Author |
Topic: Divisible by 12 (Read 907 times) |
|
NickH
Senior Riddler
Gender: 
Posts: 341
|
 |
Divisible by 12
« on: Jan 7th, 2003, 5:32am » |
 Quote  Modify
|
Let a and b be integers such that a + b + ab is divisible by 12. What is the most we can deduce about a and b?
|
|
 IP Logged |
Nick's Mathematical Puzzles
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
 |
Re: Divisible by 12
« Reply #1 on: Jan 7th, 2003, 8:31am » |
Quote Modify
|
I don't know what the most is..
But at least a, and b are integer (trivial deduction)
a-b is divisible by 12
a (and likewise b) is either 6*i, or 6*i -2 for some integer i
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Cyrus
Newbie
 
Gender:
Posts: 27
|
|
Re: Divisible by 12
« Reply #2 on: Jan 7th, 2003, 9:34am » |
Quote Modify
|
I also don't know what the MOST is, but I've come up with a few constraints:
1) a & b must be integers and must be even
2) a + b must be a multiple of 4
3) a - b must be either 0 or a multiple of 12
4) ab must be a multiple of 4
This is another problem though, where I used my usual approach:
1) I came up with a few (5) possible solutions.
2) I then made some assumptions/constraints based on those solutions.
3) I tested my constraints by trying other numbers to see if I could find a pair of numbers that proved my constraints wrong.
The problem with my approach, is clear even to me. How can I be SURE this is the OPTIMAL/maximum amount of constraints? And unless I tested infinite numbers through my constraints how can I be sure that my constraints are even correct at all??
|
|
IP Logged |
|
|
|
Cyrus
Newbie
Gender:
Posts: 27
|
|
Re: Divisible by 12
« Reply #3 on: Jan 7th, 2003, 9:38am » |
Quote Modify
|
on Jan 7th, 2003, 8:31am, towr wrote:
a-b is divisible by 12
a (and likewise b) is either 6*i, or 6*i -2 for some integer i |
|
Oh, I just now read your post towr, and I like that 2nd point
"a must be either 6x or 6x-2", so that should be added to my list. See I missed one already.
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Divisible by 12
« Reply #4 on: Jan 7th, 2003, 10:14am » |
Quote Modify
|
it's easy enough to prove a and b are even, and thus a*b is a multiple of 4.. After that I don't know..
a+b+ab=12N for some integer a,b,N
(a+1)(b+1)-1=12N
(a+1)(b+1)=12N-1 , 12N-1 is always odd, so a+1 and b+1 are odd, so a and b have to be even..
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Divisible by 12
« Reply #5 on: Jan 7th, 2003, 10:23am » |
Quote Modify
|
here are some example solutions to work with:
-30 -30 -14 -14 0 12 18 -30
-30 -18 -14 -2 0 24 18 -18
-30 -6 -14 10 4 -20 18 -6
-30 6 -14 22 4 -8 18 6
-30 18 -12 -24 4 4 18 18
-30 30 -12 -12 4 16 18 30
-26 -26 -12 0 4 28 22 -26
-26 -14 -12 12 6 -30 22 -14
-26 -2 -12 24 6 -18 22 -2
-26 10 -8 -20 6 -6 22 10
-26 22 -8 -8 6 6 22 22
-24 -24 -8 4 6 18 24 -24
-24 -12 -8 16 6 30 24 -12
-24 0 -8 28 10 -26 24 0
-24 12 -6 -30 10 -14 24 12
-24 24 -6 -18 10 -2 24 24
-20 -20 -6 -6 10 10 28 -20
-20 -8 -6 6 10 22 28 -8
-20 4 -6 18 12 -24 28 4
-20 16 -6 30 12 -12 28 16
-20 28 -2 -26 12 0 28 28
-18 -30 -2 -14 12 12 30 -30
-18 -18 -2 -2 12 24 30 -18
-18 -6 -2 10 16 -20 30 -6
-18 6 -2 22 16 -8 30 6
-18 18 0 -24 16 4 30 18
-18 30 0 -12 16 16 30 30
-14 -26 0 0 16 28
|
| « Last Edit: Jan 7th, 2003, 11:20am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
NickH
Senior Riddler
Gender:
Posts: 341
|
|
Re: Divisible by 12
« Reply #6 on: Jan 8th, 2003, 5:29am » |
Quote Modify
|
towr, I got your original result, namely that a and b are either both divisible by 6, or both leave a remainder of 4 when divided by 6.
As you indicated, write a + b + ab = (a + 1)(b + 1) - 1.
Then (a + 1)(b + 1) = 1 (modulo 12.)
Hence a + 1 and b + 1 are coprime with 12.
Considering the multiplication table, mod. 12, the following are the only possibilities for a + 1 and b + 1:
a + 1 = 1 => b + 1 = 1
a + 1 = 5 => b + 1 = 5
a + 1 = 7 => b + 1 = 7
a + 1 = 11 => b + 1 = 11
Therefore a = b = 0, 4, 6, or 10 (mod 12.)
Equivalently, a = b = 0 or 4 (mod. 6.)
Nick
|
|
IP Logged |
Nick's Mathematical Puzzles
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print