How do I start?

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

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

Ijuststartedusinglatex.SorryI\quad just\quad started\quad using\quad latex.\quad Sorry

#Algebra

Note by Ceesay Muhammed
6 years, 5 months ago

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1 vote

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Comments

BadEnglishInthequestion.SorryBad English In the question. Sorry

Ceesay Muhammed - 6 years, 5 months ago

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

Marc Vince Casimiro - 6 years, 5 months ago

Since we need to check the divisibility by 100100, we'll only concentrate on the last two digits of the numbers.

The only possible last two digits are 00, 01, 02, ..........., 98, and 99\text{00, 01, 02, ..........., 98, and 99}, a total of 100100 last two digits.

If the last two digits of any 22 numbers are the same, their difference will be divisible by 100100.

Hence, we know that the last two digits of all numbers are different. Now, we divide the numbers into pairs that add up to 100100. The pairs are (01,99)(01,99), (02,98)(02,98), ........., (49,51)(49,51). And we can also use 00 and 5050 only once.

Total number of pairs is 4949 and 22 extra numbers. This makes a total of 5151 numbers, but we need 5252.

Hence, we must either use two numbers from any pair (making the sum divisible by 100100), or we need to use a number twice (making the diifference divisible by 100100).

QED\Large \mathbb{QED}

Satvik Golechha - 6 years, 5 months ago

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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.

Satvik Golechha - 6 years, 5 months ago

Q.E.D. is an understatement :D

Marc Vince Casimiro - 6 years, 5 months ago

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

Satvik Golechha - 6 years, 5 months ago
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