- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,678

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,678

- Nov 29, 2013

- 172

Hello.

Again.

There will be an easier way, not?

Solution:

[tex]1 \le a \le 8[/tex]

[tex]0 \le b \le 7[/tex]

[tex]4 \le c \le 1005[/tex]

Supposing: a>b

Regards.

Edit. Misprint

- Nov 29, 2013

- 172

Hello.Hello.

Again.

There will be an easier way, not?

Solution:

[tex]1 \le a \le 8[/tex]

[tex]0 \le b \le 7[/tex]

[tex]4 \le c \le 1005[/tex]

Supposing: a>b

Regards.

Edit. Misprint

Thank you very much.

Evidently, I did not understand well the question.

In addition, I am going to correct another misprint.

Solution:

[tex]1 \le a \le 8[/tex]

[tex]0 \le b \le 7[/tex]

[tex]0 \le c \le 1005[/tex]

[tex]k=1999[/tex]

Regards.

Edit.

- Thread starter
- Admin
- #4

- Feb 14, 2012

- 3,678

Thanks for participatingHello.

Thank you very much.

Evidently, I did not understand well the question.

In addition, I am going to correct another misprint.

Solution:

[tex]1 \le a \le 8[/tex]

[tex]0 \le b \le 7[/tex]

[tex]0 \le c \le 1005[/tex]

[tex]k=1999[/tex]

Regards.

Edit.

- Nov 29, 2013

- 172

Hello.Thanks for participatingmente oscura, but I'm sorry, your answer isn't correct.

Excuse me.

Allow me to take part again. I have checked my calculations and have a mistake.

The longest line:

[tex]For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006} [/tex]

[tex]For : \ a=2 \ , \ b=0 \rightarrow{k_2=245}[/tex]

Added:

[tex]For : \ a=3 \ , \ b=1 \rightarrow{k_3=2}[/tex]

[tex]k_1+k_2+k_3=1253[/tex]

Regards.

- Thread starter
- Admin
- #6

- Feb 14, 2012

- 3,678

Sorry, please try again.Hello.

Excuse me.

Allow me to take part again. I have checked my calculations and have a mistake.

The longest line:

[tex]For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006} [/tex]

[tex]For : \ a=2 \ , \ b=0 \rightarrow{k_2=245}[/tex]

Added:

[tex]For : \ a=3 \ , \ b=1 \rightarrow{k_3=2}[/tex]

[tex]k_1+k_2+k_3=1253[/tex]

Regards.

- Nov 29, 2013

- 172

Sorry, please try again.

[tex]0 \le c \le 1005[/tex]

[tex]c: \ 0,1,2,...,1004,1005[/tex]

It corresponds:

[tex]k: \ 1,3,5,...,2009,2011[/tex]

[tex]For : \ a=2 \ , \ b=0 \rightarrow{k_2=250}[/tex]

[tex]0 \le c \le 249[/tex]

[tex]c: \ 0,1,2,...,248,249[/tex]

It corresponds:

[tex]k: \ 16,24,32,...,2000,2008[/tex]

Solution:

[tex]k_1+k_2=1256[/tex]

Regards.

- Moderator
- #8

- Feb 7, 2012

- 2,702

If $(a,b) = (1,0)$ then we get $1+2c = k$. That generates all the odd numbers, 1 to 2011, total $1006$ numbers.

If $(a,b) = (2,0)$ then we get $16+8c = k.$ That generates all the multiples of 8 except for 8 itself. Total $250$ numbers.

After that, there will be nothing new. If $a$ and $b$ are both even or both odd then we get multiples of 8. If they have opposite signs then we get odd numbers. So the overall total is $1256.$

- Thread starter
- Admin
- #9

- Feb 14, 2012

- 3,678

Yes, that is correct and well done! By the way, I hope you don't mind me asking you to keep plugging away at this problem.

[tex]0 \le c \le 1005[/tex]

[tex]c: \ 0,1,2,...,1004,1005[/tex]

It corresponds:

[tex]k: \ 1,3,5,...,2009,2011[/tex]

[tex]For : \ a=2 \ , \ b=0 \rightarrow{k_2=250}[/tex]

[tex]0 \le c \le 249[/tex]

[tex]c: \ 0,1,2,...,248,249[/tex]

It corresponds:

[tex]k: \ 16,24,32,...,2000,2008[/tex]

Solution:

[tex]k_1+k_2=1256[/tex]

Regards.

Thank youmente oscura's latest version.

If $(a,b) = (1,0)$ then we get $1+2c = k$. That generates all the odd numbers, 1 to 2011, total $1006$ numbers.

If $(a,b) = (2,0)$ then we get $16+8c = k.$ That generates all the multiples of 8 except for 8 itself. Total $250$ numbers.

After that, there will be nothing new. If $a$ and $b$ are both even or both odd then we get multiples of 8. If they have opposite signs then we get odd numbers. So the overall total is $1256.$

- Nov 29, 2013

- 172

Not. On the contrary. I am very grateful, of that one has offered more opportunities, to correct my "miscalculations". Thank you.... By the way, I hope you don't mind me asking you to keep plugging away at this problem.

Regards.