I am at a loss here, how would i decrypt this back?Code:sum1 = ((num % 10) / 1); rm1 = ((sum1 + 7) % 10); sum2 = ((num % 100) / 10); rm2 = ((sum2 + 7) % 10); sum3 = ((num % 1000) / 100); rm3 = ((sum3 + 7) % 10); sum4 = ((num % 10000) / 1000); rm4 = ((sum4 + 7) % 10); printf("%d %d %d %d\n", rm1, rm2, rm3, rm4);



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But that's just my opinion.