Suppose that x is an integer that satisfies the following congruences:
x == 1 (mod 3)
x == 2 (mod 5)
x == 1 (mod 7)
What is the remainder when x is divided by 105?
please look up modular artihmetic rules on mathstackexchange instead of begging for an answer it is quite elementary indeed
+118667 There probably is some simple rule but I don't know it.
I just looked at x=1 mod 7, found a list of values, and looked for the first one that made the first two equations true as well.
Then I found the mod105 of that number.
It didn't take all that long for find the answer.
Because the numbers involved are very small, just by inspection, we can come to the conclusion that the smallest x that will satisfy the 3 congruences is 22, since:
22 mod 3 = 1
22 mod 5 = 2
22 mod 7 = 1.
To this we can add the LCM[3, 5, 7] = 105 for a genral solution:
x = 105n + 22, where n =0, 1, 2,3 .........etc. So, if you we use the smallest x, we have:
22 mod 105 = 22. Or:
[105 + 22] =127 mod 105 =22
