(40 + 30 + 48) mod 3 = (1+0+0)
(40 + 30 + 48) mod 4 = (0+2+0)
(40 + 30 + 48) mod 5 = (0+0+3)
so 40 + 30 + 48 = 118 works. As you have observed, any equivalent answer mod 60 will work. Depending on your needs, the smallest answer is either 58 or 2.
As you seemed to already know, this is the Chinese Remainder Theorem. The trick is to make a sum so that all but one addend is zero for each modulo. I have difficulty imagining anything simpler.
It's also more help than we ought to give in homework. I justify it on the grounds of your claim to have changed the numbers and the usefulness of a pedagogical example.
