Combinatorics and number theory

james_bond · #1 $Old$ December 25th, 2008, 02:09 AM

10 integers are given. We know that there is no way of choosing some of them whose sum is divisible by 10. What can be the remainders of the 10 integers when divided by 10?

awkward · #2 $Old$ December 26th, 2008, 02:40 PM

Quote:

Originally Posted by james_bond $View Post$

10 integers are given. We know that there is no way of choosing some of them whose sum is divisible by 10. What can be the remainders of the 10 integers when divided by 10?

Hi james bond,

No such collection of integers can exist.

Let's say the integers are $x_1, x_2, \dots , x_{10}$ . Consider the sums $s_n = \sum_{i=1}^{10} x_i$ for $n = 1, 2, \dots , 10$ . If any of the sums is divisible by 10 we're done, so let's suppose none of them are. Then the possible remainders of the sums are 1, 2, ..., 9. Since there are 10 sums and 9 possible remainders, at least two of the sums, say $s_j$ and $s_k$ , with $j < k$ , have the same remainder, by the Pigeonhole Principle. Then $\sum_{i=j+1}^k x_i$ is divisible by 10.