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?

Guest Oct 9, 2020

#1**0 **

please look up modular artihmetic rules on mathstackexchange instead of begging for an answer it is quite elementary indeed

Guest Oct 9, 2020

#2**+1 **

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.

Melody Oct 9, 2020

#3**+1 **

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

Guest Oct 9, 2020