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Math Expert V
Joined: 02 Sep 2009
Posts: 58452
1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =  [#permalink]

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Difficulty:   5% (low)

Question Stats: 89% (01:16) correct 11% (01:11) wrong based on 74 sessions

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1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =

A. 6,030,053
B. 6,030,054
C. 6,030,055
D. 6,030,056
E. 6,030,057

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Math Expert V
Joined: 02 Aug 2009
Posts: 7984
Re: 1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =  [#permalink]

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1
2
Bunuel wrote:
1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =

A. 6,030,053
B. 6,030,054
C. 6,030,055
D. 6,030,056
E. 6,030,057

hi,

the NORMAL method could be ..

Sum of squares of 1005 positive integers - Sum of squares of first 999 positive integers..

But here we can see that in ALL choices , the UNITS digit is different...
SO, finding units digit should be sufficient..

$$1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2$$
THe units digit sum would be$$0^2 + 1^2 +2^2 +3^2 +4^2 + 5^2 = 0+1+4+9+16+25..$$
so units digit = 5

ans C
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Intern  Joined: 21 Oct 2015
Posts: 45
GMAT 1: 620 Q47 V28 Re: 1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =  [#permalink]

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Interesting problem.

I think key is to notice that all the given answer choices differs in last two digits. Therefore, our entire focus should be to figure out how the given terms contribute to last two digits of total.

1000^2 -> 00
1001^1 -> 01
.
.
.
1005^2 -> 25

Total -> *55

Any other approach or better explanations most welcome.

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Current Student D
Joined: 12 Aug 2015
Posts: 2567
Schools: Boston U '20 (M)
GRE 1: Q169 V154 1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =  [#permalink]

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Interesting Question..
Here using the concept of UD => UD will be 5
so C is my choice
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Re: 1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =  [#permalink]

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_________________ Re: 1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =   [#permalink] 04 Mar 2018, 03:09
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