How To Use Inverse Modular Arithmetic? (Best solution)

Modular inverses

  1. The inverse of a number A is 1/A since A * 1/A = 1 (e.g. the inverse of 5 is 1/5)
  2. All real numbers other than 0 have an inverse.
  3. Multiplying a number by the inverse of A is equivalent to dividing by A (e.g. 10/5 is the same as 10* 1/5)

Contents

What is modular inverse used for?

Modular multiplicative inverses are used to obtain a solution of a system of linear congruences that is guaranteed by the Chinese Remainder Theorem.

What is modular inverse of a number?

A modular inverse of an integer ( modulo ) is the integer such that. A modular inverse can be computed in the Wolfram Language using PowerMod[b, -1, m]. Every nonzero integer has an inverse (modulo ) for a prime and not a multiple of.. For example, the modular inverses of 1, 2, 3, and 4 (mod 5) are 1, 3, 2, and 4.

How do you find modular exponentiation?

Modular exponentiation can be performed with a negative exponent e by finding the modular multiplicative inverse d of b modulo m using the extended Euclidean algorithm. That is: c = be mod m = de mod m, where e < 0 and b ⋅ d ≡ 1 (mod m). Modular exponentiation is efficient to compute, even for very large integers.

What is multiplicative inverse in modular arithmetic?

The modular inverse of a number refers to the modular multiplicative inverse. For any integer a such that (a, p) = 1 there exists another integer b such that ab≡ 1 (mod p). The integer b is called the multiplicative inverse of a which is denoted as b = a1.

Can you divide in modular arithmetic?

Can we always do modular division? The answer is “NO”. In modular arithmetic, not only 4/0 is not allowed, but 4/12 under modulo 6 is also not allowed. The reason is, 12 is congruent to 0 when modulus is 6.

How do you do inverse mod on a calculator?

For every number x from this set, calculate a * x mod m, i.e., the remainder from the division of a * x by m. The modular multiplicative inverse of a modulo m is the value of x for which this remainder is equal to 1.

What is the value of mod 7?

15 MOD 10 is equal to 5 (because 15 – (15 DIV 10) equals 5 — the remainder of the division is 5) 20 MOD 7 = 6. 21 MOD 7 = 0.

What is the inverse of 7 mod 11?

7x≡1≡12≡23≡34≡45≡56(mod11). Then from 7x≡56(mod11), we can cancel 7, obtaining x≡8(mod11). Hence, −3 is the inverse of 7(mod11).

What is the inverse of 19 mod 141?

Therefore, the modular inverse of 19 mod 141 is 52.

How do you reduce modular arithmetic?

In modular arithmetic, when we say “reduced modulo,” we mean whatever result we obtain, we divide it by n, and report only the smallest possible nonnegative residue. The next theorem is fundamental to modular arithmetic. Let n≥2 be a fixed integer. If a≡b (mod n) and c≡d (mod n), then a+c≡b+d(modn),ac≡bd(modn).

Modular multiplicative inverse

Given two numbers ‘a’ and’m,’ compute the modular multiplicative inverse of ‘a’ under the modulo of’m’ by multiplying them together. The modular multiplicative inverse is a positive integer ‘x’ that has the property that. an x = 1 an x = 1 an x = 1 an x = 1 an x = 1 an x = 1 (mod m) The value of x should be in the range of integer modulo m, i.e., inside the integer modulo m range. (It should be noted that x cannot be zero since a*0 mod m would never be 1). The multiplicative inverse of “a modulo m” occurs if and only if a and m are relatively prime (i.e., if gcd(a, m) = 1).

Examples: a = 3, m = 11 are the input values.

(under 11).

Input values: a = 10, m = 17, 12 as a final result Because (10*12) mod 17 = 1, 12 is the modulo inverse of the number 10.

Method No.

For each integer x, determine whether (a*x) percent m equals 1.

C++

Includeiostreamusingnamespacestd;intmod Inverse(inta,intm)intmain()

Java

Import java.io.*;classGFGpublicstaticvoidmain(String args);classGFGpublicstaticvoidmain(String args)

Python3

DefmodInverse(a, m):forxinrange(1, m): DefmodInverse(a, m):forxinrange(1, m): In the case of ( (((a percent m)*(x percent m) percent m) percent m=1): returnxreturn-1a=3m=11print(modInverse(a, m)) returnxreturn-1a=3m=11 print(modInverse(a, m))

C

UsingSystem;classGFGpublicstaticvoidMain()}

PHP

The following code: phpfunctionmodInverse($a,$m) $a = 3 and $m = 11;echomodInverse($a,$m); phpfunctionmodInverse($a,$m);

Javascript

The following script function modInverse(a, m): let a = 3; let m = 11; document.write(modInverse(a, m));/script The following script function Time Complicatency: O (m). Method No. 2 (Works when m and a are coprime) The goal is to employ Extended Euclidean algorithms, which take two numbers ‘a’ and ‘b’ and discover their gcd, as well as find ‘x’ and ‘y’ such thatax + by = gcd, to solve the problem (a, b) In order to find the multiplicative inverse of ‘a’ under’m,’ we substitute b = m in the preceding formula.

Taken together, ax + my 1 (mod m) gives usax + my 1 (mod m).

The following is an example of how the aforesaid algorithm is implemented.

C++

The following script function modInverse(a, m): let a = 3; let m = 11; document.write(modInverse(a, m)); Complications in respect to time: O (m). The second approach is described below (Works when m and a are coprime) In this case, the idea is to useExtended Euclidean algorithms that take two integers ‘a’ and ‘b’ and find their gcd, as well as find the values of x and y such that ax + by = gcd (a, b) Put b = m in the preceding formula to determine the multiplicative inverse of the letter an under the letter m.

In the case of two integers y, we getax + my m (mod m).

(mod m) We may then deduce from the Extended Euclid Algorithm that ‘x’ is the multiplicative inverse of the variable “a.” The method described above has been implemented in the following manner:

C

Includestdio. hintgcdExtended(inta,intb,int* x,int* y);voidmodInverse(inta,intm)intx1, y1;intgcdExtended(inta,intb,int* x,int* y)intx1, y1;intgcd = gcdExtended(b percent a, a,x1,y1);*x = ()

PHP

≅phpfunctionmodInverse($a,$m)} funcgcdExtended($a,$b,$x,$y)$x1;$y1;$gcd= gcdExtended($b percent $a,$a,$x1,$y1);$x=$y1- (int)($b/$a) *$x1;$y=$x1;$gcd= gcdExtended($b percent $a,$a,$x1,$y1) $a= 3;$m= 11;modInverse($a,$m); modInverse($a,$m); Iterative implementation of the modular multiplicative inverse:

C++

Includebits/stdc++.husingnamespacestd;intmod returnx;inverse(inta,intm)if(x0)x += m0;intmain ()

C

Includestdio.hintmod returnx;inverse(inta,intm)if(x0)x += m0;intmain ()

Java

ClassGFGif(x0)x += m0;returnx;publicstaticvoidmain(String args);publicstaticvoidmain(String args)

Python3

The following are the results of DefmodInverse(a, m):m0=my=0x+1if(m=1):return0while(m1=1):return0 while(a1):q=a/mm=a percent yy=tt=x-q*ytif(x0):x=x+m0returnxa=3m=11 DefmodInverse(a, m):m0 = my= print(“The modular multiplicative inverse is”,modInverse(a, m)) print(“The modular multiplicative inverse is”,modInverse(a, m))

C

UsingSystem;classGFGif(x0)x += m0;returnx;publicstaticvoidMain();classGFGif(x0)x += m0;publicstaticvoidMain();

PHP

≅phpfunctionmodInverse($a,$m) if($x0)$x+=$m0;return$x;} $a= 3;$m= 11;echo”Modular multiplicative inverse isn”,modInverse($a,$m);echo”Modular multiplicative inverse isn”,modInverse($a,$m);echo”Modular multiplicative inverse isn”,modInverse($a,$m);echo”Modular multiplicative inverse isn”,modInverse($a,$m);echo”Modular multiplicative inverse is

Javascript

Modification of the script function modInverse (a, m) returnx;let a = 3;let m = 11;document if (x0)x += m0, returnx;let m = 11;document. written as write(‘The modular multiplicative inverse is $’);/scriptOutput The inverse of a modular multiplicative module is four. Time Complicatency: o o o o o o o o o o o o o o o o o (Log m) Method No. 3 (Works when m is prime) If we know that m is a prime number, we may also apply Fermat’s little theorem to get the inverse of the prime number. an m-1 1 an m-1 1 an m-1 1 an m-1 1 (mod m) The result is a -1 am-2 when both sides are multiplied by one (mod m) The following is an example of how the aforementioned concept was put into action.

C++

Includeiostreamusingnamespacestd;intgcd(inta,intb);intpower(intx, unsignedinty, unsignedintm);voidmod;intgcd(inta,intb);intpower(intx, unsignedinty, unsignedintm);voidmod Inverse(inta,intm)intpower(intx, unsignedinty, unsignedintm)intgcd(inta,intb)intmain(intx, unsignedinty, unsignedintm) ()

Java

Import java.io.*;classGFGstaticintpower(intx, inty, intm)staticintgcd(inta, intb)publicstaticvoidmain(String args)staticintgcd(inta, intb)staticintgcd(inta, intb)staticintgcd(inta, intb)staticintgcd(inta,

Python3

DefmodInverse(a, m):g=gcd DefmodInverse(a, m) (a, m) Print the message “Inverse does not exist” if (g!=1) is true. else: The modular multiplicative inverse is “,power(a, m-2, m)). print(“Modular multiplicative inverse is “,power(a, m-2, m)). defpower(x, y, m):if(y=0):return1p=power(x, y/2, m) defpower(x, y, m) defpower(x, y, m) (x*p) percent mp=(p*p) percent mif(y percent 2=0):returnpelse:return((x*p) percent mp):return Define GCD(a, b) as follows: if (a=0): returnbreturngcd(b percent a, a)a=3m=11modInverse; defgcd(a, b) as follows: (a, m)

C

UsingSystem;classGFG}staticintpower(intx,inty,intm)staticintgcd(inta,intb)publicstaticvoidMain()}

PHP

≅phpfunctionmodInverse($a,$m)} functionpower($x,$y,$m)functiongcd($a,$b)$a= 3;$m= 11;modInverse($a,$m); functionpower($x,$y,$m)functiongcd($a,$b)$a= 3;$m= 11; modInverse($a,$m);

Javascript

ScriptfunctionmodInverse(a, m) ScriptfunctionmodInverse(a, m) power to perform a function (x, y, m) Let a = 3, m = 11, and functiongcd(a, b)let a = 3;modInverse(a, m);/scriptOutput The inverse of a modular multiplicative module is four. Time Occupational Complexity:O (Log m) We’ve gone through three different approaches to finding the multiplicative inverse modulo m. 1) The naive method, or, O (m) 2) The GCD algorithm of Euler was extended, and the result was O (Log m) 3) Fermat’s Little Theorem, often known as O (Log m) Computing the modular multiplicative inverse is a critical step in the RSA public-key encryption technique, which is used to protect sensitive information.

Please leave a comment if you see something that is inaccurate or if you wish to offer further information regarding the subject matter covered here.

Modular multiplicative inverse – Wikipedia

In mathematics, and notably in the field of arithmetic, amodular multiplicative inverseof an integer is defined as an integerx such that the productaxis is congruent to 1 with regard to the modulus of the integer. According to the usual notation ofmodular arithmetic, this congruence is expressed as, which is a shorthand method of presenting the fact thatmdivides (evenly) the quantityax by the integermis 1, or, to put it another way, the residual after dividingax by the integermis 1. Assuming that there is an inverse modulom for a congruence, there exist an unlimited number of solutions to this congruence, which constitute a congruence class with regard to the modulus in which it is defined.

Assuming that the congruence class containingw is denoted by the notation, this may be written as follows:where the symbol signifies the multiplication of equivalence classes modulom, the congruence class containingw is denoted by the notation Written in this manner, the analogy with the conventional idea of amultiplicative inversein the set ofrationalorreal numbersis easily depicted, with the numbers being replaced by congruence classes and the binary operation being altered in the proper way.

In the same way that the comparable operation on the real numbers has a basic application, this operation has a fundamental use in solving linear congruences of the type It is also possible to use modular multiplicative inverses for practical purposes in the field of cryptography, for example, in public-key cryptography and the RSA method.

One advantage of computer implementation of these applications is that there is a very fast method (the extended Euclidean algorithm) that can be utilized for the computation of modular multiplicative inverses, which is advantageous for computer implementation of these applications.

Modular arithmetic

Two numbers, a and b, that are congruent modulomifmdivides their difference for a given positive integerm are referred to as being congruent modulomifmdivides their difference. This binary relationship is symbolized by the symbol In this case, the equivalency relation is on the set of integers, Z, and the equivalence classes are referred to as congruence classes modulomorresidue classes modulom, respectively. Suppose we want to indicate the congruence class that contains the integera, then A Linear congruence is a type of modular congruence that has the shape of a line Linear congruences, in contrast to linear equations over the reals, may have zero, one, or numerous solutions, depending on the context.

  1. It follows that if d is the greatest common divisor of the numbers a and m, then the linear congruenceaxb(modm)has solutions only when and only if ddividesb.
  2. When an integer is multiplicatively inversed in a modular way with regard to its modulus, it is a solution of the linear congruence.
  3. coprime).
  4. Although this is commonly used to indicate that ax 1 (modm) has a solution, it might be considered an abuse of notation since it could be misunderstood as the reciprocal of ax.1 (modm) (which, contrary to the modular multiplicative inverse, is not an integer except whenais 1 or -1).
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Integers modulom

Modulom is a congruence relation that divides the set of integers into mcongruence classes, which are a subset of each other. The following is an example of how to define addition and multiplication operations on thesemobjects: To either add or multiply two congruence classes, first select a representative (in any way) from each class, then perform the usual integer operation on the two representatives, and finally take the congruence class that the result of the integer operation lies in as the result of the operation on the congruence classes.

To add or multiply two congruence classes, first select a representative (in any way) from each class, then perform the usual integer operation on the two representatives, and finally take the congruence class that the result of These definitions areandrepresent the operations on congruence classes in symbols, withandrepresenting the operations on congruence classes in symbols.

Classifications of themcongruence that have these two defined operations together form what is known as the ring of integers modulom.

Traditional names for the congruence classes of the integers modulom were residue classes modulo m, which referred to the fact that all components in a congruence class have the same remainder (i.e., “residue”) whenm is divided bym.

This is demonstrated by the division methods, which demonstrate that the set of integers, when combined, create a full system of residues modulom, also known as theleast residue system modulom.

When dealing with arithmetic difficulties, it is sometimes more convenient to work with a complete system of residues and to speak in the language of congruences, while at other times it is more convenient to approach the problem from the perspective of the congruence classes of the ring.

Multiplicative group of integers modulom

Not every element of a full residue system modulomhas a modular multiplicative inverse; for example, the number zero never has a modular multiplicative inverse. An areduced residue system is formed by deleting all of the components of a complete residue system that are not relatively prime tom and leaving only the ones that are relatively prime tom. All of the elements of a reduced residue system have modular multiplicative inverses. The number of elements in a reduced residue system is, whereis theEuler totient function, i.e., the number of positive integers smaller thanm that are relatively prime tom, and is the number of elements in a reduced residue system.

Similarly to how the product of two units is a unit, the units of a circle form a group, which is denoted by RifRis the name of the ring and is commonly denoted byRifRis the name of the circle.

It has particular significance in terms of order (size),.

In this situation, the multiplicative group of integers modulopform an acyclic group of orderp 1 and the multiplicative group of integers modulopform an acyclic group of orderp 1.

Example

Consider the following example, which makes use of the modulus 10 to demonstrate the aforementioned definitions. Two numbers are congruent mod 10 if and only if the difference between them is divisible by 10. For example, since 10 divides 32 12 = 20, and since 10 divides 111 1 = 110, two integers are congruent mod 10. With regard to this modulus, the following are some of the ten congruence classes: It is impossible to find a solution to the linear congruence4 x 5 (mod 10)because all of the numbers that are congruent to 5 (i.e., those in) are odd, whereas4 xis always even in the case of 4 x.

  • It is the case that gcd(4, 10) = 2 and that 2 doesn’t divide 5, but does divide 6.
  • As a matter of fact, the number 7 satisfies this congruence (i.e., 21 1 = 20).
  • In particular, every integer inwill fulfill the congruence because these integers have the form7 + 10 rfor some integerrandis definitely divisible by 10.
  • This congruence contains only one congruence class of solutions, which is the solution to the congruence.
  • In order to determine the product of congruence classesand, take an element from, say 25, an element from, say 2, and observe that their product (25)(2) = 50 is in the congruence class.
  • The term “addition” is defined in a similar manner.
  • It is possible to have a full residue system modulo 10 if the set contains each integer in a separate congruence class modulo 10.
  • It is possible to have a reduced residue system modulo 10.
  • This indicates that these four congruence classes are members of a group, in this instance the cyclic group of order four, which has either the number 3 or the number 7 as a generator (multiplicative).

The ring’s group of units is formed by the congruence classes that are represented. Congruence classes with modular multiplicative inverses are precisely those that fall into this category.

Computation

With the use of the extended Euclidean technique, it is possible to find the modular multiplicative inverse of amodulom. This algorithm finds the greatest common divisor (gcd) of two numbers, saya and m, using the Euclidean principle. Because of the multiplicative inverse modulom, this gcd must be one in this case. It is possible to solve the final of numerous equations generated by the procedure for this gcd by using this method. Then, using a technique known as “back substitution,” it is possible to construct an expression that connects the original arguments and this gcd.

To put it another way, this is that has been determined, which is the modular multiplicative inverse ofa The extended Euclidean algorithm is a more efficient version of the method that, by the use of auxiliary equations, decreases the number of runs through the process from two to only one (back substitution may be thought of as traveling through the algorithm in reverse).

In large O notation, this method runs in timeO(log(m) 2), assuming|a|m, and is thought to be significantly faster and more efficient than it.

Using Euler’s theorem

Euler’s theorem, which is an alternative to the extended Euclidean algorithm, may be used to compute modular inverses as an alternate method. When the conditional probability of aiscoprimetom is fulfilled, that is, if gcd(a,m) = 1, then whereisEuler’s totient function. Because abelongs to the multiplicative group if and only ifaiscoprimetom, this follows as a logical consequence. As a result, a modular multiplicative inverse may be obtained in a straightforward manner: The particular situation wheremis a prime and a modular inverse is provided by is known as It is often slower than the extended Euclidean algorithm, although it is occasionally employed when a modular exponentiation implementation for a certain problem is already available.

This is a particularly remarkable advantage of this approach.

Multiple inverses

When using the Euclidean method and three multiplications for each new input, it is feasible to compute the inverses of multiple numbersa I modulo a commonm with a single execution of the algorithm and modulo a commonm. It is the essential concept to build the product of all thea I invert that, and then multiply bya jfor allji to leave just the desireda1i remaining. The algorithm is (all arithmetic is performed modulom): (all arithmetic is performed modulom)

  1. Calculate the prefix products for the word alli
  2. Computeb1nusing any algorithm that is currently available
  3. Calculate forifromndown to 2, forifromndown to 2

In order to take advantage of parallel computing, it is possible to perform the multiplications in a tree structure rather than linearly.

Applications

A modular multiplicative inverse may be found in many algorithms that rely on the idea of modular arithmetic, and this can be useful in many situations. Examples include the use of modular arithmetic in cryptography, where it enables some operations to be carried out more rapidly and with less storage needs, but other operations become more difficult to carry out. Both of these characteristics may be utilized to one’s benefit. Encryption and decryption of a message are performed using a pair of integers that are multiplicative inverses with respect to a modulus that has been carefully chosen in the RSA method.

Because it is thought to be computationally impossible to distinguish the concealed number from the public number, the system is able to function to protect the user’s personal information.

The following is one possible solution:

  1. A modular multiplicative inverse may be found in many algorithms that rely on the idea of modular arithmetic, and finding it has numerous applications in computer science and mathematics. Examples include the use of modular arithmetic in cryptography, where it enables some operations to be completed more rapidly and with less storage needs, while other processes become more complex as a result. Taking use of both of these characteristics is a good idea. Encryption and decryption of a message are accomplished using a pair of integers that are multiplicative inverses with respect to a modulus that has been carefully chosen in the RSA method. Two numbers are utilized in this technique: one is made public and may be used in a quick encryption procedure
  2. The other, which is used in a decryption procedure, is kept confidential. Because it is thought to be computationally impossible to distinguish the concealed number from the public number, the system is able to function to protect the user’s identity. A distinct type of precise division problem may be illustrated by the odd word-size number problem in computer science, where you have a list of odd word-size numbers that are individually divisible byk and you desire to divide them all byk. As an example, the following is one possible resolution:

Division is a slower operation than multiplication on many machines, particularly those lacking hardware support for division. As a result, this technique can result in a significant speedup on some devices. Although the first step is time-consuming, it only has to be completed once. The Chinese Remainder Theorem guarantees that a solution to a system of linear congruences will be found using modular multiplicative inverses. This is achieved by employing modular multiplicative inverses. The systemX 4(mod 5)X 4(mod 7)X 6(mod 11)has similar solutions because the numbers 5, 7, and 11 are pairwise coprime.

For example, X= 3 (7 11) 4 + 6 (5 11) 4 + 6 (5 7) 6 = 3504 and in its unique reduced formX= 3504 39 (mod 385)because 385 is the LCMof 5, 7, and 11 The modular multiplicative inverse also plays an important role in the formulation of the Klosterman sum, as can be seen below.

See also

  • Inversive congruential generator- a pseudo-random number generator that employs modular multiplicative inverses
  • Rational reconstruction (mathematics)
  • Inversive congruential generator- a pseudo-random number generator that employs modular multiplicative inverses

Notes

  1. Rosen 1993, p. 132
  2. Schumacher 1996, p. 88
  3. Stinson, Douglas R. (1995),Cryptography / Theory and Practice, CRC Press, pp. 124–128, ISBN0-8493-8521-0
  4. TrappeWashington 2006, pages. 164–169
  5. Moriarty, K.
  6. Kaliski, B.
  7. Jonsson, J.
  8. Rusch, A. (1998), Cryptography / Theory and Practice (2016). It is referred to as “PKCS1: RSA Cryptography Specifications Version 2.2.” Internet Engineering Task Force RFC 8017. Internet Engineering Task Force. Internet Engineering Task Force Obtainable on January 21, 2017
  9. Other notations are also employed, such asandm
  10. And Rossen 1990, p. 32
  11. Shoup, Victor (2005), A Computational Introduction to Number Theory and Algebra, Cambridge University Press, Theorem 2.4, p. 15, ISBN9780521851541
  12. Rossen 1993, p. 121
  13. Thomas Koshy.Elementary number theory with applications, 2nd edition.ISBN978-0-12-372487-8. P. 346
  14. Brent, Richard P
  15. Zimmermann, Paul (2006), A Computational Introduction to (December 2010). “2.5.1 Several inversions occur at the same time” (PDF). Arithmetic using a Computer in the Modern Era. 67–68. ISBN 978-0-521-19469-3
  16. TrappeWashington 2006, p. 167
  17. TrappeWashington 2006, p. 165
  18. TrappeWashington
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References

  • Kenneth Ireland and Michael Rosen are the authors of this article (1990), An Introduction to Modern Number Theory in the Classical Tradition (2nd ed.), Elementary Number Theory and Its Applications, Springer-Verlag, ISBN 0-387-97329-X
  • Rosen, Kenneth H. (1993),Elementary Number Theory and Its Applications (3rd ed.), Carol Schumacher’s book, Addison-Wesley, ISBN 978-0-201-57889-8. (1996). Fundamental Concepts in Abstract Mathematics (Chapter Zero) Trappe, Wade, and Washington, Lawrence C. (2006), Introduction to Cryptography with Coding Theory(2nd ed. ), Prentice-Hall, ISBN 978-0-13-186239-5
  • Trappe, Wade, and Washington, Lawrence C. (2006), Introduction to Cryptography with Coding Theory(2nd ed.), Prentice-Hall, ISBN 978-0-13-186239-5
  • Trappe, Wade, and Washington, Lawrence C. (2006), Introduction to Cryptography with

External links

  • Weisstein, Eric W., “Modular Inverse.” Modular Inverse. A solved example of calculating the modulo multiplicative inverse using Euclid’s Algorithm is provided by Guevara Vasquez, Fernando on MathWorld
  • He can be found here.

How to find the inverse modulo $m$?

The extended Euclidean algorithm$ $or a specific form of Euclid’s algorithm that computes inverses modulo primes, which I refer to as Gauß’s algorithm$, are only a few of examples of approaches that may be used. money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money money When employing modular fraction arithmetice, the computations are frequently less complicated.

  1. for approximately$20$motley working examples, check here, here, and hereandherefor a variety of techniques (and see the sidebar “Linked” questions lists there for many more).
  2. $!bmod 31!:, $!bmod 31!:,dfracequiv dfrac,dfracequiv dfrac,dfracequiv dfrac,dfracequiv dfrac,dfracequiv 5cdot 8equiv 9$!bmod 31!:,dfracequiv dfrac,dfracequiv dfrac,dfracequiv dfrac,dfracequiv 5cdot 8equiv 9$!
  3. $!!bmod 31, n equiv, dfrac, k 7 equiv, dfrac, 31!:, 7.
  4. 7equiv 9pmod 7equiv 9pmod $ Alternatively,$ by Euler Repeated squaring allows $rm (a,m)=1 to be rapidly computed as a Rightarrow a pmod,$ to be quickly computed as a pmod,$.

Note$ $The latter produces a simple closed form for the CRT (Chinese Remainder Theorem)$quad$if$rm,(m,n)=1$then$rmquad begin rm x! equiv rm a! (mod m)$then$rmquad end iff x!equiv a!+b!+m!+m!+m!+m (mod mn) $ See also the Peirce decomposition for a more broad overview.

Modular Inverse

Algebra Applied Mathematics is a branch of mathematics that is used in a variety of applications. Calculus and Statistical Analysis Discrete Mathematics is a branch of mathematics that deals with discrete events. The Mathematics of the Foundations Geometry The Background and Terminology Number Theory is a branch of mathematics that studies the relationships between numbers. Probability and Statistics are two terms that are used interchangeably. Recreational Mathematics is a term that refers to the study of mathematics for fun.

  • This is a new concept in MathWorld.
  • Examples are the modular inverses of the numbers 1, 2, 3, and 4 (mod 5) and the inverses of the numbers 1, 3, and 4.
  • In reality, a nonzerointegerhas a modular inverse moduloiffandarerelativelyprime, whereas a zerointegerhas a modular inverse moduloiffandarerelativelyprime.
  • The triangle above (OEISA102057) has the modular inverses of(mod) for the numbers 2, 3, and 4, respectively.
  • If andare relatively prime, there are integersandsuch that, and such integers may be obtained using the Euclidean technique.
  • Taking this equationmodulo into consideration, it follows that; i.e.
  • Hence,.

See also

Congruence, Congruence Equation, and Linear Congruence Equation are all terms that are used to describe congruence. Nick Hobson(author’s link) has contributed to this entry with a number of contributions. Reid Nichol provided some of the content for this entry.

Explore with Wolfram|Alpha

In “The On-Line Encyclopedia of Integer Sequences,” N. J. A. Sloane describes SequenceA102057 as “The On-Line Encyclopedia of Integer Sequences.”

Referencedon Wolfram|Alpha

Modular Inverse is a term that refers to the reverse of a modular structure.

Cite this as:

Hobson, Nick; Nichol, Reid; and Weisstein, Eric W. “ModularInverse.” FromMathWorld-A Wolfram Web Resource. Hobson, Nick; Nichol, Reid; and Weisstein, Eric W. “ModularInverse.”

Subject classifications

Allow a and x to be integers. The identity element is said to be a modular inverse ofa when an algebraic operation performed onxandayreturns the identity element. Depending on whether the operation in issue is addition or multiplication, we may distinguish between two types of modular inverses: additive and multiplicative, which are described below. Let’s discuss about these two concepts in further depth:

Modular additiveinverse

Addition has the identification element 0 as its identity element. So we argue thatxis anadditive inverseofamodulomifx and0are congruent modulom: x and 0 are congruent modulom. a + x mod m = a + x mod m Everyaandm has anadditive inverse ofamodulo, and this is true for allaandm. It may be represented by any number of the form a + k * m, where k is an integer. a + k * m The inverse of a division bym is frequently sought after in the range, that is, in the set of remainders of division bym.

How can you find the inverse of the modular additive by hand? It’s really simple! In order to determine the additive modular inverse ofamodulom, you can follow the procedures outlined below.

  1. Write down the letter a
  2. Write down the numbers generated by continually adding or subtracting mtoa
  3. And write down the letters b and c. Identify the digit that falls between the range of 0 andm – 1

Let’s see how well it works out for you.

  • As an example, consider the problem of determining the additive inverse of4modulo30. The following are the numbers in the form–4 + 30kare:., -4, 26, 56,. As a result, in the set, the additive inverse of4modulo30is26
  • In the set, 2. Let us try to find the additive inverse of 44modulo13 in the following manner: These are the numbers in the form 44 + 13kare:., -44, -31, -18, -5, 8, 22., -44, -31, -5, 8, 22. The additive inverse of44modulo13is8 in the set as a result of this.

Modular multiplicative inverse

Remember that the first element of multiplication is the identity element. Sox is the multiplicative inverse of amodulomifa * x and 1 are equivalent modulom: x = 1 * amodulomifa * x A*x mod M is equal to 1 mod A*x mod M. It is not necessarily true that the multiplicative modular inverse exists, in contrast to additive inverses. If it does exist, on the other hand, all numbers of the formx + k * m meet the congruency requirements. In particular, in such situations, you can always find a solution (and it will always be the same one!) inside the range.

This assertion may be verified by observing that forx x (mod 6)the operation2 * x (mod 6) does not result in1.

Such numbers are referred to as coprime numbers.

Although the entire theory of modular inverses may appear a little esoteric at first glance, you may be thinking, “Why would anyone be interested in modular multiplicative inverses?” If you’re interested in cryptography, you should be aware that the computation of modular multiplicative inverses is critical in many applications, including the RSA encryption technique in particular.

It goes without saying that using our inverse modulo calculator is the most efficient technique of obtaining multiplicative modular inverses!

Modular Multiplicative Inverse – Algorithms for Competitive Programming

Module multiplicative inverseof an integer (a) is an integer (x) such that(a cdot x)is congruent to(1 )modular some modulus (m). To put it another way, we want to find an integer (x) that is large enough that We shall also use the letter (a) to signify the letter (x). However, it is important to note that the modular inverse is not always present. Consider the following example: (m = 4) and (a = 2). By examining all of the potential values modulo(m), it should become evident that we will be unable to discover (a) that will meet the aforementioned equation.

In this article, we show two ways for determining whether or not the modular inverse exists, as well as one approach for determining whether or not the modular inverse exists for all integers in linear time.

Finding the Modular Inverse using Extended Euclidean algorithm

Consider the following equation (in which the variables (x) and (y) are unknowns): This is a Linear Diophantine equation in two variables with a linear coefficient. In the case when (gcd(a, m) = 1), the equation has a solution that may be discovered using the extended Euclidean method, as demonstrated in the linked article. It should be noted that (gcd(a, m) = 1) is also the requirement for the existence of the modular inverse. Now, if we take the modulo (m)of both sides, we can get rid of the (m cdot y) and the equation is reduced to the following: As a result, the modular inverse of (a) is (x).

The resultingxfrom the extended Euclidean method may be negative, and the resultingxpercent mmmay likewise be negative, therefore we must first addmto make it positive before we can proceed.

Finding the Modular Inverse using Binary Exponentiation

Another approach for determining the modular inverse is to apply Euler’s theorem, which asserts that the following congruence holds true if (a) and (m) are both relatively prime numbers: Euler’s Totient function is denoted by the symbol phi. Again, it should be noted that the fact that (a) and (m) were both relative primes was also a need for the existence of the modular inverse. In the case where (m) is a prime number, this reduces to Fermat’s little theorem: When we multiply both sides of the preceding equations by (a), we get the following result:

  • For an arbitrary (but coprime) modulus (m), the following is true: (a equiv a mod m)
  • For a prime modulus (m), the following is true: (a equiv a mod m)

We can easily determine the modular inverse from these findings by utilizing the binary exponentiation approach, which works in (O(log m)) time and is extremely fast. However, even though this approach is simpler to grasp than the method given in the preceding paragraph, in the scenario when (m) is not a prime integer, we must compute the Euler phi function, which requires factorization of (m), which may be difficult to do successfully. If the prime factorization of (m) is known, then the complexity of this technique is (O(log m)), which is the smallest possible.

Finding the modular inverse for every number modulo(m)

Specifically, we wish to compute the modular inverse for every number in the range, which is as follows: (). The techniques presented in the preceding sections can be used to achieve a solution of complexity (O(m log m)), as shown in the figure below. In this paper, we provide a more efficient approach with complexity (O(m)). In the case of this specific procedure, however, we demand that the modulus (m) be a prime number. This is denoted by (text) as the modular inverse of (i). The following equation is true for (i1) in this case: = – [leftlfloor fracrightrfloor cdot textbmod m] [leftlfloor fracrightrfloor cdot textbmod m] As a result, the implementation is straightforward: inv=1;for(inti=2;im;++i) inv=m-(m/i)*inv percent m; inv=1;for(inti=2;im;++i) inv=m-(m/i)*inv percent m;

Proof

We have the following: Taking both sides modulo(m) gets the following results: In order to simplify the problem, multiply the two sides by(icdot (m mod I which produces the following result:

Practice Problems

  • A one-unit machine
  • Hackerrank – the longest increasing sequence arrays
  • And UVa 11904 – the one-unit machine Codeforces 300C – The Art of Creating Beautiful Numbers
  • Codeforces 622F – The Sum of the k-th Powers
  • The Sum of the k-th Powers
  • Codeforces 717A – Festival Organization
  • Codeforces 896D – Nephren Runs a Cinema
  • Codeforces 717A – Festival Organization
  • Codeforces 717A –
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Modular Inverse Calculator (A^-1 Modulo N) – Online InvMod

Look find a tool to use. The Modular Multiplicative Inverse is a tool for computing the modular inverse of a number in a modular manner. When an integer N modulo m is multiplied together, the modular multiplicative inverse of that integer is an integer n such that the inverse of N modulo m equals n. Results Modular Multiplicative Inverse-dCodeTag(s):ArithmeticModular Multiplicative Inverse-dCodeTag(s):Arithmetic Codes that are shared and mored Code is free, and its tools are a helpful resource in a variety of situations such as games, math, geocaching, puzzles, and challenges to solve on a daily basis!

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Modular Inverse Calculator

A multiplicative inverse is the inverse of amodular inverse, which is the alternative term for it.

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Integer multiplicative inverse via Newton’s method

My intention was to include computing the multiplicative inverse through Newton’s iteration in a post, but to my surprise, I couldn’t locate a straightforward explanation that was freely available on the internet. (We take a break from this article in final cut to show you something similar from Daniel Lemire1) This 2012 paper2by Jean-Guillaume Dumas is fairly succinct and easy to grasp when used in a formal presenting setting. Please keep in mind that I will be using mod as an operator rather than for congruence and that I will only be discussing power-of-two modulus (linked paper covers generalization).

Given an odd integer $a$, we wish to find an integer $x$ that has the following properties: [[beginleft(axright) bmod 2beta = 1 ][[beginleft(axright) bmod 2beta = 1 ][[beginleft(axright) bmod 2beta = 1 ][[beginleft(axright) bmod 2beta = 1 ][[beginleft(axright) bmod 2beta = 1 ][[be When dealing with reals and floating point numbers, we may compute the inverse of $a$ if we start with a decent initial guess $x 0$ and apply the following recurrence (Newton’s method3) a sufficient number of times: [[beginx = x n left(2-ax n right) labelend ][[beginx = x n left(2-ax n right) labelend ][[beginx = x n left(2-ax n right) labelend ][[beginx = x n left(2-ax n right) label To put it another way, to put it bluntly Hensel’s (lifting) lemma4expands Newton’s approach to modular arithmetic, where (in our example) if we have the value $x n$ such that:then after applying the recurrence $eqref $ we’d have:then after applying the recurrence $eqref $ we’d have:then after applying the recurrence $eqref $ we’d have:then Because of this, the amount of bits doubles for each application.

We must compute a starting value $x 0$ that is accurate to a certain number of bits ($k$).

Starting with three bits and applying the recurrence over and over again would result in the following: $left(6,12,24,48,96,ldotsright)$ If we’re working in a 32-bit register with any odd 32-bit input, we’d require four iterations to complete the task: staticinlineuint32 tmod inverse 1(uint32 ta)uint32 tx,t;x=a;/3 bitsx*=2-a*x;/6 bitsx*=2-a*x;/ 12x*=2-a*x;/ 24x*=2-a*x;/ 48 / retaining bottom 32returnx;uint32 tmod inverse 1(uint32 ta Take note that if we used four bits for our initial value, we’d end up with the following results after each succeeding step: $left(8,16,32,64,128,ldotsright) We’d only need three iterations if we used $ bits.

In other words, if we can obtain an extra bit for less money than the cost of one step, we’ll be ahead of the game.

(For the serial comment below, a*aa-1 is an independent variable.) the x*=2-a*x;/8x*=2-a*x;/ 16x*=2-a*x;/ 32returnx;/ the x*=2-a*x;/8x*=2-a*x; Another paper6 describes a method for determining an initial value that is good to 5 bits: uint32 tmod inverse 3(uint32 ta) The following code: uint32 tx;x=3*a2;/5 bitsx=2-a*x;/ 10x*=2-a*x;/ 20x*=2-a*x;/ 40 – 32 low bitsreturnx; Dumas continues with the derivation (see Section 3.3) to generate algorithm 3, which when revised looks like this: (See Section 3.3) staticinlineuint32 tmod inverse 4(uint32 ta) {uint32 tu=2-a;uint32 ti=a-1;i*=i;u*=i+1;i*=i;u*=i+1;i*=i;u*=i+1;i*=i;u*=i+1;returnu;} Because the first three variations are nearly identical in structure, it appears to be a toss-up in terms of performance in terms of architecture.

The final point is one that I find extremely fascinating.

Renamed, and with operations that can take place at the same time on the same line as one another.

The variables i3 and i4 are equal to i2 and i3t3, respectively.

Especially if there is a known bit-bound on the input numbers, this may be quite intriguing! We have an option between three, four, or five initial bits, and we must complete at least one stage.

Modular division via the multiplicative inverse of the denominator

Following that, we’ll look at inverting modular-arithmetic functions. Up:Contents Previous:Multiplication via a modular formula Modular division, on the other hand, is more difficult. As a starting point, we consider the concept of division as the polar opposite of multiplication. We learned that in mod 7 arithmetic, multiplying 5 by 6 results in the number 2. As a result, 2 divided by 6 should result in the number 5. In reality, this is correct. This, however, will not always be the case. What is the result of dividing 10 by 5 in modulo 15?

  • On the other hand, since 5 times 5 equals 10 in the case above, it is possible that the answer is 5.
  • Instead of allowing for numerous responses, we would state that there is no response in this instance.
  • In both cases, the idea of amultiplicative inverse will assist us in obtaining a solution when one exists and in making predictions when there is no response available.
  • Consider the following scenario: we are working with a modulus of 7.
  • (For the same reason, the multiplicative inverse of 5 is the number three.) By constructing a multiplication table, we may discover the multiplicative inverses.
  • The multiplicative inverse of 1 is 1, the multiplicative inverse of 2 is 4 (and vice versa), the multiplicative inverse of 3 is 5 (and vice versa), and the multiplicative inverse of 6 is 6.
  • It should be noted that the number zero does not have a multiplicative inverse.

After learning how to divide 1 by the different integers modulo 7, we can divide 1 bybis, the multiplicative inverse ofb, and so on.

According to the requirements of this course, the answer is as follows: to divideabyb, just divide 1 byband and multiply the result bya.

(modulo 7).

As a result, we multiply 5 by 2, which gives us 3.

Should we just take it as a given?

On the left-hand side, we have the number 5.

More Mod 7 examples may be found in the section below.

Some moduli do not have multiplicative inverses for some integers, and this is true for some moduli.

There is no number that, when multiplied by 12, yields a value that is congruent with the number 15 to one when divided by 12.

This phenomena is discussed in further detail in the next section.

It would make sense to defineto be 9 in this case, for example, because.

Following that, we’ll look at inverting modular-arithmetic functions. Up:Contents Previous:Multiplication via a modular formula Lisa Eckstein is a writer who lives in New York City. 1996-10-21 22:56:24 EDT Mon, Oct 21

Which mod m numbers have multiplicative inverses?

The following section contains information about this paper. Up:Contents Inverting modular-arithmetic functions was the previous step. An important theorem is that each integer between 0 and 1 has a modmmultiplicative inverse if it is substantially prime to a number between 0 and -1 In this case, the only number between 0 andm-1 that does not have a modmmultiplicative inverse is 0. If m is prime, the only number between 0 and -1 that does not have a modmmultiplicative inverse is 0. Take the case of takem =7.

  • Take, for example, the number m =15.
  • There are no mod 15 multiplicative inverses for any of these integers.
  • It is important to remember from the Number Theory handout that is the number of numbers from zero to one hundred and one that are relatively prime to one hundred and one.
  • Similarly to what was described in the Number Theory handout, the difference between andpqis rather negligible for bigpandq.
  • Lisa Eckstein is a writer who lives in New York City.

How to write matlab code for modular multiplicative inverse in cryptography –

Greetings, Professionals. Can somebody provide guidance on how to build Matlab code for the following example? Make the following calculation in cryptography: a=1/b mod m to find the modular multiplicative inverse. Here, b and m are known, while an is to be discovered.

Answers (3)

A multiplicative inverse modulo some integer p indicates thatmod(x*xinv,p) = 1 when the number p is multiplied by the number x. If the equality holds, then x has a mutiplicative inverse modulo p, which is the case if the equivalence holds. For example, we will see that mod(2 * 4,7) = = 1 when we divide by 2. Because of this, 4 is the mutiplicationative inverse of 2, modulo 7. The fact that all numbers except zero have a multiplicative inverse is demonstrated by the fact that the modulus (7 in my case) is prime.

There is no integer x such that the function mod(3*x,12) will produce the number one.

Three and twelve are not relatively prime – they have an integer component higher than one in common.

After considering all of above, how can we compute the modular multiplicative inverse in MATLAB using the following formula: It isn’t all that tough, after all.

Using the example above, determine the multiplicative inverse of 2, mod 7.= gcd (2,7) G = 1D = -3 G = 1D = -3 G = 1D = -3 G = 1D = -3 G = 1D = -3 G = 1D = -3 G = 1D = -3 G = 1D = -3 G = 1D = -3 G = 1D = -3 As long as G is one, the second argument will be the multiplicative inverse of the first.

If this is the case, then D will be a multiplicative inverse of A.

It is powered by the GCD engine.

Yes, without a doubt.

The Extended Euclidean algorithm is the key to solving this problem.

But, once again, GCD is sufficient.

Normally, I refrain from posting apparent answers and instead recommend that you study the Getting Started section of the manual.

However, I’m perplexed by this particular question.

It is expressed in rather odd terms, but I believe the aim is to find an integer “a” such that mod(a*b,m)=1 is what is being sought here. This should be effective: is equal to gcd(b,m);ifdiv=1a is equal to mod(c1,m);elseend

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