Number Theory: What is the least residue [math]\mod N = 95[/math] of [math]3^{1.1 \cdot 10^{43}}[/math]?
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This is a practice problem. Since [math]5 \cdot 19[/math] are prime factors of [math]95[/math] I tried to break it into two congruence equations and use CRT, but I can't seem to work this out. By fermat's little theorem we have [math]3^5[/math] congruent to [math]3 \mod (5)[/math], but squaring that multiple times cannot be the solution because then we have to square the RHS of the equation, which is some ridiculous thing. Also [math]3^{19}[/math] congruent to [math]3 \mod 19[/math], which doesn't help much (I think?) and [math]3^{18} [/math] congruent to [math]1 \mod 19[/math] which also doesn't help much (I think...). Some help would be appreciated.
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Answer:
You want to use the generalization of Fermat's Little Theorem sometimes known as Euler's Theorem. 3 and 95 are relatively prime, so 3 generates the entire multiplicative group (mod 95). Calculate the order of this group to get the multiplicative order of 3 (mod 95).
Daniel McLaury at Quora Visit the source
Other answers
3^4=1 ( mod 5) and 3^18=1 ( mod 19). Here, lcm (4,18)=36 whence 3^36=1 ( mod 95). The idea is to find the remainder of 1.1*10^43 upon division by 36. For that, 11=11 ( mod 36) and 10^42=28 ( mod 36) [ It isn't that hard ], whence 11*10^42 =28*11=20 ( mod 36) . Thus 1.1*10^43= 36k+20 3^36=1 ( mod 95) => 3^(36k)=1 (mod 95) => 3^(36k+20)=3^20 ( mod 95). It isn't much difficult to show that 3^20=66 (mod 95) and thus making 3^(1.1*10^43)=66 ( mod 95)
Murtaza Wani
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