User Tag List

05292010, 08:14 AM #11

05292010, 08:20 AM #12
Alright, new question...
Is it possible to place all the numbers from 1 to 12 along the circumference of a circle so that for every three numbers a,b and c ,such that, b is next to a, and c is next to b, and the number b squared  ac is divisible by 13?

05292010, 08:22 AM #13
Do you see my first post where I wrote down that series? The number you replace becomes part of a series. It starts with a (all 3 numbers are a), the second number becomes 2a1. The third becomes 3a2, the fourth 4a3. You should see a pattern with the numbers. Take 4a3. If you substitute "b" in the place of 4, and b1 in the place of 3, (which works for all expressions in this series), you get:
b*a  (b1) = 2010
You can then simplify the expression to:
b*ab=2009
Then factor out b:
b(a1)=2009
Knowing that b and a are both integers, it should be pretty obvious that you're simply looking for the factors of 2009. Since we're looking for "a", you need to add 1 to the factors for your answer."We grow up thinking that beliefs are something to be proud of, but they're really nothing but opinions one refuses to reconsider. Beliefs are easy. The stronger your beliefs are, the less open you are to growth and wisdom, because "strength of belief" is only the intensity with which you resist questioning yourself. As soon as you are proud of a belief, as soon as you think it adds something to who you are, then you've made it a part of your ego."

05292010, 08:36 AM #14
I need clarification. Is each set of 3 numbers exclusive to all other sets? Or does each number belong to 3 separate sets?
Here's an example:
Let's say you have 1 4 9 5 7 8
Would you only be concerned with 1 4 9? Then 5 7 8? Or would you also be concerned with 4 9 5 and 9 5 7?"We grow up thinking that beliefs are something to be proud of, but they're really nothing but opinions one refuses to reconsider. Beliefs are easy. The stronger your beliefs are, the less open you are to growth and wisdom, because "strength of belief" is only the intensity with which you resist questioning yourself. As soon as you are proud of a belief, as soon as you think it adds something to who you are, then you've made it a part of your ego."

05292010, 08:42 AM #15

05292010, 08:45 AM #16
Then the answer is no, because 3^2 (and 2^2 and 1^2) is less than 13 (assuming that "divisible by 13" means you are looking for an integer).
Edit: wait, I was assuming positive integers."We grow up thinking that beliefs are something to be proud of, but they're really nothing but opinions one refuses to reconsider. Beliefs are easy. The stronger your beliefs are, the less open you are to growth and wisdom, because "strength of belief" is only the intensity with which you resist questioning yourself. As soon as you are proud of a belief, as soon as you think it adds something to who you are, then you've made it a part of your ego."

05292010, 09:00 AM #17

05292010, 09:09 AM #18
12 2 9 8 10 6 1 11 4 5 3 7
"We grow up thinking that beliefs are something to be proud of, but they're really nothing but opinions one refuses to reconsider. Beliefs are easy. The stronger your beliefs are, the less open you are to growth and wisdom, because "strength of belief" is only the intensity with which you resist questioning yourself. As soon as you are proud of a belief, as soon as you think it adds something to who you are, then you've made it a part of your ego."

05292010, 01:36 PM #19
I think this is the answer. This is similar to Fibonacci Sequence (), except we don't necessarily start with F(0)=0 and F(1)=1.
From Wikipedia Fibonacci number  Wikipedia, the free encyclopedia
So start with 2010/X = 1.618033 ==> X=1242, then fill in the rest. 1.618033 is the ratio of the difference between each number of N, using this will make everything add up correctly in the appropriate stacks, for lack of better terminology.
You will get the sequence
2010 1242 768 474 293 181 112 69 43 26 16 10 6 4 2 2 0.
So one answer is 2 2 2, that will start this sequence. Any other combination will not get us to 2010. But then we also have a trivial answer of 1005 1005 1005, which is going off the seed values and stopping right after, which is too short to even start seeing the ratio appear.
Wikipedia has some sort of pictures to explain the ratio relationship by showing parts of the sequence as representing a spiral, but it didn't seem clear to me. I think the summation was a lot clearer and the spiral is good for just visualizing the idea, I guess.
I like math
Edit: Wait, no. That's not right.
Edit: Ming, is it even possible to solve this problem? Outside of a being 1020. I'm not 100% sure, but if I had to stack my life on it, I would say you proposed a trick question that has no answer.

05292010, 02:12 PM #20
Similar Threads

[NT] How would you solve this problem?
By Little_Sticks in forum The NT Rationale (ENTP, INTP, ENTJ, INTJ)Replies: 16Last Post: 06142010, 09:34 PM 
help me figure out this math problem
By Evan in forum The Fluff ZoneReplies: 5Last Post: 07042008, 10:08 PM 
Can somebody please help me with this math problem?
By disregard in forum The BonfireReplies: 5Last Post: 02102008, 04:54 PM