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Find an integer  x such that 0 <= x < 527 and x^37 = 490 (mod 527).

 Nov 28, 2020
 #1
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I don't think that there is an interger x that satisfies x^37=490. The closest integer would probably be 1. Hope this helps!!

 Nov 28, 2020
 #2
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x^37 mod 527=490, solve for x

 

a=1; c=a^37 % 527; if(c==490, goto3, goto4);printc, a; a++;if(a<1000, goto1, 0)

 

x =527c + 56, where c =0, 1, 2, 3........etc.

 

The smallest x = 56,  and  56^37 mod 527 =490

 Nov 28, 2020

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