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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
Expert Reply
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A
B
C
D
E
Question Stats:
58% (01:36) correct
41% (01:21) wrong based on 12 sessions
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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) x2+2xy+y2
(A) x
(B) 3xy
(C) x+2y−1
(D) xy+x+1
(E) x2+2xy+y2
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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) x2+2xy+y2
(A) x
(B) 3xy
(C) x+2y−1
(D) xy+x+1
(E) x2+2xy+y2
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) 02+2(0)(0)+02=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
