Thread: Combinatorics and number theory

1. Combinatorics and number theory

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?

2. Originally Posted by james_bond
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 $\displaystyle x_1, x_2, \dots , x_{10}$. Consider the sums $\displaystyle s_n = \sum_{i=1}^{10} x_i$ for $\displaystyle n = 1, 2, \dots , 10$. 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 $\displaystyle s_j$ and $\displaystyle s_k$, with $\displaystyle j < k$, have the same remainder, by the Pigeonhole Principle. Then $\displaystyle \sum_{i=j+1}^k x_i$ is divisible by 10.

3. Originally Posted by awkward
Consider the sums $\displaystyle s_n = \sum_{i=1}^{10} x_i$ for $\displaystyle n = 1, 2, \dots , 10$.
Should this be:

$\displaystyle s_n = \sum_{i=1}^{n} x_i$ for $\displaystyle n = 1, 2, \dots , 10$?

4. Originally Posted by Grandad
Should this be:

$\displaystyle s_n = \sum_{i=1}^{n} x_i$ for $\displaystyle n = 1, 2, \dots , 10$?