You have 2021 beanbags and 7 boxes numbered 1 through 7. You place your first beanbag in box 1, your second in box 2, all the way to 7th in box 7. Then, you start going backwards; placing your 8th beanbag in box 6, 9th beanbag in box 5, etc. all the way until you put your 13th beanbag in box 1. Your 14th beanbag will be in box 2. This process is iterated until all beanbags are put into boxes. What box will the 2021st beanbag be in?
Thanks in advance for the help!
+36916 First pass place 7 bags each subsequent pass places 6 bags
2021 - 7 = 2014
2014 / 6 = 2014/6 = 335 passes then 4 bags are left over
First pass goes from 7 to 1 ( this is the first pass of '6' after the initial 7)
335th pass goes from 7 to 1
then 4 more bags go to boxes 2 3 4 5 I think.
Sorry I think you misunderstood the problem statement. Thank you for your answer though!
Notice that it doesn't go from 7 to 1, it actually goes back and forth. Once you put a beanbag in box 7, then next beanbag you put will be in box 6, and then box 5, and so on. Once you reach box 1 again, you go to box 2, and so on. It keeps looping...
+36916 That is what I posted it goes 1 - 7 then backwards to 1 then forwards to 7 then the 2021st bag winds up in box 5 ( by my calcs)
+128475 Note that we will have the following
As EP said ....after the first pass each subsequent pass places 6 bags
So
[2021 - 7] / 6 = 335 additional passes after the first pass plus 4 additional bags
Let the first pass = pass 0
And notice that after 5 more passes we end up in box 1
pass 0 1 2 3 4 5
1 2 3 4 5 6 7 / 6 5 4 3 2 1 / 2 3 4 5 6 7 / 6 5 4 3 2 1 / 2 3 4 5 6 7 / 6 5 4 3 2 1 /
We have 335 additional passes after the first pass....so there will be 335/5 = 67 "red" cycles with 30 boxes in each cycle
And at the start of the 68th cycle we will place 4 more bags in boxes 2 3 4 5
So...we end up in box 5.....just as EP found 
Proof
7 + 30*67 + 4 = 2021 boxes filled
