INT Function

D

Dan (WA)

I need the remainder results of a multiplication formula. The Function INT
should provide this results -- Excel Function Help for INT provide the
formula =A2-INT . I have tried this sever times but the results is alway the
same #Name... Is excel broken or what am i missing?
 
D

David Billigmeier

The INT() function truncates a decimal number down to it's integer part (i.e.
=INT(9.9 will return 9)

This doesn't sound like the function you want to use if you're looking for
remainders. Try using MOD()
 
F

Fred Smith

You need to supply a parameter to the INT function. To get the portion to the
right of the decimal, use:

=a2-int(a2)

You can also get the same thing in one step with the Mod function:

=mod(a2,1)
 
H

Harlan Grove

Fred Smith wrote...
You need to supply a parameter to the INT function. To get the portion to the
right of the decimal, use:

=a2-int(a2)

You can also get the same thing in one step with the Mod function:

=mod(a2,1)
Click to expand...
....

Use the INT formula. MOD is broken when quotients are large, e.g.,

=2^30/3-INT(2^30/3)

correctly returns 1/3 (approximately, accurate to only 7 decimal
digits), while

=MOD(2^30/3,1)

returns #NUM!.
 
R

Ron Rosenfeld

I need the remainder results of a multiplication formula. The Function INT
should provide this results -- Excel Function Help for INT provide the
formula =A2-INT . I have tried this sever times but the results is alway the
same #Name... Is excel broken or what am i missing?
Click to expand...

Actually, the formula in HELP is

=A2-INT(A2)

If you type in what you posted above

=A2-INT

you will get a #NAME? error.


An alternative formula, giving the same result, would be:

=MOD(A2,1)




--ron
 
R

Ron Rosenfeld

Actually, the formula in HELP is

=A2-INT(A2)

If you type in what you posted above

=A2-INT

you will get a #NAME? error.


An alternative formula, giving the same result, would be:

=MOD(A2,1)




--ron
Click to expand...

However, as Harlan just pointed out (and which he has pointed out in the past,
but which I have forgotten), MOD does not work with large numbers, so INT would
be safer.


--ron
 
D

Dan (WA)

Thanks -- To all those who Help me see the errors of my ways.. So simple just
needed a kick start..
Dan
 
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