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13 May 2020, 07:58
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A
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If x and y are integers, is \(3x^4 + 4y\) even?
(1) \(x^3\) is even
(2) \(y^{2x} + 3\) is even
Are You Up For the Challenge: 700 Level Questions
(1) \(x^3\) is even
(2) \(y^{2x} + 3\) is even
Are You Up For the Challenge: 700 Level Questions
13 May 2020, 08:49
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IMO, A based on following analysis:
Is \(3x^4 + 4y\) even?
Which means question reduces to: Is x is even? Since, 4y will always be even.
1. \(x^3\) is even this implies x is even, as only even number multiple results in even irrespective of exponential power. Which means, \(3x^4\) is even, and hence, A is sufficient.
2. \(y^2x + 3\) is even ----> \(y^2x \) will be odd, since Odd + Odd is even. Hence, Y is odd and x can be either be odd or even. Hence, this is not sufficient.
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Is \(3x^4 + 4y\) even?
Which means question reduces to: Is x is even? Since, 4y will always be even.
1. \(x^3\) is even this implies x is even, as only even number multiple results in even irrespective of exponential power. Which means, \(3x^4\) is even, and hence, A is sufficient.
2. \(y^2x + 3\) is even ----> \(y^2x \) will be odd, since Odd + Odd is even. Hence, Y is odd and x can be either be odd or even. Hence, this is not sufficient.
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13 May 2020, 08:57
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If x and y are integers, is 3x^4+4y even?
(1)x^3 is even
so, x= even (2^3=8 =even)
so,x^4 = even
3x^4 = even
now, if y = even, 4y = even
also, if y = odd, 4y = even
so, always 3x^4+4y even (even+even= even ) sufficient
(2) y^2x+3 is even
so, y^2x= odd
or, case I : y= even x= even y^2x= even ( it is not the case)
case II : y= even x= odd y^2x= even ( it is not the case)
case III : y= odd x= even y^2x= odd ( it is the case) (1^4=1 odd)
case IV : y= odd x= odd y^2x= odd ( it is the case) (1^2=1 odd)
so, x may be even or odd
so,x^4 may be even or odd
so,3x^4 may be even or odd
so,3x^4 + 4y may be even or odd (4y always even) insufficient
correct answer A
(1)x^3 is even
so, x= even (2^3=8 =even)
so,x^4 = even
3x^4 = even
now, if y = even, 4y = even
also, if y = odd, 4y = even
so, always 3x^4+4y even (even+even= even ) sufficient
(2) y^2x+3 is even
so, y^2x= odd
or, case I : y= even x= even y^2x= even ( it is not the case)
case II : y= even x= odd y^2x= even ( it is not the case)
case III : y= odd x= even y^2x= odd ( it is the case) (1^4=1 odd)
case IV : y= odd x= odd y^2x= odd ( it is the case) (1^2=1 odd)
so, x may be even or odd
so,x^4 may be even or odd
so,3x^4 may be even or odd
so,3x^4 + 4y may be even or odd (4y always even) insufficient
correct answer A
