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Sequence L1, L2, ...L10, and M1 to M10, where L is integers

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getzgetzu
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Sequence L1, L2, ...L10, and M1 to M10, where L is integers [#permalink] New post   27 Nov 2005, 03:05
Sequence L1, L2, ...L10, and M1 to M10, where L is integers and M is remainder of L divided by 2. What is the sum of M?

1) Sum of L = 100
2) Six terms of L are even



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Re: Sequence L1, L2, ...L10, and M1 to M10, where L is integers [#permalink] New post   27 Nov 2005, 03:24
getzgetzu wrote:
Sequence L1, L2, ...L10, and M1 to M10, where L is integers and M is remainder of L divided by 2. What is the sum of M?

1) Sum of L = 100
2) Six terms of L are even



Notice, value of M can only be 0 or 1
1.sum of L is even ----> the number of odd terms in the L sequence is even ..that is to say, the number of odd terms can be 0,2,4,6,8,10 ---> the sum of M sequence can be 0,2,4,6,8,10 --->insuff

2. ----> 4 terms of L are odds ...each odd term forms a term M of M sequence with the value of 1 ---> the sum of M sequence is 4*1=4 --->suff
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Re: Sequence L1, L2, ...L10, and M1 to M10, where L is integers [#permalink] New post   27 Nov 2005, 03:31
The OA is B. Are you Honghu's twin? Your explantions are beyond the fancy of the loudest expression....
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Re: Sequence L1, L2, ...L10, and M1 to M10, where L is integers [#permalink] New post   26 Dec 2005, 14:21
IMO its D)

1) when 2 numbers are div by the same div, the remainders are r1 and r2, then when the sum of these 2 numbers is div by the same div, the remainder is r1+r2. sum of L is 100. when 100 is div by 2, the sum of M is the remainder of the division.

2) suff as expl.
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Re: Sequence L1, L2, ...L10, and M1 to M10, where L is integers [#permalink] New post   26 Dec 2005, 17:22
I get that the second statement, 2) Six terms of L are even is sufficient..
Little confused about the first one.



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Re: Sequence L1, L2, ...L10, and M1 to M10, where L is integers [#permalink]
26 Dec 2005, 17:22
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