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Chinese Remainder Theorem.

This was something I learned recently. A friend who is also a math professor showed me this.

This problem is as follows: Now, suppose we have a number of objects. We don't know how many objects we have, but we know that if we divide this number into groups of two equal groups, there will be one left over. We also know that if we divide this number into three equal groups, we have two left over. Thirdly, if the number is divided into five equal groups, we always have three left over. The question is: What is the smallest number of objects this could be?

The solution my friend showed me was beyond my understanding at first. I am starting to understand this a little more though it still is interesting to me.

Here is his unusual method to solve this:
Ok, we are dividing these numbers into groups of 2, 3 and 5. We will need to deal with each of these three scenarios one at a time.
Starting with the first scenario, we’ll deal with dividing this number into 2 groups. Let’s find a number that when divided into two group will have a remainder of 1. If you think about it, every number that we divide by two will have a remainder of 0 or 1. Only odd numbers will have a remainder of 1. So, you could pick any odd number, but these will not meet the criteria of the other numbers (3 and 5)
So, the way my friend taught me to deal with the number 2, is to multiply the other digits (3 and 5).
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