Basic Set Theory

Let X X be the set of integers {8,14,20,26,32,....350,356,362,368,374} and Y Y be the subset of X X such that no two elements of Y Y have a sum of 382 .

Find the maximum number of elements Y Y can have .

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The answer is 31.

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1 solution

The elements of X X are the first 62 62 terms in the arithmetic sequence where the n n th term is given by a n = 8 + 6 ( n 1 ) = 2 + 6 n a_{n} = 8 + 6(n - 1) = 2 + 6n .

Now this sequence is such that, for k < 63 k \lt 63 ,

a k + a 63 k = ( 2 + 6 k ) + ( 2 + 6 ( 63 k ) ) = 382 a_{k} + a_{63-k} = (2 + 6k) + (2 + 6(63 - k)) = 382 .

Thus we can place 1 1 element from each of 31 31 pairs that sum to 382 382 into Y Y , but any additional element placed in Y Y will necessarily yield a pair of elements that sum to 382 382 .

Thus Y Y can have a maximum of 31 \boxed{31} elements.

A great yet concise explanation ,sir.

A Former Brilliant Member - 6 years, 5 months ago

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