# How do I start?

$I\quad came\quad across\quad the\quad following\quad question:\\ Prove\quad that,\quad given\quad any\quad 52\quad integers,\quad there\quad exist\quad two\quad ofthem\quad whose\quad sum,\quad or\quad else,\quad whose\quad difference,\quad is\quad divisible\quad by\quad 100.$

$I\quad just\quad started\quad using\quad latex.\quad Sorry$

Note by Ceesay Muhammed
4 years, 10 months ago

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$Bad English In the question. Sorry$

- 4 years, 10 months ago

It's something about the pigeonhole principle. Sorry no time now, got to go to class.

- 4 years, 10 months ago

Since we need to check the divisibility by $100$, we'll only concentrate on the last two digits of the numbers.

The only possible last two digits are $\text{00, 01, 02, ..........., 98, and 99}$, a total of $100$ last two digits.

If the last two digits of any $2$ numbers are the same, their difference will be divisible by $100$.

Hence, we know that the last two digits of all numbers are different. Now, we divide the numbers into pairs that add up to $100$. The pairs are $(01,99)$, $(02,98)$, ........., $(49,51)$. And we can also use $0$ and $50$ only once.

Total number of pairs is $49$ and $2$ extra numbers. This makes a total of $51$ numbers, but we need $52$.

Hence, we must either use two numbers from any pair (making the sum divisible by $100$), or we need to use a number twice (making the diifference divisible by $100$).

$\Large \mathbb{QED}$

- 4 years, 10 months ago

@Ceesay Muhammed you don't need to latex the whole text; see the Beginner Latex Guide, it'll help a lot.

- 4 years, 10 months ago

Q.E.D. is an understatement :D

- 4 years, 10 months ago

LOL It's not even a statement... :-P

- 4 years, 10 months ago