Anonymous asked:Hi, I went through your section on Counting Problems in the Blue Book and when it finally came time to do the exercises in the Blue Book I couldn't do even one.. Do you mind if I ask you to explain some of the Blue Book counting problems to see if I actually learn how to do it? Here's the first one I couldn't do #16 468 Blue Book Second edition
I wouldn’t put this one in the same classification as the Counting Problems in my book, because you’re not counting a number of permutations or combinations here. Instead, you’re counting how many pre-determined combinations meet a certain condition.
In other words, they’re not asking you how many different combinations of 3 numbers have products under 1000, they’re asking you how many “tri-factorable” numbers do.
Still, this can still be solved using the technique I suggest at the end of my counting chapter: Whenever it’s not a COMPLETELY straightforward problem, JUST LIST!
If a tri-factorable number is the product of 3 consecutive integers, just start listing sets of consecutive integers until their products get too big.
Tri-factorable numbers (and how they’re made):
1 x 2 x 3 = 6
2 x 3 x 4 = 24
3 x 4 x 5 = 60
4 x 5 x 6 = 120
5 x 6 x 7 = 210
6 x 7 x 8 = 336
7 x 8 x 9 = 504
8 x 9 x 10 = 720
9 x 10 x 11 = 990
10x11x12 = 1320 (Oops, too high!)
So, how many tri-factorable numbers are less than 1000? 9 of them.