Sort by

recency

|

18 Discussions

|

  • + 0 comments

    This challenge is knowledge gaining and its intresting for all students who are weak in maths. I am student and hire assignment makers for my maths assignment they are professional and affordable at rates.

  • + 0 comments

    Python 3 Solution :

    from math import gcd
    primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
    fac = [1]*51
    for i in range(2, 51):
        fac[i] = fac[i-1]*i
    
    
    class CRT:
        ''' for solving a system of equations : x = a1 (mod n1)
            x = a2 (mod n2) , x = a3 (mod n3) , x = a4 (mod n4) 
            ... . It will give the x mod (LCM(n1,n2,n3,n4,..))
        '''
    
        def __init__(self, num):
            ''' num = number of equations
                a = list of a's 
                n = list of n's
            '''
            self.num = num
            self.a = [1]*num
            self.n = [1]*num
    
        def lcm(self, a, b):
            return (a*b//gcd(a, b))
    
        def GCD(self, a, b):  # solves a*x + b*y == gcd(a,b)
            x = 1
            y = 0
            x1, y1, a1, b1 = 0, 1, a, b
            while(b1):
                q = a1//b1
                x, x1 = x1, x-q*x1
                y, y1 = y1, y-q*y1
                a1, b1 = b1, a1-q*b1
            return(x, y)
    
        def solve2(self, a1, n1, a2, n2):
            ''' solving two equations x = a1 (mod n1) and x = a2 (mod n2) 
                returns the x (mod LCM(a1,a2))
            '''
            x1 = self.GCD(n1, n2)[0]
            d = gcd(n1, n2)
            if (a1-a2) % d != 0:
                return -1
            ans = -x1*(a1-a2)//d*n1+a1
            mod = self.lcm(n1, n2)
            ans = (ans % mod+mod) % mod
            return ans
    
        def solve(self):
            ''' solve the system of equations '''
            ans = self.a[0]
            N = self.n[0]
            for i in range(1, self.num):
                ans = self.solve2(ans, N, self.a[i], self.n[i])
                if ans == -1:
                    return -1
                N = self.lcm(N, self.n[i])
            return ans
    
    
    def comb(m, n, p):
        if m < n:
            return 0
        ans = fac[m]//(fac[n]*fac[m-n])
        ans %= p
        return ans
    
    
    def solve(m, n, M):
        if M == 1:
            return 0
        p = []
        for i in primes:
            if M % i == 0:
                p.append(i)
                while M % i == 0:
                    M //= i
        crt = CRT(len(p))
        crt.n = p
        for ind, i in enumerate(p):
            ans = 1
            m_copy = m
            n_copy = n
            while m_copy or n_copy:
                mm = m_copy % i
                nn = n_copy % i
                m_copy //= i
                n_copy //= i
                crt.a[ind] = (crt.a[ind]*comb(mm, nn, i)) % i
        return crt.solve()
    
    
    for _ in range(int(input())):
        m, n, M = map(int, input().split())
        print(solve(m, n, M))
    
  • + 0 comments

    Hint :

    1- Factorize M . (( M = p1 * p2 * p3 * ... * pr ))

    2 - Using Lucas theorem, compute NcR ( mod p_i ) .

    3 - Now you have r congruent equations. Each in form x = a_i ( mod n_i )

    4- Using CRT, solve them and find x. Actually x is NcR ( mod p1*p2*p3* ... * pr ) .

  • + 0 comments

    This problem will be solved by Lucas Theorem. Let us make your projects professionally with advanced cnc machining. Help from chinese remainder theorem.

  • + 0 comments

    Thanks a lot for sharing the solution of this complex question. It is tricky. I don't know how to solve this puzzle and tried the way you have shared here on solving the problem. It was simple. Thanks for explaining it so simple. http://www.cushioncutengagementring.com/two-stone-diamond-ring-designs/a-look-at-two-stone-rings/