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pb_india
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y1, ... , y10 - remainders after dividing by 2 - that means a sequence of 0 and 1. 0 for each even x and 1 for each odd.

1. we have 6 even numbers out of 10 - that means 4 odd. We can count out even numbers from the sum y1+..+y10 because even numbers give us 0es, we will have four ones in there so the sum is 4.

2. x1 + ... + x10 = 100

Pick x1 = x2 = ... = x9 = 2, x10= 82
y1 = y2 = ... = y10 = 0, sum is 0
Now pick x1 = x2 = ... = x8 = 2, x9 = 1, x10 = 83
y1 = y2 = ... = y8 = 0, y9 = 1, y10 = 1, sum is 2
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MA
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guys, do we know that all numbers in the sequences are integers?
i think it should be E.
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MA
guys, do we know that all numbers in the sequences are integers?
i think it should be E.


If the problem states that Yn is the remainder of Xn/2, then how can non-integer numbers have remainders? As far as I know, only integers can have remainders. MA, could you please correct be if I'm wrong.

1st statement is sufficient. The remaining 4 are odd, therefore, the remainders will be 1 and the sum 4.

2nd statement is insufficient. Say, Xn=2p+Yn, where p is a whole number. We can express all Xs by this equation and summing up all those 10 equations we'll left with 2 unknowns, p and sum of Ys. Insufficient

Answer is A. Please post OA.
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