1,104 reputation
625
bio website math.stackexchange.com/users/…
location Antarctica
age 78
visits member for 1 year, 9 months
seen Jan 31 at 5:07

Self-learning mathematics. Many thanks to those who make time to help others on the site.


Dec
28
accepted If an integer is divisible by 8 and 15, then the integer also must be divisible by which of the following?
Dec
28
accepted How is this a property of Pascal's triangle?
Dec
26
awarded  Analytical
Dec
26
asked How is this a property of Pascal's triangle?
Dec
20
comment If an integer is divisible by 8 and 15, then the integer also must be divisible by which of the following?
Ah! I figured it out just now. Thank you
Dec
20
comment If an integer is divisible by 8 and 15, then the integer also must be divisible by which of the following?
What else does it imply? (I actually thought about it, but I'm quite hesitant to state what I think).
Dec
20
asked If an integer is divisible by 8 and 15, then the integer also must be divisible by which of the following?
Dec
20
awarded  Supporter
Dec
20
comment Pigeon-hole Principle: Does this proof have a typo?
Ah yes, that's what I figured. I was thinking that if it were less than, say a radius of 0.01, then you could definitely arrange the dots so that the circumcircles don't contain them.
Dec
19
asked Pigeon-hole Principle: Does this proof have a typo?
Dec
11
awarded  Editor
Dec
11
awarded  Scholar
Dec
11
accepted I can't figure out this combinatorics problem… Or at least why my solution doesn't work.
Dec
11
revised I can't figure out this combinatorics problem… Or at least why my solution doesn't work.
added 62 characters in body
Dec
9
comment I can't figure out this combinatorics problem… Or at least why my solution doesn't work.
@JoshKeneda Why is it that (1/10)*(1/9) can't represent 8, THEN 1? It seems like if I wanted to pick an 8, and then a 1, the probability would be the same.
Dec
9
comment I can't figure out this combinatorics problem… Or at least why my solution doesn't work.
I'm not sure I understand your question, but in my solution, it would be for 1 case. The second and third cases would be 2,x1,x2,9 and 3,y1,y2,10, where x1,x2 are numbers between 2 and 9, and y1,y2 are numbers between 3 and 10.
Dec
9
asked I can't figure out this combinatorics problem… Or at least why my solution doesn't work.
Oct
6
awarded  Student