# Find in set builder notation the set of all positive integers a such that 137≡a (mod 5)?

I've tried looking in the text book, but i can't find the instructions for this question in my assignment. Thank you for the help in advance. :)

### 3 Answers

- Ian HLv 75 months ago
The modular equation 137 ≡ a (mod 5) is equivalent to

5n + a = 137

The least value of a which solves it is a = 2, with n = 35

(because 5*35 + 2 = 137)

But adding or subtracting any multiples of 5 to a also produces valid options.

For example a = 22, because 5*23 + 22 = 137

In everyday terms a can take any value given by 137 – 5n, where n

can be any positive or negative integer.

Now you need to convert plain English to set builder notation.

Something like “a is the set of real integers such that a = 137 – 5n, integer, n”

This is a gentle introduction to set notation

https://www.purplemath.com/modules/setnotn.htm

Scroll down to “Give a solution using a rule”

That example is sufficiently similar to guide your own effort at set building

- az_lenderLv 75 months ago
{ a : (a > 0) & (5|(a - 2)) }

where the vertical bar means "is an exact divisor of"

- ted sLv 75 months ago
137 / 5 = 27 R2====> 137 ≡ a (mod 5) = 2 + 5 n...thus a = 2 or 7 or 12 or 17 ....{ n = 0 , 1,2 , ..., 27 }