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A is a set of all natural numbers less than $500$ which can be written as sum of two or more consecutive natural numbers.How to find the maximum possible cardinality of A?

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The only numbers that can't be written as a sum of two or more consecutive natural numbers are the powers of 2. So there is no "maximum possible" cardinality of $A$. Just take out the powers of 2 that are less than 500, and the numbers 0 and 1.

Proof: If $n=k+(k+1)+...+k+(l-1)=lk+\frac{1}{2}(l-1)l$ then $2n=2lk+l(l-1)=l(2k+l-1)$. If one of these factors is even, the other one is odd. So it all comes down to which numbers can be factored as a product of an even number and and odd number, and this leads to my claim.

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Any odd number is of the form $2k+1=k+(k+1)$ so can be written as a sum of two consecutive numbers. How many of them are less than $500$? Any number that is a multiple of $3$ is of the form $3k=(k-1)+k+(k+1)$ How many of these are there? You can continue like this-sums of four successive numbers will be related to multiples of $4$ (how?) and so on. Then you have to worry about double counting. For example $15=7+8=4+5+6=1+2+3+4+5$

Alternately, you can just make a list of all the ones you can express with two numbers, all you can express with three, and so on, then throw out the duplicates.

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