Let p be a prime. Let S be a set of (p-1) integers, none of which are divisible by p. Show that some subset of S has a sum that has a remainder of 1 when divided by p.

(The sum of a set is defined as the sum of the elements of the set)

since you said let p be A prime number, there is only one that you can use. This means that the S (set) will only contain one type of digit, that being 1 less than the original p. Then, you said let S be the set of (p-1) integers, so S will be one digit repeating an infinite amount of times. So the subset can be as many of that number as the me (or whoevers solving it) wants.
therefore, i'm going to let p=3, S=2,2,2,2,2,2...
subset will be 2 and 2, making the sum of the subset 4. 4/3 has a remainder of 1.

Posted by Solomon on 2004-12-29 20:40:21