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# The mean of (54,820)^2 and (54,822)^2 =

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

=> [(54,820)^2 +(54,822)^2]/2 (using a^2 + b^2 = (a-b)^2 + 2ab)
=> [(54,82-54,820)^2 + 2*54,820*54,822]/2
=> [2^2 + 2*54,820*54,822]/2
=> [2 + 54,820*54,822]
=> [2 + (54,821-1)*(54,821+1)] (using (a+b)(a-b) = a^2-b^2)
=> [2 + 54,821^2 - 1]
=> 54,821^2 +1

and D
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
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(54820)^2 + (54822)^2 = (54821-1)^2 + (54821+1)^2
= 2 (54821)^2 + 2

So mean = (54821)^2 + 1
(D)

-----------------------
Thanks
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
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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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The mean of (54,820)^2 and (54,822)^2 = [#permalink]
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enigma123 wrote:
The mean of $$(54,820)^2$$ and $$(54,822)^2 =$$

(A) $$(54,821)^2$$

(B) $$(54,821.5)^2$$

(C) $$(54,820.5)^2$$

(D) $$(54,821)^2 + 1$$

(E) $$(54,821)^2 – 1$$

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 $$(a+1)^2$$ = $$\frac{[(a-1)^2 + (a+1)^2]}{2} = a^2 + 1$$
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
Bunuel how do you know the expression is a multiple of 4? Wouldn't I need the tens and unit digit? Not just units?
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
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$$\frac{54820^2 + 54822^2}{2}$$

$$= \frac{54820^2 + (54820 + 2)^2}{2}$$

$$= \frac{54820^2 + 54820^2 + 2 * 54820 * 2 + 2 * 2}{2}$$

$$= 54820^2 + 2 * 54820 + 1 + 1$$

$$= (54820 + 1)^2 + 1$$

$$= 54821^2 + 1$$

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
bankerboy30 wrote:
Bunuel how do you know the expression is a multiple of 4? Wouldn't I need the tens and unit digit? Not just units?

54,820^2 + 54,822^2 = even^2 + even^2 = multiple of 4 + multiple of 4 = multiple of 4.

Hope it's clear.
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
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Let a = 54820
Then, we need to find
$$\frac{a^2 + (a + 2)^2}{2}$$ => $$\frac{a^2 + a^2 + 4a + 4}{2}$$

$$\frac{2a^2+ 4a + 4}{2}$$ => $$\frac{2(a^2 + 2a + 2)}{2}$$

$$a^2 + 2a + 1 + 1$$ => $$(a + 1)^2 + 1$$

Substitute for a

$$(54820 + 1)^2 + 1$$ => $$(54821)^2 + 1$$

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

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?

Thank you very much!
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
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truongynhi wrote:
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.

Hi truongynhi

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..
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
enigma123 wrote:
The mean of $$(54,820)^2$$ and $$(54,822)^2 =$$

(A) $$(54,821)^2$$

(B) $$(54,821.5)^2$$

(C) $$(54,820.5)^2$$

(D) $$(54,821)^2 + 1$$

(E) $$(54,821)^2 – 1$$

Let 54821 = A

((A-1)^2 + (A+1)^2)/2 = A^2 + 1

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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
Let p = 54,820 . Then , 548,22 = 54,820 + 2 = p + 2

=> $$p^2 + (p + 2)^2 = p^2 + p^2 + 4 + 4p$$

=> $$p^2 + (p + 2)^2 = 2p^2 + 4 + 4p$$

Average: $$\frac{(p^2 + (p + 2)^2) }{ 2} = \frac{(2p^2 + 4 + 4p) }{ 2}$$

=> $$p^2 + 2 + 2p: p^2 + 1 + 4p + 1$$

=> $$(p+1)^2$$ + 1

=> $$(54,820 + 1)^2$$ + 1

=> $$(54,821)^2$$ + 1

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

Plug numbers-

$$2^2 + 4^2$$= $$\frac{20 }{ 2 }= 10$$

$$10 = 3^2 + 1$$

$$10^2 + 12^2 = \frac{244 }{ 2} = 122$$

$$122 = 11^2 + 1$$

54,821 is between 54,820 and 54,822. Thus, the answer is (D) $$(54,821)^2 + 1$$
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
enigma123 wrote:
The mean of $$(54,820)^2$$ and $$(54,822)^2 =$$

(A) $$(54,821)^2$$

(B) $$(54,821.5)^2$$

(C) $$(54,820.5)^2$$

(D) $$(54,821)^2 + 1$$

(E) $$(54,821)^2 – 1$$

avigutman sir can i do this question with the help of weighted average approach? i.e.
2^2 + 4^2 = 4(1) + 4(4)
the scale factor of the second one is bigger so the mean will be closer to it
only one answer states it clearly which is d
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
pdfff wrote:
avigutman sir can i do this question with the help of weighted average approach? i.e.
2^2 + 4^2 = 4(1) + 4(4)
the scale factor of the second one is bigger so the mean will be closer to it
only one answer states it clearly which is d

I don't understand, pdfff. The two terms should have equal weights here.
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Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
enigma123 wrote:
The mean of $$(54,820)^2$$ and $$(54,822)^2 =$$

(A) $$(54,821)^2$$

(B) $$(54,821.5)^2$$

(C) $$(54,820.5)^2$$

(D) $$(54,821)^2 + 1$$

(E) $$(54,821)^2 – 1$$

I will take small no's to solve the big issue....

Mean of $$10^2$$ & $$12^2$$ = $$\frac{100 + 144}{2}=\frac{244}{2}=122$$, Units digit will be $$\frac{0^2 + 2^2}{2}=2$$

Applying the same logic here as well $$\frac{0^2 + 2^2}{2}=2$$

In option (D) $$(54,821)^2$$ = Units digit as 1 , SO $$(54,821)^2 + 1$$ = Units digit as 2 (We found out earlier, hence Answer gotta be (D)
Re: The mean of (54,820)^2 and (54,822)^2 = [#permalink]
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