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:lol: No! this has got nothing to do with the knowledge of mechanical clocks, just a simple guess on what could have gone wrong when re-assembling it.Quote:
Originally Posted by rajraj
It is not a great deal mathematical either, so everyone can try.
Both the client and repairman are telling the truth, just that on the two occasions when he came to check, the time shown on the clock happens to be the correct time!
GP,Quote:
Originally Posted by GP
There is no way I can arrive at the answer using pen and paper! However I thought this would be a good opportunity to get into Modeling :lol2:
So I tried model the solution using Microsoft Solver Foundation and here is what I got:
11826^2 = 139854276
30384^2 = 923187456
Below is the OML(optimization modeling language) model I wrote to arrive at this:
I think this is one hellova tool from Microsoft! :2thumbsup: :2thumbsup: Very simple and easy to solve optimization problems using this. Others interested may try this from here:Code:Model
[
Decisions[
Integers[1,9],a,b,c,d,e,f,g,h,i
],
Decisions[
Integer[10000,31622],y
],
Constraints[
a + b*10 + c*100 + d*1000 + e*10000 + f*100000 + g*1000000 + h*10000000 + i*10000000 == y*y
Unequal[a,b,c,d,e,f,g,h,i],
],
Goals[
Minimize[a + b*10 + c*100 + d*1000 + e*10000 + f*100000 + g*1000000 + h*10000000 + i*100000000]
]
]
http://code.msdn.microsoft.com/solverfoundation
what! still not answered? :roll:
the key is that both times he was called, it was a little after 8 and a little after 7.
if you had said that the time was 8.10 and 7.05, maybe you would have got the answer by now. :D
after all a completely spoilt clock also shows correct time twice a day. :lol2:
But this partially clock would show the correct time 24 times a day :wink:Quote:
Originally Posted by NOV
Kirukans kirukkal code :wink: for getting the same resultQuote:
Originally Posted by Benny Lava
--TSQL MSSQL2005
-Code:declare @num bigint ,@min bigint ,@max bigint , @sqr bigint,@total int,@rnd varchar(9),@con varchar(9),@cnt int
set @sqr=0
set @num=10000
set @min=0
set @max=0
while(@num<100000)
begin
set @num=@num+1
set @sqr=power(@num,2)
set @rnd= ltrim(str(@sqr))
if(len(@rnd)>9) break
set @total=1
if (charindex('0',@rnd)>0) continue
set @con=''
set @cnt=1
while (@cnt<10)
begin
if(charindex(substring(@rnd,@cnt,1),@con)>0)
begin
set @total=0
break
end
else
set @con=@con+substring(@rnd,@cnt,1)
set @cnt=@cnt+1
end
if(@total=1)
begin
if(@min=0) set @min=@num
set @max=@num
end
end
select @min as Minimum,convert(bigint,power(@min,2)) Min ,@max as Maximum,convert(bigint,power(@max,2)) Max
Kirukkan
I don't know any of these languages/codes! :lol: Please give me a program in FORTRAN,APL or assembly language in 1960s/70s mainframes ! :lol: :lol: :lol:
For you fortran uncompiledQuote:
Originally Posted by rajraj
Code:Program PUZZLE
INTEGER Num, min, max,total,cnt,sqr
CHARACTER(LEN=9) rnd, con
sqr=0
num=10000
min=0
max=0
do while(num<100000)
num=num+1
sqr=num*num
Write( rnd, '(i10)' ) sqr
if(len(rnd)>9) break
total=1
if (index('0',rnd)>0) continue
con=''
cnt=1
do while (cnt<10)
if(index(rnd(cnt,1),con)>0) then
total=0
break
else
con=con//rnd(cnt,1)
end if
cnt=cnt+1
end do
if(total=1) then
if(min=0)then
min=num
end if
max=num
end if
end do
print *, min
print *, max
End program
kirukan: You get an A+ ! :) I don't have access to a mainframe to verify your program. I don't have a teaching assistant to do it! Still, you get an A+ ! :lol:
GP has enough clues to fix the clock! :)
Here is a simple one:
Which is the odd number here?
25 35 48 63 80
is it 25?
everythign else goes by the rule of (n+1) x (n+3) :oops: vendikkai innum saapadanum-pOla