Harlan Grove,
I tried it and the returns in blank cells I,J,K were #DIV/0!. to clean it
up I added:
=IF(CONCATENATE($I7279,$J7279,$K7279)="","",IF(AND (ABS(COUNTIF($I7279:$K7279,C7279)/SUMPRODUCT(COUNTIF($I7279:$K7279,$C7279:$E7279))-0.5)<0.5,COUNTIF($C7279:C7279,C7279)<=COUNTIF($I72 79:$K7279,C7279)),C7279,""))

I soon discovered some similar exceptions such as the last post (before this
one) I made to Biff. It seems if there are double numbers (ie 988) on the
left C,D,E and only one digit on the right in either I,J or K, it wants to
return one or two digits, usually in G and/or H.
You did eliminate the extra 5 that was appearing so not a total loss.
7 5 5 7 5 x 7 5 7
Perhaps a new function to introduce into excel would be =NOIFANDORBUTIF()
lol
Luke


"Harlan Grove" wrote:

"Biff" wrote...
....
However, according to the OP, the correct output should be:

_ | 8 | 8
....
4 | 5 | _
....
The entries that are 4|5|5 and _|4|5 are kicking my butt! Specifcally, and
this is where I'm stuck, the last 5 should not appear in column H.
....

vs

_ | 8 | 8
....
4 | 5 | 5

The records (cols C through K) with my results are

2 | 8 | 8 || _ | 8 | 8 || 8 | _ | 8
....
4 | 5 | 5 || 4 | 5 | 5 || _ | 4 | 5

and the records with the supposed desired results are

2 | 8 | 8 || _ | 8 | 8 || 8 | _ | 8
....
4 | 5 | 5 || 4 | 5 | _ || _ | 4 | 5


My question is what makes the 8s in D2 and E2 different from the 5s in D9
and E9? That there are two 8s in I2:K2 but only one 5 in I9:K9? If so,

F2:
=IF(AND(ABS(COUNTIF($I2:$K2,C2)/
SUMPRODUCT(COUNTIF($I2:$K2,$C2:$E2))-0.5)<0.5,
COUNTIF($C2:C2,C2)<=COUNTIF($I2:$K2,C2)),C2,"")

Fill F2 into G2:H2, then fill F2:H2 into F3:H15. The results then look like

_ | 8 | 8
1 | _ | 5
_ | _ | _
8 | 6 | _
9 | 0 | 2
_ | _ | _
_ | _ | _
4 | 5 | _
_ | _ | _
_ | _ | _
_ | _ | _
_ | _ | _
_ | _ | _
_ | _ | _