[wolfgang]

I made a puzzle which has no solution (as far as I tried),

Is it possible to solve it?

Draw a pentagonal star(see attached file).

Give Numbers(1 to 10) for each point, each number should come only once.

so that the sum of four numbers on each straight line is equall to sum of each 4 numbers on the other lines.

[dark_magician_92]

i m not mentioning the reason at this moment but yes i think its not possible .

[Guest]

Not Possible - However, if you use the numbers 0 - 9 it would be...

[Guest]

<p>I'm taking the other side and thinking that its not possible.</p>

<p>This is my reasoning,</p>

<p> </p>

<p>First the sums for all the sides would require a double count of each point, since 2 sides passes through each point.</p>

<p>This would mean that sum of all the sides would equal 110, (2*(1+2+...+10)) . Thus, each side would need to sum to 22, since there are 5 sides.</p>

<p>Possible combinations would be:</p>

<p>10 + 9 + 2 + 1 = 22</p>

<p>9 + 8 + 3 + 2 = 22</p>

<p>9 + 8 + 4 + 1 = 22</p>

<p>8 + 7 + 6 + 1 = 22</p>

<p>8 + 7 + 5 + 2 = 22</p>

<p>8 + 7 + 4 + 3 = 22</p>

<p>7 + 6 + 5 + 4 = 22</p>

<p> </p>

<p>Since there are 5 sides, there needs to be 5 usable equations out of the 7 shown. However, that isn't the case, since 5 of these equations require the number 8. And since one number can only appear twice in the 5 sides, this would reduce the number of possible equations down to 4, which isn't enough for 5 sides.</p>

<p> </p>

<p> </p>

[curr3nt]

10 9 2 1

10 8 3 1

10 7 4 1

10 6 5 1

10 7 3 2

10 6 4 2

10 5 4 3

9 8 4 1

9 7 5 1

9 8 3 2

9 7 4 2

9 6 5 2

9 6 4 3

8 7 6 1

8 7 4 3

7 6 5 4 

[Guest]

Def impossible

from your triagle, the rules are:

a+b+d+e = c+d+f+g = c+b+j+i = e+f+h+i = a+j+h+g

I made a database with 10 tables (a-j) .

Each table has a single column i to hold a value 1-10.

select a.i a,b.i b,c.i c,d.i d, e.i e,f.i f,g.i g,h.i h,i.i i,j.i j

from a,b,c,d,e,f,g,h,i,j

where

a.i <> b.i

and

a.i <> c.i and b.i <> c.i

and

a.i <> d.i and b.i <> d.i and c.i <> d.i

and

a.i <> e.i and b.i <> e.i and c.i <> e.i and d.i <> e.i

and

a.i <> f.i and b.i <> f.i and c.i <> f.i and d.i <> f.i and e.i <>f.i

and

a.i <> g.i and b.i <> g.i and c.i <> g.i and d.i <> g.i and e.i <>g.i and f.i<>g.i

and

a.i <> h.i and b.i <> h.i and c.i <> h.i and d.i <> h.i and e.i <>h.i and f.i<>h.i and g.i <> h.i

and

a.i <> i.i and b.i <> i.i and c.i <> i.i and d.i <> i.i and e.i <>i.i and f.i<>i.i and g.i <> i.i and h.i<>i.i

and

a.i <> j.i and b.i <> j.i and c.i <> j.i and d.i <> j.i and e.i <>j.i and f.i<>j.i and g.i <> j.i and h.i<>j.i and i.i <> j.i

and /* first sum must == the other four*/

a.i+ b.i+ d.i+ e.i = c.i+ d.i+ f.i+ g.i

and

a.i+ b.i+ d.i+ e.i = e.i+ f.i+ h.i+ i.i

and

a.i+ b.i+ d.i+ e.i = c.i+ b.i+ j.i+ i.i

and

a.i+ b.i+ d.i+ e.i = a.i+ j.i+ h.i+ g.i

;

Empty set (3.07 sec)

no solutions found

[Guest]

<p>@spartan: Draw the picture and allot the numbers for each point. So that we can know which number comes where..</p>