Verifying a proof of a combinatorics problem

Question

today I attempted this problem and have a proof I am not sure is correct.

Show that given any 52 integers, there exist two of them whose sum, or else whose difference is divisible by 100

Firstly it is evident that if the given integers have 2 or more integers that are equal to each other then we are done $\because$ $$ n - n=0\\ \mod(0,100) = 0 $$

Notice that $$ \forall n \in \mathbb{Z}\\ \mod(n,100) = \{0\dots99\}$$

Now I use the pigeonhole principle to realise that $$ \lceil 99/52 \rceil = 2$$

Therefore there must be two numbers congurent to each other and we are done.

Answer 1

Consider the integers modulo 100 . Then Any integer $x$ of them must has $x \mod 100 \in \{0,1,2,\cdots,99\} $. If there are two integers $x,y$ such that $x = y \mod 100$ , then their difference $x-y = 0 \mod 100 $. Hence we are done.

Now let's assume the $52$ integers are all different under modulo 100. We want to determine if there are two integers such that their sum is $0$ under modulo $100$. There are 50 situations such that $x+y = 0 \mod 100$ , i.e. $(x,y)\in \{(1,99),(2,98),(3,97),\cdots,(49,51),(50,50) \}$ all in modulo 100. Since we assume all integers are different . $(50,50)$ is not allowed. Now we have 49 pairs $\{(1,99),(2,98),(3,97),\cdots,(49,51) \}$. And we have $52$ integers . So by the pigeonhole principle . There should be $2$ integers fall in a same pair. And their sum is $0$ under modulo $100$.

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Edit: The entir pairs to make sum is $0$ under modulo $100$ should be $\{(0,0),(1,99),(2,98),(3,97),\cdots,(49,51),(50,50)\}$. The $52$ integers must have 2 integers fall in the same pair. And their sum is $0$ under modulo $100$.

Answer 2

Moderately alternative description of the answer.

The congruency classes $\pmod{100}$ can be expressed as

$S = \{-49, -48, \cdots, -1, 0, 1, 2, 3, \cdots, 49, 50\}$.

Now consider the set $T$ of the absolute value of the actual elements in the set $S$:

$T = \{|-49|, |-48|, \cdots, |-1|, |0|, |1|, |2|, |3|, |49|, |50|\}.$

In fact, the set $T$ only has $(51)$ distinct elements.

Per the constraints of the problem, with both $(x + y)$ and $(x - y)$ being interrogated $\pmod{100}$, if any two elements $x,y$ are both represented by the same element in $T$, then you will have that [$\pmod{100}$]:

• either $x \equiv y \implies (x - y) \equiv 0$

• or $x \equiv -y \implies (x + y) \equiv 0$.

Since $T$ has only $51$ distinct elements, in any group of $52$ integers, there must be $2$ of those integers that are represented by the same element in $T$.