be even or odd, so eliminated
Option C - \(x + 2y - 1\)
If both \(x\) and \(y\) are even, then [even] + 2[even] - 1 = ([even] + 2[even]) - 1 = even - 1 = odd
If both \(x\) and \(y\) are odd, then [odd] + 2[odd] - 1 = [odd] + [even] - 1 = ([odd] + [even]) - 1 = odd - 1 = even
so eliminated
Option D - \(xy + x + 1\)
If both \(x\) and \(y\) are even, [even x even] + [even] + 1 = [even] + [even] + 1 = ([even] + [even]) + 1 = [even] + 1 = odd
If both \(x\) and \(y\) are odd, [odd x odd] + [odd] + 1 = [odd] + [odd] + 1 = ([odd] + [odd]) + 1 = [even] + 1 = odd
D is possible
Apply same concept for E, though since its a single correct you dont really need to go further.
Hence, Answer is D
