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# 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.

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

Note by Ceesay Muhammed
2 years 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 $$100$$).

$$\Large \mathbb{QED}$$ · 2 years ago

Q.E.D. is an understatement :D · 2 years ago

LOL It's not even a statement... :-P · 2 years ago

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

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

$$Bad English In the question. Sorry$$ · 2 years ago