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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.

Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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10 Dec 2012, 02:04

2

This post received KUDOS

Shrink the monster technique. The two numbers are two consecutive even numbers raised to 2. Shrink the monster and think of baby numbers such as 0 and 2.

\(mean=\frac{{0^2 + 2^2}}{2}=2\)

Substitute you baby numbers to the answer choices.

Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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10 Dec 2012, 03:42

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Ans: Replace 54820 by x and 54822 by (x+2) and then solve by algebra. Mean of x^2 and (x+2)^2 is x^2+2x+2 which can be written as x^2+2x+1+1. So the answer is (D)
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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24 Jul 2014, 06:56

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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14 Mar 2016, 01:04

Bunuel wrote:

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.

Answer: D.

Hi,

Concerning last digits, I only know that we can add or multiply last digits of various numbers to determine the last digits of their sum or product. Can somebody tell me the rule for division of last digits (as presented by Bunuel)?

Another question: when and only when a multiple of 4 is divided by 2, the last digit of the resulting quotient will always be 2? Is there any other rules to memorize?

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.

Concerning last digits, I only know that we can add or multiply last digits of various numbers to determine the last digits of their sum or product. Can somebody tell me the rule for division of last digits (as presented by Bunuel)?

Quote:

Another question: when and only when a multiple of 4 is divided by 2, the last digit of the resulting quotient will always be 2? Is there any other rules to memorize?

Thank you very much!

Hi,

Quote:

Another question: when and only when a multiple of 4 is divided by 2, the last digit of the resulting quotient will always be 2? Is there any other rules to memorize?

This is not CORRECT.. In the given context, last digit 4 and multiple of 4 are interlinked, so just stating one that multiple of 4 when divided by 2 gives 2 will be wrong

reason-

- 8 is a multiple of 4 and will give last digit as 4 and NOT 2..

what is meant here is we have seen that the last digit of SUM is 4 and this SUM is multiple of 4.. A number Multiple of 4, with last digit 4 is 04,24, 44,64,84 as last two digits, so when div by 2, in each case last digit is 2.. the number cannot be ending in 14,34 etc as Property of 4 is last two digits should be div by 4 if the entire number is to be div by 4.. _________________

Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]

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20 Dec 2016, 05:54

One more cyclicity question. Fisrt, we should find out the unit digit of the mean of 54820^2 and 54822^2. To figure it out we need find out last digits of both numbers. Last number of 54820^2 will be 0, last number of 54822^2 will be 4. Mean will be (0+4)/2 = 2, thus unit digit of the mean number will be 2. Analyzing all options, it is clear that only one option gives us this value, Option D 54821^2+1

gmatclubot

Re: The mean of (54,820)^2 and (54,822)^2 =
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20 Dec 2016, 05:54

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