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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?

 Oct 9, 2020
 #1
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please look up modular artihmetic rules on mathstackexchange instead of begging for an answer it is quite elementary indeed

 Oct 9, 2020
 #2
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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.

 Oct 9, 2020
 #3
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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

 Oct 9, 2020

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