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09 Sep 2019, 01:40
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Is \(b^2\) divisible by 4, if a, b, and c are positive integers where a is an odd number and \( c = 4 + a^2+b^2\)?
(1) 1 + 2m = c – a + 1
(2) 8n - 3 = c, where n is an integer
(1) 1 + 2m = c – a + 1
(2) 8n - 3 = c, where n is an integer
09 Sep 2019, 05:20
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Is \(b^2\) divisible by 4, if a, b, and c are positive integers where a is an odd number and c = 4 + \(a^2\) + \(b^2\)?
As b is a positive integer, \(b^2\) will be divisible by 4, when b is even.
c = 4 + \(a^2\) + \(b^2\)..........\(c=4+(odd)^2+b^2............c=(Odd)+b^2\), Thus c and b are opposite in property...
(1) \(1+2m = (c – a + 1)\)
\(c-a=2+2m\)=even............c-odd=even, so c is ODD
Since c is odd, b will be even, and, therefore, \(b^2\) is divisible by 4..
Sufficient
(2) 8n - 3=c
so c=even-odd=odd
Since c is odd, b will be even, and, therefore, \(b^2\) is divisible by 4..
Sufficient
D
As b is a positive integer, \(b^2\) will be divisible by 4, when b is even.
c = 4 + \(a^2\) + \(b^2\)..........\(c=4+(odd)^2+b^2............c=(Odd)+b^2\), Thus c and b are opposite in property...
(1) \(1+2m = (c – a + 1)\)
\(c-a=2+2m\)=even............c-odd=even, so c is ODD
Since c is odd, b will be even, and, therefore, \(b^2\) is divisible by 4..
Sufficient
(2) 8n - 3=c
so c=even-odd=odd
Since c is odd, b will be even, and, therefore, \(b^2\) is divisible by 4..
Sufficient
D
09 Sep 2019, 22:36
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I did it in a similar way.
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