Hi guys ,

I thought i could do it easily using Brute Force SQL but it gave a tuff time

First i tried with whole numbers but it didnt work but then i found this site (I got the puzzle in a whatsapp group). Once here i saw the answer in decimals.

Of course then i solved it easily with Decimal brute force

use dba

drop table tableA
drop table TableAandC


create table tableA (a bigint)



declare @icount int
set @icount = -50000
while @icount < 50000
begin

insert into Tablea select @icount
set @icount = @icount +1
end

select * From tablea a, tablea b, tablea c , tablea d
where a.a + b.a = 8 and a.a + c.a = 13 and b.a + d.a = 8 and c.a- d.a = 6


--TRY to do it in parts first find all combinations that can have a+b = 8 and a+ c = 13 then check for the rest two


select * From tablea a, tablea c
where a.a + c.a = 13





select A.a AS A , C.A as C into TableAandC From tablea a, tablea c
where a.a + c.a = 13


--Nothing found with whole numbers
-- So lets try with decimals


/**SECOND APPROACH ***/
use dba

drop table tableA


create table tableA (a money , b money,c money,d money)



declare @icount money
set @icount = -10.00
while @icount < 100.00
begin

insert into Tablea select @icount , 8-@icount , 13-@icount , 13-@icount - 6
set @icount = @icount +0.1
end

select * From tablea a
where a.a + b = 8 and a + c = 13 and b+ d = 8 and c- d = 6

The above query gives correct answer. The code will work in SQL SERVER