Posted July 16, 2013 Another puzzle from that Australian site, this time more mathy: 0 Share this post Link to post Share on other sites

0 Posted July 20, 2013 n^{n-1}/2 = 1; n = 2sqrt(n)-1 = 2; n = 9sqrt(n)+1 = 3; n = 4n^{2} = 4; n = 2(3n)!-n = 5; n = 1(3(n!))!+n = 6; n = 02srqt(n)+1 = 7; n = 9n*1^{n} (=n if n<infinity) = 8; n = 82+n = 9; n = 7n^{2}+1 = 10; n = 3n^{2}+(n-1)! = 11; n = 32n+|n| (=3n, n>0, or =n, n<0) = 12; n = 4(n^{3}+n)/10 = 13; n = 52n = 14; n = 710(n+1)/n = 15; n = 22n-(sqrt(n)-1)! = 16; n = 92n+1 = 17; n = 8(2n)^{n}+n = 18; n = 2(n-1)!-n = 19; n = 5(2n)!-n^{2} = 20; n = 2n*sqrt(9) = 21; n = 7n+20 = 22; n = 2[10/n]^{n/2}*n-sqrt(n) = 23; n = 4(11-n)! = 24; n = 7n^{2} = 25; n = 5 (a(b·c^{d}·sqrt{ef-g+sqrt[(((h·i-j)·k·l+m)·n+o)·p] ·q(r·s)^{t}} -u+v/sqrt(w))-x)y(2(9·4^{2}·sqrt{10-9+sqrt[(((8·7-3)·3·4+5)·7+2)·9] ·8(2·5)^{2}} -7+2/sqrt(4))-7)5(2(144 ·sqrt{ 1 + sqrt[(( 641 )·7+2)·9] · 800 } - 6 )-7)5(2(144 ·sqrt{ 1 + 201 · 800 } - 6 )-7)5(2(144 · 401 - 6 )-7)5(2(144·401-6)-7)5577345Now we just need seven more of those to make the whole tongue-twister. 0 Share this post Link to post Share on other sites

0 Posted July 16, 2013 I think the answer is Ichigo 0 Share this post Link to post Share on other sites

0 Posted July 17, 2013 What are we supposed to do? Or is that part of the problem? My guess is 3. 0 Share this post Link to post Share on other sites

0 Posted July 17, 2013 What are we supposed to do? Or is that part of the problem? Yes. 0 Share this post Link to post Share on other sites

0 Posted July 17, 2013 I think it should be 25. Here's why: Basic assumption: Results of all expressions should be integers Simplifying all the expressions, it is clear that the number should be a perfect square. Next, in 2 or 3 expression with the decimals, the single digit should divide 10, or double digit number should divide 100, a triple digit should divide 1000... and so on Finally, the last expression yields an integer if the number is 16 or 25. Using all the above combinations, 25 seems to be the correct answer 0 Share this post Link to post Share on other sites

0 Posted July 17, 2013 I don't think 25 is the answer -- the first expression would evaluate to 25^{24}/2 which is not an integer. I see 25 sets of "small" expressions. For the first few of them at least, for expression #N, I can solve for a value x which when placed in each of the boxes will make the expression evaluate to N. After those 25 small expression, there's that big expression with 25 boxes which might be the solutions to the previous 25 small equations. 0 Share this post Link to post Share on other sites

0 Posted July 17, 2013 Simplifying all equations without any solutions to make things easier: n^{n-1}/2 sqrt(n)-1 sqrt(n)+1 n^{2} (3n)!-n (3(n!))!+n 2srqt(n)+1 n*1^{n} (=n if n<infinity) 2+n 2n+1 2n+(n-1)! 2n+|n| (=3n, n>0, or =n, n<0) 2n*.n+.n (=.n*(2n+1)) 2n (n+1)/.n 2n-(sqrt(n)-1)! 2n+1 2n^{n}+n (n-1)!-n 2n!-1 n*sqrt((n-.n)/.n) n(1+2/.n) sqrt(.n^{-}^{n})*n-sqrt(n) (nn/n-n)! n^{2} 0 Share this post Link to post Share on other sites

0 Posted July 19, 2013 Simplifying all equations without any solutions to make things easier: n^{n-1}/2 sqrt(n)-1 sqrt(n)+1 n^{2} (3n)!-n (3(n!))!+n 2srqt(n)+1 n*1^{n} (=n if n<infinity) 2+n 2n+1 2n+(n-1)! 2n+|n| (=3n, n>0, or =n, n<0) 2n*.n+.n (=.n*(2n+1)) 2n (n+1)/.n 2n-(sqrt(n)-1)! 2n+1 2n^{n}+n (n-1)!-n 2n!-1 n*sqrt((n-.n)/.n) n(1+2/.n) sqrt(.n^{-}^{n})*n-sqrt(n) (nn/n-n)! n^{2} I thought it was more of a fill-in thing, that squares 0 Share this post Link to post Share on other sites

0 Posted July 20, 2013 n^{n-1}/2 = 1; n = 2sqrt(n)-1 = 2; n = 9sqrt(n)+1 = 3; n = 4n^{2} = 4; n = 2(3n)!-n = 5; n = 1(3(n!))!+n = 6; n = 02srqt(n)+1 = 7; n = 9n*1^{n} (=n if n<infinity) = 8; n = 82+n = 9; n = 7n^{2}+1 = 10; n = 3n^{2}+(n-1)! = 11; n = 32n+|n| (=3n, n>0, or =n, n<0) = 12; n = 4(n^{3}+n)/10 = 13; n = 52n = 14; n = 710(n+1)/n = 15; n = 32n-(sqrt(n)-1)! = 16; n = 92n+1 = 17; n = 8(2n)^{n}+n = 18; n = 2(n-1)!-n = 19; n = 5(2n)!-n^{2} = 20; n = 2n*sqrt(9) = 21; n = 7n+20 = 22; n = 2[10/n]^{n/2}*n-sqrt(n) = 23; n = 4(11-n)! = 24; n = 7n^{2} = 25; n = 5 a(b·c^{d}sqrt{ef-g+sqrt[(((h·i-j)·k·l+m)·n+o)·p] ·q(r·s)^{t}} -u+v/sqrt(w)-x)y2(9·4^{2}sqrt{10-9+sqrt[(((8·7-3)·3·4+5)·7+3)·9] ·8(2·5)^{2}} -7+2/sqrt(4)-7)52(144 sqrt{ 1 + sqrt[(( 641 )·7+3)·9] · 800 } - 12 )5 But that gives me 577352.1557..... so I think I goofed somewhere. 0 Share this post Link to post Share on other sites

0 Posted July 20, 2013 Soooo close... check #15 Also the final answer is supposed to be a word, which might be a little hard to see, so here's a hint: !dlrow 01134 0 Share this post Link to post Share on other sites

0 Posted July 20, 2013 [spoiler= ]n^{n-1}/2 = 1; n = 2sqrt(n)-1 = 2; n = 9sqrt(n)+1 = 3; n = 4n^{2} = 4; n = 2(3n)!-n = 5; n = 1(3(n!))!+n = 6; n = 02srqt(n)+1 = 7; n = 9n*1^{n} (=n if n<infinity) = 8; n = 82+n = 9; n = 7n^{2}+1 = 10; n = 3n^{2}+(n-1)! = 11; n = 32n+|n| (=3n, n>0, or =n, n<0) = 12; n = 4(n^{3}+n)/10 = 13; n = 52n = 14; n = 710(n+1)/n = 15; n = 22n-(sqrt(n)-1)! = 16; n = 92n+1 = 17; n = 8(2n)^{n}+n = 18; n = 2(n-1)!-n = 19; n = 5(2n)!-n^{2} = 20; n = 2n*sqrt(9) = 21; n = 7n+20 = 22; n = 2[10/n]^{n/2}*n-sqrt(n) = 23; n = 4(11-n)! = 24; n = 7n^{2} = 25; n = 5 a(b·c^{d}sqrt{ef-g+sqrt[(((h·i-j)·k·l+m)·n+o)·p] ·q(r·s)^{t}} -u+v/sqrt(w)-x)y2(9·4^{2}sqrt{10-9+sqrt[(((8·7-3)·3·4+5)·7+2)·9] ·8(2·5)^{2}} -7+2/sqrt(4)-7)52(144 sqrt{ 1 + sqrt[(( 641 )·7+2)·9] · 800 } - 12 )52(144 sqrt{ 1 + 201 · 800 } - 12 )52(144 401 - 12 )5577320 0 Share this post Link to post Share on other sites

0 Posted July 20, 2013 ...your math. Your numbers are all correct, but the answer is not...I don't know how you got a "-12" from that last part... You're doing so well I wanted you to get the final answer, lol, that's why I couched the hint in a way that I thought would be obvious to like, programming ppl 0 Share this post Link to post Share on other sites

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Another puzzle from that Australian site, this time more mathy:

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