What is the multiplicative inverse of 5 in Z_11, Z_12, and Z_13? You can do a tr

Question

What is the multiplicative inverse of 5 in Z_11, Z_12, and Z_13? You can do a trial-and-error search using a calculator or a PC. With this simple problem we want now to stress the fact that the inverse of an integer in a given ring depends completely on the ring considered. That is, if the modulus changes, the inverse changes. Hence, it doesn't make sense to talk about an inverse of an element unless it is clear what the modulus is. This fact is crucial for the RSA cryptosystem, which is introduced in Chap. 7. The extended Euclidean algorithm, which can be used for computing inverses efficiently, is introduced in Sect. 6.3.

Explanation / Answer

Answer :

By trial - and - error search

The multiplicative inverse of 5 in Z11 , Z12 and Z13 .

The formula for multiplicative inverse of a in Zn is   => (a*x)(mod n) = 1( mod n )    (or)

(a*x) – 1 ( mod n ) = 0 (mod n)

Here x is the multiplicative inverse of given number a.

Multiplicative inverse of 5 in Z11 is

in Z11 , the multiplicative inverse of 5 is 9 , using above formula

(5*x)mod 11 => (5*9) mod 11 = 45 ( mod 11 ) = 1 (mod 11)      (or)

since 5*9 => 45 – 1 (mod11) = 0 mod 11

Multiplicative inverse of 5 in Z12 is

in Z12 , the multiplicative inverse of 5 is 17 , using above formula

(5*x)mod 12 => (5*17) mod 12 = 85(mod 12) = 1(mod 12)      (or)

since 5*17 => 85 -1 (mod12) = 0 mod 12.

Multiplicative inverse of 5 in Z13 is

in Z13 , , the multiplicative inverse of 5 is 8 using above formula

(5*x)mod 13 => (5*8) mod 13 = 40(mod 13) = 1(mod 13)      (or)

since 5*8 => 40 – 1 (mod13) = 0 mod 13.

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