Dear all,
Don't know why when I input the numeric 179510085292681233, it always rounds
up to 179510085292671000.
I'm using Excel 2002 SP3 version.


Kent

Excel works to 15 significant figures, so any *number* will be rounded to 15
figures. If you need more digits, enter as text, either by formatting the
cell as text before you enter the number, or by preceding the number with an
apostrophe.
--
David Biddulph

"Kent" wrote in message
...
Dear all,
Don't know why when I input the numeric 179510085292681233, it always
rounds up to 179510085292671000.
I'm using Excel 2002 SP3 version.


Kent

"David Biddulph" <groups [at] biddulph.org.uk wrote:
If you need more digits, enter as text

.....If you intend to use the "number" as text, for example a credit card
identifier.

If you enter 179510085292681233 as text, then use it in an arithmetic
expression, you will not get any more precision.


Kent wrote:
Don't know why when I input the numeric 179510085292681233,
it always rounds up to 179510085292671000.
I'm using Excel 2002 SP3 version.

That is odd. I cannot speak for Excel 2002 SP3, but....

First, Excel 2003 SP3 __truncates__, not rounds, the first 15 significant
digits when entered manually. If I manually type 12345678901234567890, the
resulting value is the same as the constant 12345678901234500000.

Second, the first 15 significant digits of 179510085292681233 are
179510085292681. So the resulting value should be the same as the constant
179510085292681000, regardless of whether Excel 2002 SP3 rounds or truncates
the first 15 significant digits.

Oh, perhaps Kent simply has a typo, and he is misusing the word "round", as
so many people do in these forums.


----- original message -----

"David Biddulph" <groups [at] biddulph.org.uk wrote in message
...
Excel works to 15 significant figures, so any *number* will be rounded to
15 figures. If you need more digits, enter as text, either by formatting
the cell as text before you enter the number, or by preceding the number
with an apostrophe.
--
David Biddulph

"Kent" wrote in message
...
Dear all,
Don't know why when I input the numeric 179510085292681233, it always
rounds up to 179510085292671000.
I'm using Excel 2002 SP3 version.


Kent

I see, thank you David & Joe

Kent




"Joe User" <joeu2004 ...
"David Biddulph" <groups [at] biddulph.org.uk wrote:
If you need more digits, enter as text

....If you intend to use the "number" as text, for example a credit card
identifier.

If you enter 179510085292681233 as text, then use it in an arithmetic
expression, you will not get any more precision.


Kent wrote:
Don't know why when I input the numeric 179510085292681233,
it always rounds up to 179510085292671000.
I'm using Excel 2002 SP3 version.

That is odd. I cannot speak for Excel 2002 SP3, but....

First, Excel 2003 SP3 __truncates__, not rounds, the first 15 significant
digits when entered manually. If I manually type 12345678901234567890,
the resulting value is the same as the constant 12345678901234500000.

Second, the first 15 significant digits of 179510085292681233 are
179510085292681. So the resulting value should be the same as the
constant 179510085292681000, regardless of whether Excel 2002 SP3 rounds
or truncates the first 15 significant digits.

Oh, perhaps Kent simply has a typo, and he is misusing the word "round",
as so many people do in these forums.


----- original message -----

"David Biddulph" <groups [at] biddulph.org.uk wrote in message
...
Excel works to 15 significant figures, so any *number* will be rounded to
15 figures. If you need more digits, enter as text, either by formatting
the cell as text before you enter the number, or by preceding the number
with an apostrophe.
--
David Biddulph

"Kent" wrote in message
...
Dear all,
Don't know why when I input the numeric 179510085292681233, it always
rounds up to 179510085292671000.
I'm using Excel 2002 SP3 version.


Kent