You mean 50 (the modulus).

For small moduli and , adding the modulus until cancellation is probably the best way to solve . For this particular congruence, add the modulus once to 20 to get:

Since , you can 'divide' by 7 to get your solution.

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For a more general case where , as long as , then we can always use the Euclidean algorithm to find one solution modulo and use it to find all solutions modulo m. Here's an example in this wiki entry: Linear congruence theorem