What is the sum of the smallest and second-smallest positive integers a satisfying the congruence 27a = 17 (mod 42)?

Guest Jan 23, 2022

#1**-5 **

Step 1 is to make this a little more clear:

\(27a = 42x + 17\)

Since the smallest value of x must be 0, then the base equation would be 27a = 17. 27 is 3^3, meaning that the sum of the digits of the number must be divisible by 3. But because 42 has a digit sum of 6 which is divisible by 3, but 17 has a digit sum of 8, then no matter how many values of x you try and input, or how many 42's you add to the 17, you will always get a number that has a digit sum that isn't divisible by 3. Therefore you can't get 'a' to be a whole number.

But 'a' could be 17/27, and 27/59.

17/27 + 27/59 = **1732 / 1593.**

proyaop Jan 23, 2022