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Find the inverse of:.20mod79,3mod62,21mod91,5mod23

Short Answer

Expert verified

Inverse of the given numbers is obtained.

Step by step solution

01

Inverse of 20 mod 79

GCD(20,79)79=(20×3,19)20=(19×1,1)GCD(20,79)=1soinverseof20mod79So,Bybackwardsubstitution,1=20-1×19=20-1×(79-3×20)=20-79+3×20=4×20-1×79.So,20-1=4mod79.

02

Inverse of 3 mod 62

GCD(3,62)62=(3×30,2)3=(2×1,1)GCD(3,62)is1so,inverseof3mod62is1=3-2=3-(62-20×3)=3-62+20×3=21×3-62×1So,3-1=21mod62.

03

Inverse of 21 mod 91

GCD(21,91)91=21×4,721=7×3,0So,HereGCD(21,91)is0soinverseisnotpossibleasperEuclidTheorem

04

Inverse of 5 mod 23

GCD(5,23)23=(5×4,3)5=(3×1,2)3=(2×1,1)GCD(5,23)isequalto1soinverseof5mod231=3-2=3-(5-3)=3×2-5×1=(23-4×5)×2-5×1=23×2-9×5So,5-1=-9=14mod23.

Hence, inverse of given numbers is obtained.

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Most popular questions from this chapter

Prove that the grade-school multiplication algorithm (page 24), when applied to binary numbers, always gives the right answer.

The grade-school algorithm for multiplying two n-bit binary numbers x and y consist of addingtogethern copies of r, each appropriately left-shifted. Each copy, when shifted, is at most 2n bits long.
In this problem, we will examine a scheme for adding n binary numbers, each m bits long, using a circuit or a parallel architecture. The main parameter of interest in this question is therefore the depth of the circuit or the longest path from the input to the output of the circuit. This determines the total time taken for computing the function.
To add two m-bit binary numbers naively, we must wait for the carry bit from position i-1before we can figure out the ith bit of the answer. This leads to a circuit of depthΟ(m). However, carry-lookahead circuits (see
wikipedia.comif you want to know more about this) can add inΟ(logn)depth.

  1. Assuming you have carry-lookahead circuits for addition, show how to add n numbers eachm bits long using a circuit of depth Ο(lognlogm).
  2. When adding three m-bit binary numbers x+y+z, there is a trick we can use to parallelize the process. Instead of carrying out the addition completely, we can re-express the result as the sum of just two binary numbersr+s, such that the ith bits of r and s can be computedindependently of the other bits. Show how this can be done. (Hint: One of the numbers represents carry bits.)
  3. Show how to use the trick from the previous part to design a circuit of depthΟ(logn)for multiplying two n-bit numbers.

Suppose you want to compute the nth Fibonacci number Fn , modulo an integer p. Can you find an efficient way to do this?

How many integers modulo113 have inverses?(Note:113=1331)

Calculate 2125mod127using any method you choose. (Hint: 127 is prime.)

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