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

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

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

Note by Ceesay Muhammed
2 years, 2 months ago

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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 \(100\)).

\(\Large \mathbb{QED}\) Satvik Golechha · 2 years, 2 months ago

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@Satvik Golechha Q.E.D. is an understatement :D Marc Vince Casimiro · 2 years, 2 months ago

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@Marc Vince Casimiro LOL It's not even a statement... :-P Satvik Golechha · 2 years, 2 months ago

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@Satvik Golechha @Ceesay Muhammed you don't need to latex the whole text; see the Beginner Latex Guide, it'll help a lot. Satvik Golechha · 2 years, 2 months ago

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It's something about the pigeonhole principle. Sorry no time now, got to go to class. Marc Vince Casimiro · 2 years, 2 months ago

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\(Bad English In the question. Sorry\) Ceesay Muhammed · 2 years, 2 months ago

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