# Trick for modular division ( (x1 * x2 …. xn) / b ) mod (m)

Given integers x1, x2, x3……xn, b and m, we are supposed to find the result of ((x1*x2….xn)/b)mod(m).

Example 1 :
Suppose that we are required to find (55C5)%(1000000007) i.e ((55*54*53*52*51)/120)%1000000007

Naive Method :

1. Simply calculate the product (55*54*53*52*51)= say x,
2. Divide x by 120 and then take its modulus with 1000000007

Using Modular Multiplicative Inverse :
Above method will work only when x1, x2, x3….xn have small values.
Suppose we are required to find the result where x1, x2, ….xn fall in the range of ~1000000(10^6). So we will have to exploit the rule of modular mathematics which says :
(a*b)mod(m)=(a(mod(m))*b(mod(m)))mod(m)

Note that the above formula is valid for modular multiplication. Similar formula for division does not exist.
i.e (a/b)mod(m) != a(mod(m))/b(mod(m))

1. So we are required to find out modular multiplicative inverse of b say i and then multiply ‘i’ with a .
2. After this we will have to take the modulus of the result obtained.
i.e ((x1*x2….xn)/b)mod(m)=((x1*x2….xn)*i)mod(m)= ((x1)mod(m) * (x2)mod(m) *…. (xn)mod(m) * (i)mod(m))mod(m)

Note : To find modular multiplicative inverse we can use Extended Eucledian algorithm or Fermat’s Little Theorem.

Example 2 : Let us suppose that we have to find (55555C5)%(1000000007) i.e ((55555*55554*55553*55552*55551)/120)%1000000007.

 `// To run this code, we need to copy modular inverse  ` `// from below post. ` `// https://www.geeksforgeeks.org/multiplicative-inverse-under-modulo-m/ ` ` `  `int` `main() ` `{ ` `    ``// naive method-calculating the result in a single line ` `    ``long` `int` `naive_answer = ((``long` `int``)(55555 * 55554 *  ` `                ``55553 * 55552 * 55551) / 120) % 1000000007;  ` ` `  ` `  `    ``long` `int` `ans = 1; ` ` `  `    ``// modular_inverse() is a user defined function ` `    ``// that calculates inverse of a number ` `    ``long` `int` `i = modular_inverse(120, 10000007); ` ` `  `    ``// it will use extended Eucledian algorithm or ` `    ``// Fermat’s Little Theorem for calculation. ` `    ``// MMI of 120 under division by 1000000007 ` `    ``// will be 808333339 ` `    ``for` `(``int` `i = 0; i < 5; i++)  ` `        ``ans = (ans * (55555 - i)) % 1000000007; ` `     `  `    ``ans = (ans * i) % 1000000007; ` `    ``cout << ``"Answer using naive method: "` `<< naive_answer << endl; ` `    ``cout << ``"Answer using multiplicative"` `         ``<< ``" modular inverse concept: "` `<< ans; ` ` `  `    ``return` `0; ` `} `

```Input :
Output :Answer using naive method: -5973653
Answer using multiplicative modular inverse concept: 300820513
```

It is clear from the above example that the naive method will lead to overflow of data resulting in incorrect answer. Moreover, using modular inverse will give us the correct answer.

Without Using Modular Multiplicative Inverse :
But it is interesting to note that a slight change in code will discard the use of finding modular multiplicative inverse.

## C++

 `#include ` `using` `namespace` `std; ` `int` `main() ` `{ ` `    ``long` `int` `ans = 1; ` `    ``long` `int` `mod = (``long` `int``)1000000007 * 120; ` `    ``for` `(``int` `i = 0; i < 5; i++)  ` `        ``ans = (ans * (55555 - i)) % mod;     ` `    ``ans = ans / 120; ` `    ``cout << ``"Answer using shortcut: "` `<< ans; ` `    ``return` `0; ` `} `

## Java

 `class` `GFG { ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``long` `ans = ``1``; ` `        ``long` `mod = (``long``)``1000000007` `* ``120``; ` `         `  `        ``for` `(``int` `i = ``0``; i < ``5``; i++)  ` `            ``ans = (ans * (``55555` `- i)) % mod;  ` `             `  `        ``ans = ans / ``120``; ` `        ``System.out.println(``"Answer using"` `                    ``+ ``" shortcut: "``+ ans); ` `    ``} ` `} ` ` `  `// This code is contributed by smitha. `

## Python3

 `ans ``=` `1` `mod ``=` `1000000007` `*` `120` ` `  `for` `i ``in` `range``(``0``, ``5``) : ` `    ``ans ``=` `(ans ``*` `(``55555` `-` `i)) ``%` `mod  ` `     `  `ans ``=` `int``(ans ``/` `120``) ` ` `  `print``(``"Answer using shortcut: "``, ans) ` ` `  `# This code is contributed by Smitha. `

## C#

 `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``long` `ans = 1; ` `        ``long` `mod = (``long``)1000000007 * 120; ` `         `  `        ``for` `(``int` `i = 0; i < 5; i++)  ` `            ``ans = (ans * (55555 - i)) % mod;  ` `             `  `        ``ans = ans / 120; ` `        ``Console.Write(``"Answer using "` `                    ``+ ``"shortcut: "``+ ans); ` `    ``} ` `} ` ` `  `// This code is contributed by smitha. `

## PHP

 ` `

Output :

```Answer using shortcut: 300820513
```

Why did it work?
This will work only in case when the denominator is a factor of numerator i.e. when a % b = 0 following the rule:
If b | a, then we can write (a/b) % p as (a % p*b)/b.
This rule proves useful for small values of b.

Let us consider a = x1*x2*x3…….xn
We have to find ans = (a/b)%1000000007

1. Let result of a%(1000000007*b) be y. To avoid overflow, we use modular multiplicative property. This can be represented as
a = (1000000007*b)q + y where y < (1000000007*b) and q is an integer
2. Now dividing LHS and RHS by b, we get
y/b = a/b -(1000000007*b)*q/b
= a/b -1000000007*q < 1000000007 (From 1)
Therefore, y/b is equivalent to (a/b)mod(b*1000000007). 🙂 My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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