More random questions

[Salvadoramerican]

Getting in some topics i'm weak in
Tuff teef has a dozen jawbreakers. How many ways can he put them into 6 bags so that each bag has at least one 1 jawbreaker in them? ans.462

Odd numbers from 1 to 17 are to be place in this magic box where the sum of the diagonals, columns, and rows are the same find x?
-Gonna need your imaginations here their is a 3x3 box with 9 squares
-the first row from the top contains 13 in the center box
-the second row contains 17 in the first box and 1 in the last box
-the bottom row contains x in the third box

help is appreciated thanks in advance

[nsguy1350]

I think the first one is solved by doing [unparseable or potentially dangerous latex formula]. You do 12 choose 2 to get the possible combinations, and then you divide by 2 because you only fill one spot in each bag, out of two. Shaky on this, so back-up would be appreciated.

[unparseable or potentially dangerous latex formula]
Therefore x = 15.
Works! The rows, diagonals, and columns always sum up to the median x the number of elements is a row, I believe. This is equivalent to last + first + median, which in this case is (1 + 9 + 17) = 27. I figured out the 5 because of the sum requirement coming from the 9 and 13, then I thought that since 1 and 5 are pretty low, a 15 probably goes in that corner, and it was easy from there.

[Salvadoramerican]

thanks for your help on the second one, it looks like the first one is still a work in progress though i think its 11C5, hey have you found any method to tackle the lucky number, only thing i know is that their odd and/or prime and that long method. I'm looking at Christian Goldbach's conjecture focus on the lucky number, It kind of works though it doesn't give all the lucky numbers though.

[nsguy1350]

I thought about it, and I don't know any way besides maybe memorization (but memorizing to 700+ isn't good), or making a lucky number program in your calculator. Mayeb memorize the first few, up to 100, then program into the calculator for the rest.
And yes, the first one is weird!

[Salvadoramerican]

good idea about the program i know just the thing, also this guy kind of explain the first one he said line up the 12 jawbreakers in a line with a gap between them,there
are then 11 gaps of which we choose 5 in which to place a separator.
The gaps can be chosen in C(11,5) = 462 ways. if you understood that please explain i kind of get though not sure where he got 5 unless he meant that their is to be one jawbreaker left over.

[Salvadoramerican]

i just figure something out all lucky numbers are either divisble of 3 or have a remainder of 1 when divided by 3, i notice that when i wrote them all out