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# Multiplying multiple cells of data at once

I am trying to multipy 150 individual cells by the same constant, 2.2

I am able to do it one cell at a time but can not figure out how to highlight 100+ cells and do them all at the same time. I'm a newbie, so thanks for a simple lesson?

4 Replies

# Re: Multiplying multiple cells of data at once

Depends how your data is structured. Let assume it's in A1:A100. In B1 enter the formula =A1*2.2

After when stay on B1 double click the dot on the border around the cell at bottom right.

Alternatively after entering the formula select B1:B100 and Ctrl+D.

# Re: Multiplying multiple cells of data at once

When I highlight one cell, the formula bar shows: 617.554245
The 617 is the date, in this case the cost. I can type in the formula line =617.554245*2.2 and when I hit return the result is correct. When I try and highlight a number of cells, that doesn't work?

# Re: Multiplying multiple cells of data at once

@Segalassoc  wrote: ``I am trying to multipy 150 individual cells by the same constant, 2.2``. And later: ``I can type in the formula line =617.554245*2.2 and when I hit return the result is correct. When I try and highlight a number of cells, that doesn't work``.

I wonder if the following is what you want to do.

1. Enter 2.2 into any cell.  Then select that cell, and press ctrl+c to copy.

2. Select the 150 cells (A1:A150), right-click and click Paste Special, select Multiply and click OK.

(If the cells are not a contiguous range, you can use ctrl+Click to select them.)

If you had 617.554245 in a cell, the cell value will become 1358.619339, not the formula =617.554245*2.2.

Is that acceptable?  Or do you insist on replacing the constant 617.554245 with the formula.

# Re: Multiplying multiple cells of data at once

First step

Double click on dot at the bottom right of B1. Result is

Another way - after first step select range in column B

Press Ctrl+D. Result will be the same.

If you'd like multiply numbers in column A substituting them with the result, when as @Joe User  advised.