# Modular multiplicative inverse

Your task is to given two integer numbers, a and b calculate the modular multiplicative inverse of a modulo b, if it exists.

The modular inverse of a modulo b is a number c such that ac ≡ 1 (mod b). This number is unique modulo b for any pair of a and b. It exists only if the greatest common divisor of a and b is 1.

## Input and Output

Input is given as either two integers or a list of two integers. Your program should output either a single number, the modular multiplicative inverse that is in the interval 0 < c < b, or a value indicating there is no inverse. The value can be anything, except a number in the range (0,b), and may also be an exception. The value should however be the same for cases in which there is no inverse.

0 < a < b can be assumed

## Rules

• The program should finish at some point, and should solve each test case in less than 60 seconds
• Standard loopholes apply

## Test cases

Test cases below are given in the format, a, b -> output

1, 2 -> 1
3, 6 -> Does not exist
7, 87 -> 25
25, 87 -> 7
2, 91 -> 46
13, 91 -> Does not exist
19, 1212393831 -> 701912218
31, 73714876143 -> 45180085378
3, 73714876143 -> Does not exist


# Scoring

This is code golf, so the shortest code for each language wins.

This and this are similar questions, but both ask for specific situations.

• It follows from Fermat's Little Theorem that the multiplicative inverse of a, if it exists, can be computed efficiently as a^(phi(b)-1) mod b, where phi is Euler's totient function: phi(p0^k0 * p1^k1 * ...) = (p0-1) * p0^(k0-1) * (p1-1) * p1^(k1-1) * ... Not saying it leads to shorter code :)
– ngn
Aug 24, 2017 at 16:08
• @Jenny_mathy Taking additional input is generally disallowed. Aug 24, 2017 at 16:32
• I count six answers that seem to be brute forcing, and unlikely to run all test cases in 60 seconds (some of them give a stack or memory error first). Aug 24, 2017 at 23:28
• @ngn : You've conflated Fermat's Little Theorem (FLT) with Euler's improvement to it. Fermat did not know about the Euler phi function. Further, FLT and Euler's improvement only apply if gcd(a,b) = 1. Finally, in the form you have written it, "a^(\phi(b)-1) mod b" is congruent to 1, not a^(-1). To get a^(-1), use a^(\phi(b)-2) mod b. Aug 25, 2017 at 4:49
• @EricTowers Euler's is a consequence. Regarding "gcd(a,b)=1" - I did say "if it [the inverse] exists". Are you sure about phi(b)-2?
– ngn
Aug 25, 2017 at 6:45

# Mathematica, 14 bytes

Obligatory Mathematica builtin:

ModularInverse


It's a function that takes two arguments (a and b), and returns the inverse of a mod b if it exists. If not, it returns the error ModularInverse: a is not invertible modulo b..

# JavaScript (ES6), 797362 61 bytes

Returns false if the inverse does not exist.

It uses the extended Euclidean algorithm and solves all test cases almost instantly.

f=(a,b,c=!(n=b),d=1)=>a?f(b%a,a,d,c-(b-b%a)/a*d):b<2&&(c+n)%n


### Test cases

f=(a,b,c=!(n=b),d=1)=>a?f(b%a,a,d,c-(b-b%a)/a*d):b<2&&(c+n)%n

console.log(f(1, 2)) // -> 1
console.log(f(3, 6)) // -> Does not exist
console.log(f(7, 87)) // -> 25
console.log(f(25, 87)) // -> 7
console.log(f(2, 91)) // -> 46
console.log(f(13, 91)) // -> Does not exist
console.log(f(19, 1212393831)) // -> 701912218
console.log(f(31, 73714876143)) // -> 45180085378
console.log(f(3, 73714876143)) // -> Does not exist

• Why is it not possible to write the name of function f, as in f(c,a,b=0,d=1,n=a)=>c?f(a%c,c,d,b-(a-a%c)/c*d,n):a<2&&(b+n)%n ?
– user58988
Aug 24, 2017 at 17:47
• @RosLup f(x,y) is always parsed as a function call, except if it is explicitly preceded by the function keyword. An anonymous arrow function, on the other hand, is declared as (x,y)=>something and f=(x,y)=>something assigns the function to the f variable. Aug 24, 2017 at 18:01

# Python 2, 34 bytes

f=lambda a,b:a==1or-~b*f(-b%a,a)/a


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Recursive function that gives True for print f(1,2), which I believe to be acceptable, and errors for invalid inputs.

We are trying to find $$\x\$$ in $$\a\cdot x\equiv 1\pmod{b}\$$.

This can be written as $$\a\cdot x-1=k\cdot b\$$ where $$\k\$$ is an integer.

Taking $$\\mod{a}\$$ of this gives $$\-1\equiv k\cdot b\pmod{a}\$$. Moving the minus gives $$\-k\cdot b\equiv1\pmod{a}\$$, where we have to solve for $$\k\$$.

Seeing how it resembles the initial scenario, allow us to recurse to solve for $$\k\$$ by calling the function with $$\f(-b\%a,a)\$$ (works because Python gives positive values for modulo with a negative argument).

The program recurses for until $$\a\$$ becomes 1, which only happens if the original $$\a\$$ and $$\b\$$ are coprime to each other (ie there exists a multiplicative inverse), or ends in an error caused by division by 0.

This value of $$\k\$$ can be substituted in the equation $$\a\cdot x-1=k\cdot b\$$ to give $$\x\$$ as $$\\frac{k\cdot b+1}{a}\$$.

# Jelly, 2 bytes

æi


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This uses a builtin for modular inverse, and returns 0 for no modular inverse.

# Jelly, 7 bytes

R×%⁸’¬T


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Outputs empty set (represented as empty string) on no modular inverse. Runs out of memory on TIO for the largest test-cases, but should work given enough memory.

How it Works

R×%⁸’¬T
R        Generate range of b
×       Multiply each by a
%⁸     Mod each by b
’    Decrement (Map 1 to 0 and all else to truthy)
¬   Logical NOT
T  Get the index of the truthy element.


If you want to work for larger test-cases, try this (relatively ungolfed) version, which requires much time rather than memory:

## Jelly, 9 bytes

×⁴%³’¬ø1#


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How it Works

×⁴%³’¬ø1#
#   Get the first
ø1      one integer
which meets:
×⁴            When multiplied by a
%³          And modulo-d by b
’         Decrement
¬        Is falsy


# Mathematica, 18 bytes

PowerMod[#,-1,#2]&


input

[31, 73714876143]

# R + numbers, 15 bytes

numbers::modinv


returns NA for those a without inverses mod b.

R-Fiddle to try it!

# R, 33 bytes (non-competing)

This will fail on very large b since it actually creates a vector of size 32*b bits.

function(a,b)which((1:b*a)%%b==1)


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Returns integer(0) (an empty list) for those a without inverses mod b.

# Japt, 9 8 bytes

Takes the inputs in reverse order. Outputs -1 for no match. Craps out as the bigger integer gets larger.

Ç*V%UÃb1


Test it

• Saved 1 byte thanks to ETH pointing out an errant, and very obvious, space.
• The test input 73714876143,31 seems to produce an out-of-memory error on Firefox (and to crash Chromium). I don't think this is a valid answer. Aug 25, 2017 at 0:54
• @IlmariKaronen: I clearly pointed out that fact in my solution. We can assume infinite memory for the purposes of code golf so the memory issues and crashes do not invalidate this solution. Aug 25, 2017 at 8:51
• Unfortunately the memory issues also make it impossible to tell whether your code would actually solve the test cases in 60 seconds as stipulated by the challenge. I suspect it would not, even if there was sufficient memory available to make it not crash, but without a computer that can actually run the program for that long there's no way to tell for sure. Aug 25, 2017 at 10:09

# Python 3 + gmpy, 23 bytes

I don't think it can get any shorter in Python.

gmpy.invert
import gmpy


Try it online! (won't work if you do not have gmpy installed)

# Python 3, 49 bytes

lambda a,b:[c for c in range(b)if-~c*a%b==1][0]+1


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# Python 3, 50 bytes

lambda a,b:[c for c in range(1,b+1)if c*a%b==1][0]


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This throws IndexError: list index out of range in case there is no modular multiplicative inverse, as it is allowed by the rules.

• This fails to return a result for the input 31,73714876143 in 60 seconds (on TIO). Aug 25, 2017 at 0:58
• @IlmariKaronen Seems to finish in 56 seconds on my machine (Macbook Pro '15) Aug 25, 2017 at 4:58

# Python 2, 5149545351 49 bytes

-1 byte thanks to officialaimm
-1 byte thanks to Shaggy

a,b=input()
i=a<2
while(a*i%b-1)*b%a:i+=1
print+i


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Prints 0 when there is no solution.

• Outputs 0 for a=1 and b=2; from the test cases, it should output 1. Aug 24, 2017 at 16:03
• As a recursive algo Aug 24, 2017 at 16:03
• As Shaggy pointed out, fails for 2, 1 Aug 24, 2017 at 16:06
• This fails to return an answer in 60 seconds (on TIO) for the input 31,73714876143. Aug 25, 2017 at 0:49
• This goes into infinite loop for the input 4, 6, since there is no inverse, but b%a != 0. (For inverse to exist, it's not enough that b is not divisible by a, you need them to be coprime.) Aug 25, 2017 at 7:41

# 8th, 6 bytes

Code

invmod


Explanation

invmod is a 8th word that calculates the value of the inverse of a, modulo b. It returns null on overflow or other errors.

Usage and test cases

ok> 1 2 invmod .
1
ok> 3 6 invmod .
null
ok> 7 87 invmod .
25
ok> 25 87 invmod .
7
ok> 2 91 invmod .
46
ok> 13 91 invmod .
null
ok> 19 1212393831 invmod .
701912218
ok> 31 73714876143 invmod .
45180085378
ok> 3 73714876143 invmod .
null


# J, 28 bytes

4 :'(1=x+.y)*x y&|@^<:5 p:y'


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Uses Euler's theorem. Returns 0 if the inverse does not exist.

## Explanation

4 :'(1=x+.y)*x y&|@^<:5 p:y'  Input: a (LHS), b (RHS)
4 :'                       '  Define an explicit dyad - this is to use the special
form m&|@^ to perform modular exponentiation
y   Get b
5 p:    Euler totient
<:        Decrement
x                Get a
^          Exponentiate
y&|@             Modulo b
x+.y                   GCD of a and b
1=                       Equals 1
*                 Multiply


# Pyth, 10 bytes

3 bytes saved thanks to @Jakube.

xm%*szdQQ1


Try it here!

Returns -1 for no multiplicative inverse.

### Code Breakdown

xm%*szdQQ1      Let Q be the first input.
m      Q       This maps over [0 ... Q) with a variable d.
*szd         Now d is multiplied by the evaluated second input.
%    Q        Now the remained modulo Q is retrieved.
x        1      Then, the first index of 1 is retrieved from that mapping.


# Pyth, 15 13 bytes

KEhfq1%*QTKSK


Throws an exception in case no multiplicative inverse exists.

Try it here!

# Pyth, 15 bytes

Iq1iQKEfq1%*QTK


This adds lots of bytes for handling the case where no such number exists. The program can be shortened significantly if that case would not need to be handled:

fq1%*QTK


Try it here!

• 2 bytes saved with KExm%*QdKK1 Aug 25, 2017 at 6:51
• Or 3 bytes if you swap the order of inputs: xm%*szdQQ1 Aug 25, 2017 at 6:53
• @Jakube Thanks a lot, editing! Aug 25, 2017 at 7:26
• How does this work? Nov 20, 2018 at 20:04
• @Cowsquack I've added a completely primitive code breakdown but rn I don't have time to include a complete explanation. Hopefully it is clear enough for now but I'll try to add a more complete explanation soon. Nov 21, 2018 at 14:42

# Pari/GP, 11 bytes

a->b->1/a%b


Throws an error when there is no inverse.

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# C (gcc), 115 bytes

#define L long long
L g(L a,L b,L c,L d){return a?g(b%a,a,d-b/a*c,c):b-1?0:d;}L f(L a,L b){return(g(a,b,1,0)+b)%b;}


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Extended Euclidean algorithm, recursive version

# C (gcc), 119 bytes

long long f(a,b,c,d,t,n)long long a,b,c,d,t,n;{for(c=1,d=0,n=b;a;a=t)t=d-b/a*c,d=c,c=t,t=b%a,b=a;return b-1?0:(d+n)%n;}


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Extended Euclidean algorithm, iterative version

# C (gcc), 48 110 104 bytes

#define f(a,b)g(a,b,!b,1,b)
long g(a,b,c,d,n)long a,b,c,d,n;{a=a?g(b%a,a,d,c-(b-b%a)/a*d):!--b*(c+n)%n;}


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This should work with all inputs (that fit within a long) within 60 seconds.

Edit. I'm already abusing the n variable so I might as well assume that gcc puts the first assignment in %rax.

• Alas, this gives wrong results even for fairly small inputs due to integer overflow inside the loop. For example, f(3,1000001) returns 717, which is obviously nonsense (the correct answer is 333334). Also, even if this bug was fixed by using a wider integer type, this brute-force approach would certainly time out for some of the larger test cases given in the challenge. Aug 25, 2017 at 1:20

# APL (Dyalog Unicode), 36 38 bytes

{⌈/i×1=⍵|(i←⍳⍵)×⍵|⍺}


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Explanation:

                   ⍵|⍺} ⍝ Get ⍺ mod ⍵
(i←⍳⍵)×     ⍝ Multiply the result by all numbers up to ⍵
⍵|            ⍝ Take result mod ⍵
i×1=                ⍝ Find all numbers (1,⍵) where the mod is 1
{⌈/                      ⍝ And take the largest


Much thanks to Adam in the APL Orchard chatroom for the help with this one!

Formula obtained from this site

First iteration:

{((⍵|⍺),⍵){+/⌈/⍵×1=(¯1↑⍺)|⍵×⊃⍺}⍳⍵}

• what about larger tests? the challenge says: "The program should finish at some point, and should solve each test case in less than 60 seconds"
– ngn
Feb 17, 2020 at 16:15
• we usually count 1 char = 1 byte in apl
– ngn
Feb 17, 2020 at 16:15

# Python 3.8, 22 bytes

lambda a,b:pow(a,-1,b)


Obligatory Python 3.8+ builtin. Outputs the result, or gives the error ValueError: base is not invertible for the given modulus if there is no result.

Try it online.

Explanation:

To quote the Python 3.8 release notes:

For integers, the three-argument form of the pow() function now permits the exponent to be negative in the case where the base is relatively prime to the modulus. It then computes a modular inverse to the base when the exponent is -1, and a suitable power of that inverse for other negative exponents. For example, to compute the modular multiplicative inverse of 38 modulo 137, write:

>>> pow(38, -1, 137)
119
>>> 119 * 38 % 137
1


Modular inverses arise in the solution of linear Diophantine equations. For example, to find integer solutions for 4258𝑥 + 147𝑦 = 369, first rewrite as 4258𝑥 ≡ 369 (mod 147) then solve:

>>> x = 369 * pow(4258, -1, 147) % 147
y = (4258 * x - 369) // -147
4258 * x + 147 * y
369


(Contributed by Mark Dickinson in bpo-36027.)

# [Pyt], 1 byte

ɯ


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Takes inputs in the order b a (on separate lines). Outputs None if there is no such multiplicate inverse. Gotta love built-ins.

# Python 2 + sympy, 74 bytes

from sympy import*
def f(a,m):i,_,g=numbers.igcdex(a,m);return g==1and i%m


Try it online!

Taken from Jelly source code.

# Axiom, 45 bytes

f(x:PI,y:PI):NNI==(gcd(x,y)=1=>invmod(x,y);0)


0 for error else return z with x*z Mod y =1

# Python 2, 52 bytes

-3 bytes thanks to Mr. Xcoder.

f=lambda a,b,i=1:i*a%b==1and i or i<b and f(a,b,i+1)


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Outputs False on no solution and errors out as b gets larger.

## Embedded TIO

I'm just testing out iframes in Stack Snippets and they work absolutely fantastic.

html,body{height:100%;}iframe{height:100%;width:100%;border:none;}
<iframe src="https://tio.run/##PY3BCsIwEETv/Yq5CEndy6ZotbhfIh4SanBB09Lm4tdHU8HTvIHhzfzOjym5UqI8/SuMHp4CqfCgrd8FEfZphGJaoJeAWqLZJnu2Jd/XvEJwNUxwlmA6wrFmTzj1FdzhT4QzV@DuR7emidULTdhMA@ZFU/4@tGrLBw"></iframe>

• I'm not certain this works, can't i*a%b be 0? Aug 24, 2017 at 16:29
• Fails with "maximum recursion depth exceeded" error for input (31,73714876143). Aug 25, 2017 at 1:09

# JavaScript (ES6), 424139 38 bytes

Outputs false for no match. Will throw a overflow error as the second number gets to be too large.

x=>y=>(g=z=>x*z%y==1?z:z<y&&g(++z))(1)


# Jelly, 27 bytes

²%³
⁴Ç⁹Ð¡x⁸
ÆṪ’BṚçL\$P%³×gỊ¥


Try it online!

Uses Euler's theorem with modular exponentiation. Since Jelly doesn't have a builtin to perform modular exponentiation, it had to be implemented, and it took most of the bytes.

# Axiom, 99 bytes

w(a,b,x,u)==(a=0=>(b*b=1=>b*x;0);w(b rem a,a,u,x-u*(b quo a)));h(a,b)==(b=0=>0;(b+w(a,b,0,1))rem b)


it use the function h(); h(a,b) return 0 if error [not exist inverse] otherwise it return the z such that a*z mod b = 1 This would be ok even if arguments are negative...

this would be the general egcd() function that retunr a list of int (so they can be negative too)

egcd(aa:INT,bb:INT):List INT==
x:=u:=-1   -- because the type is INT
(a,b,x,u):=(aa,bb,0,1)
repeat
a=0=>break
(q,r):=(b quo a, b rem a)
(b,a,x,u):=(a,r,u,x-u*q)
[b,x, (b-x*aa)quo bb]


this is how use it

(7) -> h(31,73714876143)
(7)  45180085378
Type: PositiveInteger


i find the base algo in internet from https://pastebin.com/A13ybryc