A varaint of a MGMAT problem. A , B, and C have 5 bagels

That is correct. Great job.

thanks
praetorian

21, of course, is the sum of numbers from 1-6.

Which makes sense.

If person A has 0 bagels, person B can have 0,1,2,3,4,5 and person C will get whatever is left.
If person A has 1 bagels, person B can have 0,1,2,3,4 and person C will get whatever is left.
If person A has 2 bagels, person B can have 0,1,2,3 and person C will get whatever is left.
If person A has 3 bagels, person B can have 0,1,2 and person C will get whatever is left.
If person A has 4 bagels, person B can have 0,1 and person C will get whatever is left.
If person A has 5 bagels, person B can have 0 and person C will get whatever is left.

But there's still gotta be a better way...

When you say that this problem is reminiscent of an old MGMAT problem, are you referring to this thread?

https://www.gmatclub.com/phpbb/viewtopic ... ght=#17866


Because we detemined the formula for positive integers, but the addition of ZERO is the difference between that problem and this one, and I cannot figure out how to compensate/adjust for that.

no stoolfi, this problem is from MGMAT's archives.

let me see if i can send you the explanation.

great work

keep posting

thanks
praetorian

Here is another approach (virtually the same as Stoolfi's approach)

With:

5 Bagels (means 0 bagels for everyone else): 3C1=3
4 Bagels (4,1,0) = 3C1*2C1=6
3 Bagels (3,2,0) = 3C1*2C1=6
3 Bagesl (3,1,1) = 3C1=3
2 Bagels (2,2,1) = 3C1=3

Sum=21