TITLE: Verifying a proof of a combinatorics problem
QUESTION [0 upvotes]: 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.

REPLY [2 votes]: 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 $T$.
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