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Asked: 2026-06-11T00:41:52+00:00

we know that (A + B) % P = (A % P + B

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we know that

(A + B) % P = (A % P + B % P) % P
(A * B) % P = (A % P * B % P) % P

where P is a prime .

I need to calculate (A / B) % P where A,B can be very large and can overflow .

Does such kind of formula for modular arithmetic holds for (A / B) % P and (A - B) % P.

If not then please explain what the correct answer is.

I.e is it true that (A / B) % P = ((A % P) / (B % P)) % P?

I WAS TRYING TO CALULATE (N*(N^2+5)/6)%P where N can be as large as 10^15

here A=n*(n^2+5) can surely overflow for n=10^15

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1 Answer

  1. Editorial Team

    Yes, but it’s different:

    (a - b) mod p = ((a mod p - b mod p) + p) mod p

(a / b) mod p = ((a mod p) * (b^(-1) mod p)) mod p

    Where b^(-1) mod p is the modular inverse of b mod p. For p = prime, b^(-1) mod p = b^(p - 2) mod p.

    Edit:

    (N*(N^2+5)/6)%P

    You don’t need any modular inverses from this. Just simplify the fraction: N or N^2+5 will be divisible by 2 and 3. So divide them and then you have (a*b) mod P.

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