MOD

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Guest

I have a question on the funtion "MOD". When the number to be divided is more
than nine digits (like: 1,234,457,890), the funtion would return error. How
can I fix this?

Thanks.
 
The things you learn on this forum.
Steve


I have a question on the funtion "MOD". When the number to be divided is
more
than nine digits (like: 1,234,457,890), the funtion would return error..
How
can I fix this?

Thanks.
Click to expand...
 
The number of digits should not be a problem. Excel supports up to 15
digits of precision and it will accept longer representations even if
it truncates their precision to 15 digits.

On the other hand, using a number literal as formatted (i.e. with the
commas as thousand separator) WILL cause an error because commas are
confused with the argument separator.

Try entering the number in a different cell (say A1) and then using
=MOD(A1,something). You should not get an error.

HTH
Kostis Vezerides
 
Kostis,

that's what I thought and tried it - MOD(123456789,2) returns 1 as
expected, but MOD(1234567890,2) returns #NUM!, even if the number is in
a different cell.

I haven't come across this before.

Pete
 
Hmmmm,
now I understand the meaning of the second message in the thread. Well,
this is news to me too! On the other hand, the following formula:

=A1-FLOOR(A1,2)

will return the same as MOD(A1,2) and it does not suffer from this
problem, although it would be good only for positive numbers in A1.

Microsoft suggest another identity, which I am sure will also work:

MOD(n, d) = n - d*INT(n/d)

HTH
Kostis
 
Perhaps this Microsoft Knowledge Base article explains:

http://support.microsoft.com/kb/119083/EN-US/

"If the divisor argument, multiplied by 134,217,728 (or 2 raised to
the 27th power), is less than or equal to the number argument,
the MOD() function returns a #NUM! error value.

HTH,
--
Kevin James.
Tua'r Goleuni


|I have a question on the funtion "MOD". When the number to be divided is more
| than nine digits (like: 1,234,457,890), the funtion would return error. How
| can I fix this?
|
| Thanks.
 
Thank you for your method here.

Unfortunately, I have to deal with numbers more than 15 digits (such as 17
digits). Does it mean that I have no ways to manage such problems (if the
number is more than 15 digits)?

Thanks for your concern.



“vezeridâ€ç¼–写：
 
Never mind MOD
Excel doesn't let you have more than 15 significant digits

Steve
 
Excel's documented limit for numeric display is 15 digits. This is based on
the limitations of the IEEE standard for double precision representation of
numbers. Most computer software follows this standard, so if you need to do
math with 17 digit numbers, you will need to be very particular in your
choice of software. If you merely need to display 17 digits (e.g. account
numbers), then make them text (prefix with an apostrophe) instead of numeric.

Jerry
 
Steve,

You don't have to archive. Google does it for you. I use Google, newsgroups etc. to retrieve posts from years ago. I do it often. Info is still valuable.

Epinn

I don't have enought disc space to keep that much backposts :)
Steve
 
:) You must read the :)

Seriously I don't post or read via google - newsreaders work well for me..
I agree it should be a prerequisite before 90% of the posters on here post
1. check archives
2. write the question in brief, but with all relevant information
3. read the question again before posting
4....
No. I'll stop there.

I did try it out for *excel mod problem* and got some tech discussion on
the 2^27
problem in the first 10 hits

I have and would usually use these to get a bit more background on a
problem

Steve
 
J.H. wrote...
I have a question on the funtion "MOD". When the number to be divided is more
than nine digits (like: 1,234,457,890), the funtion would return error. How
can I fix this?
Click to expand...

As others have pointed out, Excel's MOD function is flawed compared to
what the hardware is capable of calculating. Instead of MOD(a,b) calls,
use (a-b*int(a/b)).
 
Unfortunately, I have to deal with numbers more than 15 digits (such as 17
digits). Does it mean that I have no ways to manage such problems (if the
number is more than 15 digits)?
Click to expand...


Hi. Don't know what you have set up, so this is real general...
Suppose you had the following text number with 21 digits in A1.
'123456789012345678901

=MyMod(A1,7)
returns the correct answer of 4.

Function MyMod(n, m)
MyMod = (CDec(n) - m * Int(CDec(n) / m))
End Function

Again, very general, but may give you some ideas if you do in fact
have numbers as text over 15 digits.

HTH
Dana DeLouis
 
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