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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.

Take 2 and 4: average= 3 Now, for \(2^2\) and \(4^2\), \(average = 10 =3^2 +1\)

Take 4 and 6: average= 5 Now, for \(4^2\) and \(6^2\): \(average= 26= 5^2+1\)

We can quickly realize that average of 54820 and 54822= 54821 So, as per the pattern derived above, average of \(54820^2\) and \(54822^2\)= \(54821^2 +1\)

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]

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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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