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# Calculated

## Question

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

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I think the answer is Ichigo

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What are we supposed to do? Or is that part of the problem?

My guess is 3.

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What are we supposed to do? Or is that part of the problem?

Yes.

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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

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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.

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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

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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

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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.

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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

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[spoiler= ]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

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...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
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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.

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