darkoshi

My alma mater's mailing list sometimes includes a mathematical brain teaser in their emails. Often, when I first look at the brain teaser, it seems like it should be easy to solve. Sometimes, it is. Other times, it takes me a long time to figure out (or I may even give up, deciding I have other pressing things to do with my time).

I wonder if I could have solved these problems quicker/easier, when I was younger. Or if they would have been the same difficult for me, back then.

Latest brain teaser:
There is a number less than 3,000 that when divided by 2 leaves a remainder of 1, when divided by 3 leaves a remainder of 2, when divided by 4 leaves a remainder of 3, when divided by 5 leaves a remainder of 4, when divided by 6 leaves a remainder of 5, and so on up to nine.
What is that number?

After writing down a lot of equations and sequences of numbers, and not getting very far with them (and beginning to worry that maybe I'm not smart enough to figure it out anymore), I finally realized the problem could be simplified as:
What number is evenly divisible by 2, 3, 4, 5, 6, 7, 8, and 9? Then subtract 1 from that number.

2, 3, 5, and 7 are prime numbers.
The other numbers can be broken down as: 4 = 2*2, 6 = 2*3, 8 = 2*2*2, 9 = 3*3

Anything divisible 8 is inherently divisible by 4 and 2.
Anything divisible 9 is inherently divisible by 3.
Anything divisible by 8*9 is inherently divisible by 6.

Therefore, the number which is evenly divisible by all the other numbers must be:
5 * 7 * 8 * 9 = 2520.

Therefore the solution to the problem is 2520 - 1 = 2519.