Solve this Math Problem? (With Proof!)

[Ming]

Three identical integers greater than 1 are written on a board. One of the numbers is rubbed off and replaced by a number which is one less than the sum of the two. This process is repeated a number of times. The number rubbed off is always chosen so that it is different from its replacement.

Adam wants to get the number 2010 from as many starting triples a, a, a as possible. One way (obviously) is to start with 2010, 2010, 2010. What are all the other ways?

 

[CrystalViolet]

Are you trying to get some one to do your home work for you ?Just teasing, LOL

 

[JAVO]

Adam should really find a more interesting and useful problem to solve.

 

[Ming]

Are you trying to get some one to do your home work for you ?Just teasing, LOL

Maybe, maybe.

I was just interested. Testing the NT's intelligence mainly.

 

[Lateralus]

I'm not sure I even understand the problem.

2a-1=2010 a=1005.5 (not an integer)
3a-2=2010 a=670.666 (not an integer)
4a-3=2010 a=503.25 (not an integer)
5a-4=2010 a=402.8 (not an integer)
6a-5=2010 a=335.8333 (not an integer)
7a-6=2010 a=288
8a-7=2010 a=252.125 (not an integer)
9a-8=2010 a=224.2222 (not an integer)
10a-9=2010 a=201.9 (not an integer)
11a-10=2010 a=183.6363 (not an integer)
12a-11=2010 a=168.4166 (not an integer)
13a-12=2010 a=155.538 (not an integer)
...

Do I really have to keep doing this?

 

[forzen]

Same here, I don't think I even understood the problem.

2009 - n, 0002 + n, 2010

1<n<2008

 

[Ming]

Alright, I'll try to explain what it means.

So you choose an integer that is greater than one, and call it a.

You repeat it 3 times, and write it on the blackboard.

You rub out one of the integers that you have written, and replace it with the one less than SUM the two remaining integers.

For example, I start with 3,3,3.

I rub out a 3.

I'm left with 3,_,3.

I fill in the blank with one less than the sum of these two. (Which is 3+3-1 = 5)

3,5,3.

From that point onwards, I take away the 3 again.

_,5,3.

I get 7,5,3. (Since 5+3-1 = 7)

And etc, etc...

Until one or more of the numbers is 2010.

The question is asking how many of a as an integer can we have? Basically how many a,a,a possibilities that are integers can we have so that one of the numbers end up to be 2010.

Get it?

 

[InsatiableCuriosity]

Is this not the basis of the Fibonaci Sequence?

 

[forzen]

Three identical integers greater than 1 are written on a board. One of the numbers is rubbed off and replaced by a number which is one less than the sum of the two. This process is repeated a number of times. The number rubbed off is always chosen so that it is different from its replacement.

Adam wants to get the number 2010 from as many starting triples a, a, a as possible. One way (obviously) is to start with 2010, 2010, 2010. What are all the other ways?

I read the bold part wrong, because if you do 3,3,3 it would give you 3,5,3.

3+3=6

One less then the sum of the two is 5.

 

[Lateralus]

The series is ba-b+1=2010 which gives you b=2009/(a-1)

It looks like you're asking for the factors of 2009 and adding 1 to them.

8, 42, 50, 288

 

[Ming]

The series is ba-b+1=2010 which gives you b=2009/(a-1)

It looks like you're asking for the factors of 2009 and adding 1 to them.

8, 42, 50, 288

You got it?

But may I ask, where does that 'b' come from?

And what the hell did you do? Kinda skipped a bit at the start

 

[Ming]

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?

 

[Lateralus]

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 2a-1. The third becomes 3a-2, the fourth 4a-3. You should see a pattern with the numbers. Take 4a-3. If you substitute "b" in the place of 4, and b-1 in the place of 3, (which works for all expressions in this series), you get:

b*a - (b-1) = 2010

You can then simplify the expression to:

b*a-b=2009

Then factor out b:

b(a-1)=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.

 

[Lateralus]

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?

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?

 

[Ming]

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?

You would be concerned with 495, 957, etc. etc. It's not exclusive.

 

[Lateralus]

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.

 

[Ming]

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.

It's possible, try again. (I'm not joking..)

 

[Lateralus]

12 2 9 8 10 6 1 11 4 5 3 7

 

[Little_Sticks]

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.

 

[forzen]

The series is ba-b+1=2010 which gives you b=2009/(a-1)

It looks like you're asking for the factors of 2009 and adding 1 to them.

8, 42, 50, 288

nice!!