Question

How to calculate the number of digits in a multiplication?

Answer 1

You can take log10 of each of the numbers being multipled, sum them, floor them, then add one to get the number of digits.

i.e. In your last example of 2*12321*1000, which is actually equal to 24642000 (you missed a 0, so it has 8 digits).

Number of Digits = ⌊log10(2)+log10(12321)+log10(1000)⌋+1=8=⌊log10(2∗12321∗1000)⌋+1

I'll begin explaining why this works with a simple observation: You can calculate the number of any power of 10 by simply taking the base-10 logarithm of it, and adding one.

For example log101=log10100=0.

log10102=log10100=2, etc.

So by adding one to each of these two examples above, we get the correct number of digits. This is just an artifact of the fact that we use base-10 to count. If we counted in base-2, we'd take log2, then add one.

Now, why do we have to floor the number and add one for any general number?

The number 12321 can be thought of as 12321=104∗1.2321, and since it has the same number of digits as 104, the extra ∗1.2321 term should be "ignored" somehow.

Since multiplications in normal space become addition after you take the logarithm, we get:

log1012321=log10(104∗1.2321)=log10104+log101.2321=4+log101.2321

Since we chose to round down to the nearest power of 10, the number we multiply 10i by will always be in the interval [1,10), and any number in this interval will have satisfy 0<log10r<1 - so the reason we floor it is just to remove this "remainder".

The final step is just to add one, as I explained above.

hope this helps...