From: b1llt on
I should have mentioned we're still on Excel 2000.
I used these and am getting a #DIV/0! result.
Any suggestions ---thanks,
Bill

"Ziggy" wrote:

> Another 2003 solution is an array formula
>
> =SUM(($K$15:$K$38=K41)*($M$15:$M$38>=1))
>
> Set with CTRL-Shift-Enter
> .
>
From: "David Biddulph" groups [at] on
There are no divide operations in that formula, so if you're seeing a
#DIV/0! result it's because you've got a #DIV/0! in the data being used by
the formula. Tackle the problem where it's being generated.
--
David Biddulph


"b1llt" <b1llt(a)discussions.microsoft.com> wrote in message
news:AF1034DE-3002-4132-A3CE-BEC4F0199798(a)microsoft.com...
> I should have mentioned we're still on Excel 2000.
> I used these and am getting a #DIV/0! result.
> Any suggestions ---thanks,
> Bill
>
> "Ziggy" wrote:
>
>> Another 2003 solution is an array formula
>>
>> =SUM(($K$15:$K$38=K41)*($M$15:$M$38>=1))
>>
>> Set with CTRL-Shift-Enter
>> .
>>

From: b1llt on
My bad! Your correct I did have it pulling a #DIV/O! into the data by mistake.
These all work great! Thanks everyone for all your help.
-B1llt

"David Biddulph" wrote:

> There are no divide operations in that formula, so if you're seeing a
> #DIV/0! result it's because you've got a #DIV/0! in the data being used by
> the formula. Tackle the problem where it's being generated.
> --
> David Biddulph
>
>
> "b1llt" <b1llt(a)discussions.microsoft.com> wrote in message
> news:AF1034DE-3002-4132-A3CE-BEC4F0199798(a)microsoft.com...
> > I should have mentioned we're still on Excel 2000.
> > I used these and am getting a #DIV/0! result.
> > Any suggestions ---thanks,
> > Bill
> >
> > "Ziggy" wrote:
> >
> >> Another 2003 solution is an array formula
> >>
> >> =SUM(($K$15:$K$38=K41)*($M$15:$M$38>=1))
> >>
> >> Set with CTRL-Shift-Enter
> >> .
> >>
>
> .
>