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

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1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =  [#permalink]

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New post 15 Apr 2016, 01:13
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

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

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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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New post 15 Apr 2016, 01:54
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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Re: 1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =  [#permalink]

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New post 15 Apr 2016, 01:55
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
Answer C.

Any other approach or better explanations most welcome.

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1,000^2 + 1,001^2 +1,002^2 + 1,003^2 + 1,004^2 + 1,005^2 =  [#permalink]

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New post 15 Apr 2016, 02:23
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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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