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Old 2010-02-17, 20:55   #3
wblipp
 
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"William"
May 2003
New Haven

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(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.
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