# Thread: Solve this Math Problem? (With Proof!)

1. ## Solve this Math Problem? (With Proof!)

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?

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

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

4. Originally Posted by FireyPheonix
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.

5. 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?

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

2009 - n, 0002 + n, 2010

1<n<2008

7. 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.

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?

8. Is this not the basis of the Fibonaci Sequence?

9. Originally Posted by 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?
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.

10. 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

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