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27 Apr 2023, 13:33
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D
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Difficulty:
95%
(hard)
Question Stats:
28% (02:27) correct
72%
(02:38)
wrong
based on 82
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If x and y are positive integers greater than 2, is \((x+2)^{y-2} + x^y * (y-2)^{x+2}\) even ?
1) \(5x + 8x^2 + 12x^3 + 9\) is odd
2) \(3y + (11 + y) + 35(3y+2)\) is even
1) \(5x + 8x^2 + 12x^3 + 9\) is odd
2) \(3y + (11 + y) + 35(3y+2)\) is even
18 Jun 2023, 07:16
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Great Q!
1)
\(5x+8x^2+12x^3+9=odd\)
\(5x+8x^2+12x^3=odd-9\)
\(5x+8x^2+12x^3=even\)
Notice that 8x^2 and 12x^2 will always be even:
\(5x+even+even=even\)
\(5x=even\)
X must be some even integer.
Now:
\((x+2)^{y-2}+x^y(y-2)^{x+2}\)
The first portion will be an even number raised to some power - even
the second portion will be an even number times whatever y is
even+even=even,
Sufficient.
2)
\(3y+(11+y)+35(3y+2)=even\)
First check if y can be odd:
If y=odd
3y=odd
11+y=even
35(3*odd+2)=odd*odd=odd
so we have
odd+even+odd
even+even=even
Y can be odd.
Can y be even?
If y=even
3y=even
11+y=odd
35(3*even+2)=even
even+even+odd=even+odd=odd
y can only be odd.
Now if y is odd, x is even -> must be even (as shown in 1 )
if y is odd, x is odd:
\((x+2)^{y-2}+x^y(y-2)^{x+2}\)
(x+odd)^odd=odd
x^y=odd
y-2=odd
odd+odd*odd=odd+odd=even
Must be even as well.
D
1)
\(5x+8x^2+12x^3+9=odd\)
\(5x+8x^2+12x^3=odd-9\)
\(5x+8x^2+12x^3=even\)
Notice that 8x^2 and 12x^2 will always be even:
\(5x+even+even=even\)
\(5x=even\)
X must be some even integer.
Now:
\((x+2)^{y-2}+x^y(y-2)^{x+2}\)
The first portion will be an even number raised to some power - even
the second portion will be an even number times whatever y is
even+even=even,
Sufficient.
2)
\(3y+(11+y)+35(3y+2)=even\)
First check if y can be odd:
If y=odd
3y=odd
11+y=even
35(3*odd+2)=odd*odd=odd
so we have
odd+even+odd
even+even=even
Y can be odd.
Can y be even?
If y=even
3y=even
11+y=odd
35(3*even+2)=even
even+even+odd=even+odd=odd
y can only be odd.
Now if y is odd, x is even -> must be even (as shown in 1 )
if y is odd, x is odd:
\((x+2)^{y-2}+x^y(y-2)^{x+2}\)
(x+odd)^odd=odd
x^y=odd
y-2=odd
odd+odd*odd=odd+odd=even
Must be even as well.
D
20 Jun 2023, 12:39
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gmatophobia
When we evaluate the parity of an expression, we can simplify the process:
* If an integer expression is raised to a positive integer power, we can simply ignore the exponent because it won’t change the parity of the base.
* If some even number is added to or subtracted from an integer, we can simply ignore the addition or subtraction because it won’t change the parity of the integer.
We know that x and y are positive integers greater than 2, so each exponent will be a positive integer, and we can rephrase the original question into:
Is x + (x)(y) = even ?
Is x(1 + y) = even ?
Is x = even OR y = odd ?
Statement One Alone:
=> 5x + 8x^2 + 12x^3 + 9 is odd
If we ignore the exponents to simplify, we have:
5x + 8x + 12x + odd = odd
25x = odd – odd
25x = even
Therefore, x must be even, and we have a definite Yes answer to our rephrased question. Statement one is sufficient. Eliminate answer choices B, C, and E.
Statement Two Alone:
=> 3y + (11 + y) + 35(3y + 2) is even
109y + 81 = even
109y = even – odd = odd
Therefore, y must be odd, and we have a definite Yes answer to our rephrased question. Statement two is sufficient.
Answer: D
