Jump to content
BrainDen.com - Brain Teasers
  • 0

Calculated


Yoruichi-san
 Share

Question

12 answers to this question

Recommended Posts

  • 0

nn-1/2 = 1; n = 2

sqrt(n)-1 = 2; n = 9

sqrt(n)+1 = 3; n = 4

n2 = 4; n = 2

(3n)!-n = 5; n = 1

(3(n!))!+n = 6; n = 0

2srqt(n)+1 = 7; n = 9

n*1n (=n if n<infinity) = 8; n = 8

2+n = 9; n = 7

n2+1 = 10; n = 3

n2+(n-1)! = 11; n = 3

2n+|n| (=3n, n>0, or =n, n<0) = 12; n = 4

(n3+n)/10 = 13; n = 5

2n = 14; n = 7

10(n+1)/n = 15; n = 2

2n-(sqrt(n)-1)! = 16; n = 9

2n+1 = 17; n = 8

(2n)n+n = 18; n = 2

(n-1)!-n = 19; n = 5

(2n)!-n2 = 20; n = 2

n*sqrt(9) = 21; n = 7

n+20 = 22; n = 2

[10/n]n/2*n-sqrt(n) = 23; n = 4

(11-n)! = 24; n = 7

n2 = 25; n = 5

(a(b·cd·sqrt{ef-g+sqrt[(((h·i-j)·k·l+m)·n+o)·p] ·q(r·s)t} -u+v/sqrt(w))-x)y

(2(9·42·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)5

577345

Now we just need seven more of those to make the whole tongue-twister.

Link to comment
Share on other sites

  • 0

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

Link to comment
Share on other sites

  • 0

I don't think 25 is the answer -- the first expression would evaluate to 2524/2 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.

Link to comment
Share on other sites

  • 0

Simplifying all equations without any solutions to make things easier:

nn-1/2

sqrt(n)-1

sqrt(n)+1

n2

(3n)!-n

(3(n!))!+n

2srqt(n)+1

n*1n (=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

2nn+n

(n-1)!-n

2n!-1

n*sqrt((n-.n)/.n)

n(1+2/.n)

sqrt(.n-n)*n-sqrt(n)

(nn/n-n)!

n2

Link to comment
Share on other sites

  • 0

Simplifying all equations without any solutions to make things easier:

nn-1/2

sqrt(n)-1

sqrt(n)+1

n2

(3n)!-n

(3(n!))!+n

2srqt(n)+1

n*1n (=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

2nn+n

(n-1)!-n

2n!-1

n*sqrt((n-.n)/.n)

n(1+2/.n)

sqrt(.n-n)*n-sqrt(n)

(nn/n-n)!

n2

I thought it was more of a fill-in thing, that squares

Link to comment
Share on other sites

  • 0

nn-1/2 = 1; n = 2

sqrt(n)-1 = 2; n = 9

sqrt(n)+1 = 3; n = 4

n2 = 4; n = 2

(3n)!-n = 5; n = 1

(3(n!))!+n = 6; n = 0

2srqt(n)+1 = 7; n = 9

n*1n (=n if n<infinity) = 8; n = 8

2+n = 9; n = 7

n2+1 = 10; n = 3

n2+(n-1)! = 11; n = 3

2n+|n| (=3n, n>0, or =n, n<0) = 12; n = 4

(n3+n)/10 = 13; n = 5

2n = 14; n = 7

10(n+1)/n = 15; n = 3

2n-(sqrt(n)-1)! = 16; n = 9

2n+1 = 17; n = 8

(2n)n+n = 18; n = 2

(n-1)!-n = 19; n = 5

(2n)!-n2 = 20; n = 2

n*sqrt(9) = 21; n = 7

n+20 = 22; n = 2

[10/n]n/2*n-sqrt(n) = 23; n = 4

(11-n)! = 24; n = 7

n2 = 25; n = 5

a(b·cdsqrt{ef-g+sqrt[(((h·i-j)·k·l+m)·n+o)·p] ·q(r·s)t} -u+v/sqrt(w)-x)y

2(9·42sqrt{10-9+sqrt[(((8·7-3)·3·4+5)·7+3)·9] ·8(2·5)2} -7+2/sqrt(4)-7)5

2(144 sqrt{ 1 + sqrt[(( 641 )·7+3)·9] · 800 } - 12 )5

But that gives me 577352.1557..... so I think I goofed somewhere.

Link to comment
Share on other sites

  • 0

[spoiler= :duh:]nn-1/2 = 1; n = 2

sqrt(n)-1 = 2; n = 9

sqrt(n)+1 = 3; n = 4

n2 = 4; n = 2

(3n)!-n = 5; n = 1

(3(n!))!+n = 6; n = 0

2srqt(n)+1 = 7; n = 9

n*1n (=n if n<infinity) = 8; n = 8

2+n = 9; n = 7

n2+1 = 10; n = 3

n2+(n-1)! = 11; n = 3

2n+|n| (=3n, n>0, or =n, n<0) = 12; n = 4

(n3+n)/10 = 13; n = 5

2n = 14; n = 7

10(n+1)/n = 15; n = 2

2n-(sqrt(n)-1)! = 16; n = 9

2n+1 = 17; n = 8

(2n)n+n = 18; n = 2

(n-1)!-n = 19; n = 5

(2n)!-n2 = 20; n = 2

n*sqrt(9) = 21; n = 7

n+20 = 22; n = 2

[10/n]n/2*n-sqrt(n) = 23; n = 4

(11-n)! = 24; n = 7

n2 = 25; n = 5

a(b·cdsqrt{ef-g+sqrt[(((h·i-j)·k·l+m)·n+o)·p] ·q(r·s)t} -u+v/sqrt(w)-x)y

2(9·42sqrt{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 } - 12 )5

2(144 sqrt{ 1 + 201 · 800 } - 12 )5

2(144 401 - 12 )5

577320

^_^

Link to comment
Share on other sites

  • 0

...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 ;)
Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...