\(I\quad came\quad across\quad the\quad following\quad question:\\ Prove\quad that,\quad given\quad any\quad 52\quad integers,\quad there\quad exist\quad two\quad ofthem\quad whose\quad sum,\quad or\quad else,\quad whose\quad difference,\quad is\quad divisible\quad by\quad 100.\)
Please help. I don't know how to start the question on the first place.
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BadEnglishInthequestion.Sorry
It's something about the pigeonhole principle. Sorry no time now, got to go to class.
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 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).
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.
Q.E.D. is an understatement :D
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LOL It's not even a statement... :-P