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Understanding the Solution for Finding the Sum of Digits of m
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[QUOTE="RUber, post: 4966258, member: 524408"] Containing all odd numbers is a consequence of the first statement, 2n+1=(n+1)^2 -n^2. This clearly shows that all odd numbers can be created as a difference of squares. For even numbers, a different identity is used, n^3=[n (n+1)/2]^2 - [n (n-1)/2]^2. This is not necessarily even, but is definitely a difference of square integers. So it can be concluded that any cubed integer is also in the set. Looking at the bounds [1,2014], the largest even cube is 12^3 = 1728. Odd numbers are 2n+1 for n = 0 to 1006, or 1007 in the set. Even numbers are the cubes of 2, 4, 6, 8, 10, and 12, or 6 in the set. This gives 1007+6 = 1013. The truth of the identities used can be shown through simple expansion. My question is how can one be sure that other even numbers should be excluded from the set? [/QUOTE]
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Understanding the Solution for Finding the Sum of Digits of m
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