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# Congruences

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

Jan 23, 2022

#1
+514
-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.

Jan 23, 2022
#2
+117100
+1

a is meant to be an integer.

Melody  Jan 23, 2022
#3
+117100
+1

\(27a=42b+17\\ 27a-42b=17\\ \)

The lowest common denominator of 27 and -42  is  3   and 3 does not go into 17.

Therefore I do not think there are any integer solutions here.

Jan 23, 2022