rocdocmac 9 Posted April 11, 2017 Report Share Posted April 11, 2017 Has this one appeared before? Quote Link to post Share on other sites

0 Solution Pickett 13 Posted April 13, 2017 Solution Report Share Posted April 13, 2017 Spoiler 166,667,166,667,000,000 Quote Link to post Share on other sites

0 bonanova 85 Posted April 11, 2017 Report Share Posted April 11, 2017 No, and after playing with it for a few minutes it seems a nice challenge. Thanks. Quote Link to post Share on other sites

0 flamebirde 196 Posted April 18, 2017 Report Share Posted April 18, 2017 (edited) If calculators can be used, then the problem is simple. Without a program or calculator, though... Spoiler Each individual term is the series from n=0 to some number X of n. So X=3 would give 3+2+1 or six atoms on the third layer. That part's simple enough. Each layer can also be defined as 1+2+3+... n where n is how many layers down the tetrahedron it is. The question is how you can add up all the layers. It would be something like (1+2+3+... 1000000)+ (1+2+3+... 999999) +... 1. Alternatively, it could become 1000000+2(999999)+3(999998)+... 1000000(1); a modified version of the classic "add up all the numbers from one to one hundred" problem. Edited April 18, 2017 by flamebirde Doubled. Quote Link to post Share on other sites

0 flamebirde 196 Posted April 19, 2017 Report Share Posted April 19, 2017 If anyone's still working on this, I also have this... Spoiler Here's the series for this problem: the sum of all numbers from 0 to some number k-1 of ((k-n)^2)n where k is the number of layers. So, the sum of all numbers from zero to one million of ((1000000-n)^2)*n is the answer to the problem. But how to do it without a calculator? Quote Link to post Share on other sites

0 flamebirde 196 Posted April 20, 2017 Report Share Posted April 20, 2017 On 4/18/2017 at 5:41 PM, flamebirde said: If anyone's still working on this, I also have this... Reveal hidden contents Here's the series for this problem: the sum of all numbers from 0 to some number k-1 of ((k-n)^2)n where k is the number of layers. So, the sum of all numbers from zero to one million of ((1000000-n)^2)*n is the answer to the problem. But how to do it without a calculator? Nope, I retract that, it should be Spoiler Series from n=0 to k of n(n+1)/2. My friend pointed out to me that it wasn't a modified version of the "add up all the numbers from one to one hundred" problem, it was literally exactly that problem. I still have no idea how I could work it out by hand. Quote Link to post Share on other sites

0 Pickett 13 Posted April 20, 2017 Report Share Posted April 20, 2017 Isn't my answer above correct? Spoiler Isn't it just the tetrahedral numbers...the formula for the series is S_{n} = ^{n(n+1)(n+2)} / _{6} So, for n=1,000,000, you have ^{((1,000,000)(1,000,001)(1,000,002))} / _{6} = ^{1,000,003,000,002,000,000} / _{6} = 166,667,166,667,000,000 Quote Link to post Share on other sites

0 Buddyboy3000 4 Posted April 21, 2017 Report Share Posted April 21, 2017 (edited) Spoiler One way to think about it is to treat the crystal like a pyramid, with some exceptions. To find the amount of dots on the bottom I did: (l)(h+1)/2 This is like the normal equation except adding the length again makes up for the part of the atoms that are cut off in the straight line. For 1 million, I got (1,000,000)(1,000,001)/2 = 500,000,500,000 With the number of atoms on the bottom, next find the equation to get the answer which is (b)(h+2)/3. Again, the h+2 is to make up for what is cut off. I would include the proof for that but is a bit complicated to explain. So, the answer is 500,000,500,000*1,000,002/3 = 166,667,166,667,000,000. This is the same answer Pickett got, so he got the right answer first. Edited April 21, 2017 by Buddyboy3000 Had a sentence repeat itself Quote Link to post Share on other sites

0 rocdocmac 9 Posted April 22, 2017 Author Report Share Posted April 22, 2017 (edited) Well done all of you! The correct answer is indeed 166 667 166 667 000 000 Edited April 22, 2017 by rocdocmac Quote Link to post Share on other sites

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