**Please Note: **
This article is written for users of the following Microsoft Excel versions: 2007, 2010, 2013, 2016, 2019, and Excel in Office 365. If you are using an earlier version (Excel 2003 or earlier), *this tip may not work for you*. For a version of this tip written specifically for earlier versions of Excel, click here: Using an Exact Number of Digits.

Henk asked if there is a way in Excel to display a number using six digits, independent of the placement of the decimal point. For instance, 0.1 would be displayed as 0.10000, 200 would be displayed as 200.000, and 25000 would be displayed as 25000.0.

Unfortunately, there is no formatting that will do the trick; all display formatting seems to be dependent on the position of the decimal point. You can format a display for a specific number of digits after the decimal point, but that number of digits will be used regardless of how many digits appear before the decimal point.

Several *ExcelTips* subscribers came up with suggestions that involve using formulas to display the number as desired. For instance, the following formula will display the value in A1 using six digits:

=FIXED(A1,IF(ABS(A1)<1,5,5-INT(LOG(ABS(A1)))),TRUE)

Other readers provided formulas that relied on converting the number to a text string and displaying it as such. Converting a number to its textual equivalent, however, has the distinct drawback of no longer being able to use the number in other formulas. (Remember—it is text at this point, not a number.) The above formula does not have that limitation.

If you wanted to, you could also use a macro to set the formatting within a cell that contains a value. The advantage to such a macro is that you don't have to use a cell for a formula, as shown above. The drawback to a macro is that you need to remember to run it on the cells whenever values within them change. The following macro is an example of such an approach:

Sub SetFigures() Dim iDecimals As Integer Dim bCommas As Boolean Dim sFormat As String Dim CellRange As Range Dim TestCell As Range bCommas = False 'Change as desired Set CellRange = Selection For Each TestCell In CellRange If Abs(TestCell.Value) < 1 Then iDecimals = 5 Else iDecimals = 5 - Int(Log(Abs(TestCell.Value)) / Log(10#)) End If sFormat = "0" If bCommas Then sFormat = "#,##0" If iDecimals < 0 Then sFormat = "General" If iDecimals > 0 Then sFormat = sFormat & _ "." & String(iDecimals, "0") TestCell.NumberFormat = sFormat Next TestCell End Sub

In order to use the macro, simply select the cells you want to format, then execute it. Each cell in the range you selected is set to display six digits, unless the number in the cell is too large or too small.

*Note:*

If you would like to know how to use the macros described on this page (or on any other page on the *ExcelTips* sites), I've prepared a special page that includes helpful information. Click here to open that special page in a new browser tab.

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This tip (10920) applies to Microsoft Excel 2007, 2010, 2013, 2016, 2019, and Excel in Office 365. You can find a version of this tip for the older menu interface of Excel here: **Using an Exact Number of Digits**.

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2015-05-13 16:41:58

patman

You have '1' in A1, '2' in A2, etc, you want it to be '000001', '000002', etc on down the column. Manually type the extra 0s in those cells, highlight them and drag them with the bottom-right handle down the column over the numbers. The numbers are now all 6-digits with leading 0s.

2015-01-16 10:13:33

Dennis Costello

4656.25 * 2510.35 = 11,688,817.1875, which on my calculator rounds to .19

2014-12-24 04:07:26

Naba Raj Bagale

2510.35

= 11688817.19

Rather the exact number should be 11688817.18

I am terrified with this case. Can you give me solution please.

2014-12-01 04:52:02

Pete

=FIXED(A1,5-INT(LOG(ABS(A1))),TRUE)

otherwise the formula suggest above

will yeald wrong results with numbers

smaller than 1 (eg. 0.0012345678)

2014-11-29 20:18:19

Bill Macky

'aaa is value, b is figures displayed,c is 1 for commas

Function MyFixed(aaa, b, c)

MyFixed = Application.WorksheetFunction.Fixed(aaa, b - Int(Log(Abs(aaa)) / Log(10)) - 1, c)

End Function

Seems to work. Actually something I've wanted for a while.

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