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

Please help. I don't know how to start the question on the first place.

\(I\quad just\quad started\quad using\quad latex.\quad Sorry\)

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## Comments

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TopNewest\(Bad English In the question. Sorry\)

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

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Since we need to check the divisibility by \(100\), we'll only concentrate on the

last two digitsof 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, theirdifferencewill 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

musteither 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 \(100\)).\(\Large \mathbb{QED}\)

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

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Q.E.D. is an understatement :D

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

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