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 May 20th, 2013, 08:49 PM #1 Member   Joined: Apr 2012 Posts: 64 Thanks: 3 find an item of series
 May 21st, 2013, 10:14 AM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: find an item of series Every member of S is of the form $2^x + 2^y + 2^z$, and S can be sorted in increasing order as follows: sort by greater z, then by greater y, and then by greater x. Let us consider the number of elements sorting by greater z: $2^0 + 2^1 + 2^2$ 1 element for z = 2 $2^0 + 2^1 + 2^3$ $2^0 + 2^2 + 2^3$ $2^1 + 2^2 + 2^3$ 3 elements for z = 3... ${k \choose 2}$ elements for z = k. Thus, up to z = k, there are a total of ${k+1 \choose 3}$ elements. The first k such that ${k+1 \choose 3} \geq 100$ is 9, so for the 100th element, z = 9. Now the 120th element is $2^7 + 2^8 + 2^9$, so we simply work backwards from there: The 113th element is $2^0 + 2^8 + 2^9$ The 112th element is $2^6 + 2^7 + 2^9$ The 106th element is $2^0 + 2^7 + 2^9$ The 105th element is $2^5 + 2^6 + 2^9$ The 100th element is $2^0 + 2^6 + 2^9$.

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