Yes Bob, adding the absolutes for the array $A$6:$E$10 was all it needed. I
works Great.
Sorry I took so long to answer.. Had to leave for work.
Thanks again
Luke

"Bob Phillips" wrote:

Is this what you mean?

=IF(COUNTIF(A$6:E$10,RIGHT(F6,2)+0),G6,"")

--
HTH

Bob Phillips

"Luke" wrote in message
...
Okay that worked but I only got a "1" where true is the case. I am
looking
for a specific result, of which I didn't mention... Just thought the
formula
would automatically do it I guess. I digress.
Let's do this:
A B C D E
F G
50 45 48 49 55 752 695
60 65 58 59 66 0 2596
70 75 68 69 77 1595 215
80 85 78 79 88 7355 795
90 95 98 89 99 7 638

If "F" column (match right 2 digits with any of A:E) then display G
Does that make since. sorry I left out the details Bob.
Thanks
Luke

"Bob Phillips" wrote:

I doubt it. MATCH doesn't like 2D.

Try this

=COUNTIF($A$6:$E$10,RIGHT(F15,2)+0)

--
HTH

Bob Phillips

"Luke" wrote in message
...
=ISNUMBER(MATCH(RIGHT(F15,2)+0,$A$6:$E$10,0))+0

The formula above worked, I thought, but now for some reason I can't
get
anything but zero. Here's a sample of sheet.

A B C D E
F
50 45 48 49 55 752
60 65 58 59 66 0
70 75 68 69 77 1595
80 85 78 79 88 7355
90 95 98 89 99 7

Any help would be great Thank you!
Luke