Originally posted by adam warlockThe question was why do the sums always equal 9. If you understand the theory behind finite Arithmetic Series you'll see that I've answered your question.
It doesn't.
First it isn't rigorous. Second I know about positional notation and how different numeral systems work (so what you said didn't bring anything new to me).
What I wanted was a rigorous proof of why this particular formula (and its generalizations) always gets n-1 as its final result.
If one number increases by one and the other decreases by one, the sum will remain the same. The initial sum is 9, so all other sums will be 9.
Can I help it if you guys aren't really bright enough to understand my explanation?
You asked why the pattern worked for 9's as I quoted in my post. I answered your question.
30 Jan 15
Originally posted by EladarEladar my answer to you wasn't a dig. It was just a straightforward answer.
The question was why do the sums always equal 9. If you understand the theory behind finite Arithmetic Series you'll see that I've answered your question.
If one number increases by one and the other decreases by one, the sum will remain the same. The initial sum is 9, so all other sums will be 9.
Can I help it if you guys aren't really bright enough ...[text shortened]... ?
You asked why the pattern worked for 9's as I quoted in my post. I answered your question.
Take this sentence of yours:
If one number increases by one and the other decreases by one, the sum will remain the same. The initial sum is 9, so all other sums will be 9.
That much is obvious to anyone and that's why I inserted the bit about the multiplication table of 9 in my first post. The thing I was looking for why such a pattern exists in the first place. And nothing of what you said answered that question.
You asked why the pattern worked for 9's as I quoted in my post. I answered your question.
Actually right in my first post I also mentioned that the formula 9*(x+1) always provided the final result of 9 because we were working in the decimal number system and that if we worked on other number systems we would get analogous results (for instance by working in base 5 number system you would always get 4 as your final answer).
Can I help it if you guys aren't really bright enough to understand my explanation?
The problem is that you didn't understand what we're talking about not that we didn't understand what you're talking about.
That much is obvious to anyone and that's why I inserted the bit about the multiplication table of 9 in my first post. The thing I was looking for why such a pattern exists in the first place. And nothing of what you said answered that question.
I explained why.
If you add two numbers together, then add 1 to one number but subtract 1 from the other you will get the same result.
Here is your two digit number number:
10x+y
x + y = 9
(x+1)+(y-1) = 9
Since adding 9 is actually the same thing as adding 1 to the tens digit and subtracting 1 from the ones digit you are doing exactly that.
9+9=18 in other words 09 + 9 = 10(0+1) + 1(9-1) AKA 18
18 + 9 = 10(1+1) + 1(8-1) AKA 27
10x+y + 9 = 10(x-1)+1(y+1)
x + y = (x-1)+(x+1)
31 Jan 15
Originally posted by EladarWhy does x+y=9 ?
Here is your two digit number number:
10x+y
x + y = 9
(x+1)+(y-1) = 9
Since adding 9 is actually the same thing as adding 1 to the tens digit and subtracting 1 from the ones digit you are doing exactly that.
9+9=18 in other words 09 + 9 = 10(0+1) + 1(9-1) AKA 18
18 + 9 = 10(1+1) + 1(8-1) AKA 27
10x+y + 9 = 10(x-1)+1(y+1)
x + y = (x-1)+(x+1)
31 Jan 15
Originally posted by wolfgang59Because we are talking about multiples of 9 and the first multiple of 9 is 9.
Why does x+y=9 ?
9= 10(0) + (1)9
That's why the in first case x+y = 9
Now look back at my earlier explanation as to why all other multiples of 9 also result in x+y=9.
31 Jan 15
Originally posted by EladarOK.
Because we are talking about multiples of 9 and the first multiple of 9 is 9.
9= 10(0) + (1)9
That's why the in first case x+y = 9
Now look back at my earlier explanation as to why all other multiples of 9 also result in x+y=9.
You are using a sort of proof by induction.
You should state the base case then the inductive step ... that
would be clearer. Though I think your inductive steps are
only crystal clear for 2 digit numbers.
Originally posted by wolfgang59The question was only asked for the first 10 multiples.
OK.
You are using a sort of proof by induction.
You should state the base case then the inductive step ... that
would be clearer. Though I think your inductive steps are
only crystal clear for 2 digit numbers.
As a matter of fact the entire topic breaks down once you hit 99 which adds up to 18, then 108 where the final two digits start adding up to 8.
03 Feb 15
Originally posted by EladarDoesn't it just suck when an idiot like Eladar is able to answer your question?
The question was only asked for the first 10 multiples.
As a matter of fact the entire topic breaks down once you hit 99 which adds up to 18, then 108 where the final two digits start adding up to 8.
15 Feb 15
Originally posted by sonhouseSometimes I get in the mood to play, sometimes I take time off. I started playing chess when I was 38 and have muddled around since.
How come you stopped making chess moves here? Your profile says you haven't moved in 152 days. Surely the players here weren't THAT boring.
16 Feb 15
Originally posted by EladarWhat stopped me this time was that my job made it difficult for me to make move daily and I don't like playing slow games. Better not to play than to make quick moves without any thought. I do that naturally anyhow, no need to play under conditions that would make it more likely.
Yeah well I'm like that. I'll go a year or two without playing then play for a year or two.