This should be simple for the pros but I've spent hours researching and can
not come up with a solution. Lets say you have 3 columns A, B & C. 10 rows
of numbers in Column A and 10 rows of numbers in column B, what is the
formula for A/B to appear in column C for all ten rows. In actuality, I have
over 1000 rows of data and I dont want to enter =SUM(A1/B1) in each row of
column C. What is the formula that will automatically compute A/B for many
rows to appear in C. Your help will be greatly appreciated.

In C1 enter
=Sum(A1/B1)

then reselect the cell.
in the lower right corner of the highlight, you should see a small square.
Double click on it and it should fill down.
--
Regards,
Tom Ogilvy


" wrote in message
...
This should be simple for the pros but I've spent hours researching and
can
not come up with a solution. Lets say you have 3 columns A, B & C. 10 rows
of numbers in Column A and 10 rows of numbers in column B, what is the
formula for A/B to appear in column C for all ten rows. In actuality, I
have
over 1000 rows of data and I dont want to enter =SUM(A1/B1) in each row
of
column C. What is the formula that will automatically compute A/B for many
rows to appear in C. Your help will be greatly appreciated.

Worked perfectly. Thanks Tom for the help!!!!!!!!!!!!


"Tom Ogilvy" wrote in message
...
In C1 enter
=Sum(A1/B1)

then reselect the cell.
in the lower right corner of the highlight, you should see a small square.
Double click on it and it should fill down.
--
Regards,
Tom Ogilvy


" wrote in message
...
This should be simple for the pros but I've spent hours researching and
can
not come up with a solution. Lets say you have 3 columns A, B & C. 10
rows
of numbers in Column A and 10 rows of numbers in column B, what is the
formula for A/B to appear in column C for all ten rows. In actuality, I
have
over 1000 rows of data and I dont want to enter =SUM(A1/B1) in each row
of
column C. What is the formula that will automatically compute A/B for
many
rows to appear in C. Your help will be greatly appreciated.