Suppose we pick 52 random intergers. Prove the sum or difference of at least one pair must be a multiple of 100.
I understand it by the ending digit of random integer is 1-9 or ends in 0. so you can minus 2-2 to end in 0. or add 6+4 to get zero. If i hvae 51 and 49, it's 100. 1+9=0 346+54 =400, but im not sure have to exactly prove it.
Imagine that we have $100$ lamps numbered $00$-$99$. As we pick numbers, one by one, we light up the lamp corresponding to the two last digits of the number we pulled, as well as $100$ minus that number.
What this signifies is that if we pull a number $n$ and its last two digits correspond a lamp already lit, then we can take the number $m$ we used to light that lamp, and say that either $m+n$ or $m-n$ is a multiple of $100$.
Most numbers, when we draw them, will light two lamps. The only exceptions are numbers that end in $00$ and $50$. So, after pulling $51$ numbers, if none of the numbers pulled corresponded to an already lit lamp, all lamps will be lit, and the $52$nd number will correspond to a lit lamp.
By adding or subtracting multiples of $100$, we may assume each number $x_i$ is in $\{0,1,\ldots, 99\}$. Let $y_i = x_i$ if $x_i \le 50$, otherwise $y_i = 100-x_i$. There are then $51$ possible values for the $y_i$: $0, \ldots, 50$. By the Pigeonhole principle, at least two of the $y_i$ are equal. The corresponding $x_i$ are either equal or their sum is $100$.
Let's sort all possible integers into 51 groups or sets. If the last two digits of an integer are either 01 or 99 we sort it in one set which we'll label as $A_{01}$. If the last two digits are either 02 or 98 we'll put them in a set labeled $A_{02} $. We'll do this for sets $A_{ij} $ which will contain all the integers ending with ij or end with the last two digits of $100-ij $.
There will be $51$ such sets from $A_{00} $ to $A_{50} $. ($A_{00} $ will contain all the digits ending with 00 and $A_{50} $ will contain those ending with 50 but the rest will have two possible pairs of last two digits.)
When you select 52 numbers there are only 51 sets they can be pull from, so at least two will come from the same set.
Those two either end with the same two last digits or will end with two digits that add to 100. So either their sum or their difference is a multiple of 100.
consider sets $$(100x+1,100x-1),(100x+2)(100x-2),...,(100x+49)(100x-49),(100x),(100x+50)$$
now we have $51$ sets &$52$ choices so by pp1 atleast one one pair comes from one box hence$$(100x+i)+(100x-i)=100(2x)$$
a multiple of 100.
