Proving A Division Property By Employing The Pigeon-Hole Pricinple

Re: Proving A Division Property By Employing The Pigeon-Hole Pricinple

Quote:

Originally Posted by

**Bashyboy**

Here is the problem: Let be a positive integer. Show that in any set of consecutive integers there is exactly one divisible by .

Here is the solution:

Let

be the integers in the sequence. The integers

,

, are distinct because

whenever

. Because there are n possible values for

and there are n different integers integers in the set, each of these values is taken exactly once. It follows that there is exactly one integer in the sequence that is divisible by n.

Something I don't quite understand: How does make the values calculated from the modulus distinct?

Since the difference of any two such number is NOT 0, they cannot be equal.

Quote:

Also, what is the motivation

for

Quote:

? Why are we allogerwed to use this fact?

You are told that " ".

(a+ j)- (a+ k)= j- k. It is greater than 0 because j is larger than k. It is less than n because j- k is less than j which is itself less than n.

Re: Proving A Division Property By Employing The Pigeon-Hole Pricinple

Quote:

Originally Posted by

**Bashyboy** How does

make the values calculated from the modulus distinct?

Quote:

Originally Posted by

**HallsofIvy** Since the difference of any two such number is NOT 0, they cannot be equal.

I would rather say the following. If j > k and (a + j) mod n = (a + k) mod n, then (a + j) - (a + k) = 0 mod n, i.e., (a + j) - (a + k) is a multiple of n. This means that either (a + j) - (a + k) = 0 or (a + j) - (a + k) ≥ n. If we exclude both of these options, this would mean that (a + j) mod n ≠ (a + k) mod n.