GIven any positive integers, show that at least two must exist whose sum or difference is divisible by 10.
25+15
25-15
There are two that exist that meet the critieria.
im sorry i meant to type in given any "seven" positive integers....
does that change the answer?
To prove that at least two positive integers must exist whose sum or difference is divisible by 10, we can use the Pigeonhole Principle.
The Pigeonhole Principle states that if there are more pigeons than pigeonholes, then at least two pigeons must go into the same hole.
In this case, we can think of the positive integers as the pigeons and the possible remainders when divided by 10 as the pigeonholes. Since there are 10 possible remainders (0 to 9) when dividing a number by 10, and an infinite number of positive integers, by the Pigeonhole Principle, there must exist at least two positive integers that have the same remainder when divided by 10.
Let's consider two positive integers, a and b, both with the same remainder r when divided by 10: a ≡ b (mod 10).
Now we need to consider two cases:
1. The remainder when the sum of a and b is divided by 10:
(a + b) ≡ (a + a) ≡ 2a (mod 10).
Since a and b have the same remainder when divided by 10, it follows that 2a would also have the same remainder when divided by 10. Therefore, the sum of a and b is divisible by 10.
2. The remainder when the difference of a and b is divided by 10:
(a - b) ≡ (a - a) ≡ 0 (mod 10).
In this case, the difference of a and b is divisible by 10 since it results in a remainder of 0.
In both cases, we have shown that at least two positive integers exist whose sum or difference is divisible by 10. Thus, the statement has been proven.