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20 Nov 2006, 06:53
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99. Is x^2+y^2 divisible by 5?
1). When x-y is divided by 5, the remainder is 1
2). When x+y is divided by 5, the remainder is 3
1). When x-y is divided by 5, the remainder is 1
2). When x+y is divided by 5, the remainder is 3
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20 Nov 2006, 07:07
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I get C. Together they are sufficient.
From A: 5p+1=x-y
From B: 5q+3=x+y
Adding the 2 equations: x = (5(p+q)+4)/2
Subtracting the 2 equations B-A: y = (5(q-p)+2)/2
Then x^2+y^2 = ((5(p+q)+4)/2)^2 + ((5(q-p)+2)/2)^2
=25(p+q)^2+4 + 25(q-p)^2+1
=25((p+q)^2 + (q-p)^2) + 5
Hence divisible by 5.
From A: 5p+1=x-y
From B: 5q+3=x+y
Adding the 2 equations: x = (5(p+q)+4)/2
Subtracting the 2 equations B-A: y = (5(q-p)+2)/2
Then x^2+y^2 = ((5(p+q)+4)/2)^2 + ((5(q-p)+2)/2)^2
=25(p+q)^2+4 + 25(q-p)^2+1
=25((p+q)^2 + (q-p)^2) + 5
Hence divisible by 5.
20 Nov 2006, 07:29
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for me it is E.
i get this:
x^2+y^2 = [25(m^2+n^2)]/2 + 5m + 15n + 5.
we can't be sure m^2 + n^2 is an even number, so we can't guarantee the first term divided by 5 is an integer.
i get this:
x^2+y^2 = [25(m^2+n^2)]/2 + 5m + 15n + 5.
we can't be sure m^2 + n^2 is an even number, so we can't guarantee the first term divided by 5 is an integer.
