Suppose the number is in A1
Then
=ROUND(A1,0) will give you the integer part
= (A1 - ROUND(A1,0)) will give you the decimal part
Multiply this by 100 and round again as in step 1

You can use ROUNDDOWN() if you do not want rounding instead of ROUND()
Let me if this is what you were looking for

"Mike H" wrote:

Hi,

Yesterday a question was posted in which the OP wanted to take the number
101.25 and extract the decimal 0.25 and convert that into an integer 25.

A one off solution is simple, for example
=MID(D1,3,LEN(D1))+0
or
=(A1-(TRUNC(A1)))*100
and of course a modulus/multiplication solution.

But none of these are generic for longer decimals so I set out to find a
generic solution for any number length.

This works perfectly for 101.25 and for any number to the left of the
decimal point
=($A$1-TRUNC($A$1))*(10^(LEN(($A$1-TRUNC($A$1)))-2))

But add any extra decimals and it can fail. For example 100.256 works
perfectly because (i think) it must have a precise decimal/binary conversion
but 101.257 falls over because the decimal portion is actually
0.257000000000005 so my formula that raises the (number*10^length of decimal
bit) fails.

I am missing something blindingly obvious so can anyone help me with a
mathmatical (not a text fiddle) to this problem that will convert any number
irrespective of the number of decimals. I haven't tried 'precision as
displayed' because I instinctively don't like it

Mike