In A5 of Sheet1, I used =INDEX(Equity!$E:$E,127+(COLUMN()-1)*13)
I dragged this across the row and it gave me references to every 13th cell
in the other sheet.
Index for a single column needs INDEX(array, row-value), where row-value is
a number referring to the POSITION not the CELL REFERENCE within the array
Note the -1 is there since I started in column 1; if I start in column E
then I need -5
(The + in your formula is not needed)
best wishes

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Bernard V Liengme
Microsoft Excel MVP
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> Here is the formula that I have inputed; however I am still getting a
> #NUM!
[quoted text clipped - 12 lines]
>
> Thank you for the help!
jordanpcpre - 18 Mar 2008 14:21 GMT
Thank you for the help Bernard. I almost have it.
My current formula in cell IRR!L7:
=INDEX(Equity!$E:$E,127+(COLUMN( )-5)*13)
I am in cell L7 of the IRR tab trying to reference cell E127 in the Equity
tab. Should the above formula reference the Equity!E127 cell correctly? I
would then like to drag this formula one column to the right and have it
reference automatically 13 rows below Equity!E127 (so IRR!M7 should reference
Equity!E140).
I would like to drag this formula to the right as far as necessary.
Thanks! We almost have it!
> In A5 of Sheet1, I used =INDEX(Equity!$E:$E,127+(COLUMN()-1)*13)
> I dragged this across the row and it gave me references to every 13th cell
[quoted text clipped - 23 lines]
> >
> > Thank you for the help!
Bernard Liengme - 18 Mar 2008 18:37 GMT
If the formula is in column L then COLUMN value is 12
So the formula computes to INDEX(Equity!$E:$E,127+(12-5)*13)
Or INDEX(Equity!$E:$E,127+7*13) or INDEX(Equity!$E:$E, 218*13)
You want INDEX(Equity!$E:$E,127+(COLUMN()-12)*13) to get
INDEX(Equity!$E:$E,127+0*13) or
INDEX(Equity!$E:$E,127)
best wishes

Signature
Bernard V Liengme
Microsoft Excel MVP
http://people.stfx.ca/bliengme
remove caps from email
> Thank you for the help Bernard. I almost have it.
>
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>> >
>> > Thank you for the help!