This problem deals with finding modulus roots.
Is there a general method for solving problems like the one described below.
Solve the following
5^(1/7)mod(47)
ie find a number x, such that x^7 = 5 (mod 47).
2007-12-26 09:20:49 · 6 answers · asked by Christine P 5 in Science & Mathematics ➔ Mathematics
The answer to that specific case is 35, but there isn't a general method for arbitrary exponent and modulus; in fact it's a generalised case of the 'RSA Problem', which is to find such a root where the modulus is a product of two primes and the exponent coprime to its totient - this is thought to be as hard as factorising the (very large) modulus.
Indeed, the fact that this problem is 'hard' is the basis for the strength of most modern (public-key) cryptography, so don't expect a nice generalised solution anytime soon...
2007-12-26 11:58:01 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The easiest way to do this is to use the theory of indices.
1). Find a primitive root mod 47. From tables, or
by trial it is 5.
2). The index of 5 with respect to base 5 is 1.
3). Take the index of both sides:
7 ind x = 1(mod 46). Here 46 = Ï(47).
4). Solve for ind x: ind x = 33(mod 46)
5). That means x = 5^33(mod 47)
6). 5^33 = 35 mod 47
7). x = 35(mod 47).
8). Check: 35^7 =5(mod 47).
2007-12-26 11:39:38 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
45.
2^7 = 128 = 42 mod 47 = -5 mod(47)
so (-2)^7 = 5 mod 47
2007-12-26 09:42:51 · answer #3 · answered by holdm 7 · 0⤊ 1⤋
18-hours of college math and I don't have the 1st clue what you're talking about. Good Luck!
2007-12-26 09:23:58 · answer #4 · answered by Anonymous · 0⤊ 2⤋
Okay, you are definitely leaving something out . . .
2007-12-26 09:24:11 · answer #5 · answered by Anonymous · 0⤊ 1⤋
you got to study
2007-12-26 09:22:54 · answer #6 · answered by kid 3 · 1⤊ 2⤋