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Answer D. Let x=54,820 and y = 54,822 = x + 2. Then average is (x^2 + y^2) / 2 = [(x^2) + (x + 2)^2] / 2 = (x^2 + x^2 + 4x + 4) /2 = x^2 +2x + 2 = (x + 1)^2 + 1 Now, sub into original numbers the average is (54,820 + 1)^2 + 1 = 54,821^2 = 1.

If I come across this question in a test, I would just take some small values to convince myself. Say \frac{(2^2 + 4^2)}{2} = 10 which can also be represented as 3^2 + 1 A couple more such examples and the pattern would be convincing. Say \frac{(4^2 + 6^2)}{2} =\frac{(16 + 36)}{2} = 26 5^2 + 1 = 26

If you insist of using algebra, average of (a - 1)^2 and (a+1)^2 = \frac{[(a-1)^2 + (a+1)^2]}{2} = a^2 + 1 Hence answer (D) _________________

Re: The average of (54,820)^2and (54,822)^2 = [#permalink]
14 Sep 2013, 10:18

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Test some small numbers: \frac{2^2+4^2}{2}=10=3^2+1 or: \frac{4^2+6^2}{2}=26=5^2+1.

APPROACH #2:

Say 54,821=x, then \frac{54,820^2+54,822^2}{2}=\frac{(x-1)^2+(x+1)^2}{2}=x^2+1=54,821^2+1.

APPROACH #3:

The units digit of 54,820^2+54,822^2 is 0+2=4. Now, since 54,820^2+54,822^2 must be a multiple of 4, then \frac{54,820^2+54,822^2}{2} must have the units digit of 2. Only answer choice D fits.

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