finding binomial co-effecient modulo prime number,Interview street challenge

Question

I have done a lot of work on this but couldnt find the answer for larger test cases

Problem statement

In mathematics, binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. C(n,k) denotes the number of ways of choosing k objects from n different objects.

However when n and k are too large, we often save them after modulo operation by a prime number P. Please calculate how many binomial coefficients of n become to 0 after modulo by P.

Input

The first of input is an integer T , the number of test cases.

Each of the following T lines contains 2 integers, n and prime P.

Output

For each test case, output a line contains the number of \tbinom nks (0<=k<=n) each of which after modulo operation by P is 0.

Sample Input

3
2 2
3 2
4 3

Sample Output

1
0
1

Constraints:

• T is less than 100
• n is less than 10^500.
• P is less than 10^9.

Attemted solution

i have completed this by using remainder theorem in binomial coeffecients

         #C(n,m) mod p
         #n=l*p+t
         #m=m*p+s
         #then C(n,r) mod p=0 when (t<s or t=0)

         inp1 = raw_input()
         N=int(inp1)
         result=[]
         for i in range(N):
            inp = raw_input()
            a, b = [long(x) for x in inp.split(' ')]
           j=0
           #n=lp+t
           t=a%b
           t=t+1
           l=a/b
           j=l*(b-t)
           result.append(j)
        for i in range(N):
           print result[i]

above condition is satisfied for small numbers

sample test case

n=18794630773460178101742670493883191390743597826923079533199667903991430393463990462500322752062869664969026409174076912867222446746310051635510258172105034070506806228555577773599018819952185016092141574603857551738968553782672643049704163674318579695215402562964641111900657274032612661770435202254364177910753450214277150377049334509093906874400306682949871260040370515062243982543271073443613028133844603853807066311479739789908983610180228625059956919930500586048799830730348503994503184106117058

p= 177080341

my output is

2296508200406431043037217853758906667313789305876262916681342008001095232916608835588093082038358975456171743147798282523487485386336553910277635713985851142371010771392102277877640275384064735398164190968282400640398659343189330639672472613876688344609533340662724884373078840434716174167873375700938597411315754265893890741730938202927739246687866166994143001482839656000969674716959277820008958538229366474207686428750934149471409162993083541475267772950721250234982686128039722553219836725588488

expected output is

18794630773460178101742635959946548665553041135822283621364103266511586625905046107130878283695016799933475657268010472422112556606280021574002866456544310584537519228161286450725015989697306855581489155139723025246780552510467580791551824827637581156204185887378181074365453150481221030356075255000460025095384537510111086396988416046942446776262625161326885418101128327416784858513888616089287333560469336094431461981368825028447505354473183546488856594449627370807707483671453574074503184106117059

Answer 1

You can look at it from the other end: How many nCr are not divisible by p? There's a rather simple formula for that.

Preliminaries:

The binomial coefficient nCr is given by

nCr = n! / (r! * (n-r)!)

so the multiplicity v_p(nCr) of p in nCr - the exponent of p in the prime factorisation of nCr - is

v_p(nCr) = v_p(n!) - v_p(r!) - v_p((n-r)!)

The multiplicity of p in n! can be easily determined, a well known way to calculate it is

v_p(n!) =  ∑ floor(n/p^k)
         k > 0

If you look at this formula considering the base-p expansion of n, you can see that

v_p(n!) = (n - σ_p(n)) / (p - 1)

where σ_p(k) is the sum of the digits of the base-p representation of k. Thus

v_p(nCr) = (n - σ_p(n) - r + σ_p(r) - (n-r) + σ_p(n-r)) / (p - 1)
         = (σ_p(r) + σ_p(n-r) - σ_p(n)) / (p - 1)

Conclusion:

nCr is not divisible by the prime p if and only if the addition of r and n-r has no carry in base p.

Because that is exactly when σ_p(a + b) = σ_p(a) + σ_p(b). A carry in the addition occurs when the sum of corresponding digits of a and b (plus possibly the carry-in if there was already a carry produced for less significant digits) is >= p, then the corresponding digit in the sum is reduced by p and the next more significant digit increased by 1, reducing the digital sum by p - 1.

We have a carry-free addition n = r + (n-r) in base p if for each digit d_k in n's base-p expansion, the corresponding digit of r is less than or equal to d_k. The admissible choices of the digits of r are independent, hence the total number is the product of the count of choices for the individual digits.

The number of nCr not divisible by the prime p is

ND(n,p) = ∏(d_k + 1)

where the d_k are the digits in n's base p expansion,

    m
n = ∑ d_k * p^k
   k=0

Since there are n+1 (nonzero) binomial coefficients nCr for a given n, the number of (nonzero) binomial coefficients nCr divisible by p is

        m
n + 1 - ∏ (d_k + 1)
       k=0

with the above base p expansion of n.

Using Michael's example n = 10, p = 3,

10 = 1*3^0 + 0*3^1 + 1*3^2

so there are (1+1)*(0+1)*(1+1) = 4 coefficients 10Cr not divisible by 3 and 10 + 1 - 4 = 7 divisible by 3.

def divisibleCoefficients(n,p):
    m, nondiv = n, 1
    while m > 0:
        nondiv = nondiv * ((m % p) + 1)
        m = m // p
    return (n + 1 - nondiv)