Can excel save me money?
January 10, 2011 3:55 PM Subscribe
In Excel (or Google Docs) is there away to divide the value in a cell by the number of cells selected? So if I have a value of x and I then highlight 10 cells, each of the 10 cells would show x divided by 10. As I select less cells, the value in each cell increases because x is being divided by less. The main reason I want to do this is to show how much I'd need to save per month to reach a specific total.
Presumably, you want the ROWS() or COLUMNS() function?
posted by schmod at 4:12 PM on January 10, 2011
posted by schmod at 4:12 PM on January 10, 2011
In Excel:
When you type
posted by fleeba at 4:22 PM on January 10, 2011 [1 favorite]
=D1/COUNT(A1:C1)When you type
=COUNT(, you can select the range of cells. D1 is the x you mention.posted by fleeba at 4:22 PM on January 10, 2011 [1 favorite]
Do you have Excel 2007? The status bar at the bottom of the window will default (I think?) to showing the average, count, and sum of the highlighted cells. If it doesn't, right click on that bar and click "Average".
posted by mhum at 4:56 PM on January 10, 2011
posted by mhum at 4:56 PM on January 10, 2011
Oh wait. Nevermind. I didn't read your question carefully enough.
posted by mhum at 4:57 PM on January 10, 2011
posted by mhum at 4:57 PM on January 10, 2011
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However, it would be very, very simple to create an equasion that referenced the values in one or more cells--instead of highlighting something, the contents of those cells would be used in the equation.
So cell A1 would be a number--let's say X. A2 would be another number--what you will be dividing by. A3 would be the result. Type 10 into A2, and X/10 would appear in A3. Replace 10 with 11, and X/11 would appear in A3. Continue changing the amount in A2 until you find the number you want.
Is that what you're looking for? To accomplish that, simply type the follwoing into A3:
=A1/A2
I don't really see how counting the number of highlighted cells makes intuitive sense as a way to tackle the problem, though, so I'm probably missing something about your question.
posted by jsturgill at 4:06 PM on January 10, 2011 [1 favorite]