Question

Question is about the modulo operator on very large numbers.

For example consider a question where the total number of permutations are to be calculated.
Consider a number of 90 digits with each of the 9 numbers (1 to 9) repeating 10 times
so 90!/(10!)^9) is to be calculated

After reading many answers on StackOverflow I used logarithms to do it.

Now consider the log value to be 1923.32877864.

Now my question is how can I display the answer (i.e. 10 ^ log10(value) ) modulo of “m”?

And is this the best method for calculating the possible number of permutations?

Edit
Got the solution 🙂

Thanks to duedl0r.

Did it the way you specified using Modular Multiplicative Inverse.Thanks 🙂

Answer 1

I’m not sure whether this is actually possible and correct, but let me summarize my comments and extend the answer from Miky Dinescu.

As Miky already wrote:

a × b ≣_m a_m × b_m

You can use this in your equality:

90! / 10!^9 ≣_m x

Calculate each term:

90!_m / 10!^9_m ≣_m x

Then find out your multiplicative inverse from 10!^9_m. Then multiplicate the inverse with 90!_m.

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update
This seems to be correct (at least for this case :)). I checked with wolfram:

(90!/10!^9) mod (10^9+7) = 998551163

This leads to the same result:

90! mod (10^9+7) = 749079870
10!^9 mod (10^9+7) = 220052161

do the inverse:

(220052161 * x) mod(10^9+7) = 1 = 23963055

then:

(749079870*23963055) mod (10^9+7) = 998551163

No proof, but some evidence that it might work 🙂