9x = ab where 1 <= x <= 9 and x is an integer tand a,b

>>9162726
Wait, so you're claiming [math] 9x = (x-1)(10-x) ~ 1 \leq x \leq 9 [/math]? Because that doesn't work, put x=1 to see. You can recast you little equation in the form [math] x^2 -2x +10= 0 [/math] which has a discriminant that's less than 0, so it has no real solutions.

>>9162745
No, im saying that a & b are the digits, they're not multiplied together.
Look
9*5 = (5-1) (10-5)
= 45

Or
9*7 = (7-1)(10-7)
= 63
See?

Not really, for the one digit integers, multiplying them by 9 give us that integer multiplied by 10 minus the integer (10-x), because any number (from 1 to 9) it's obvious that the first integer is the integer minus one, for the 10-x we have the same reasoning

>>9162827
So say 9x=10a+b for natural numbers a,b

>>9162827
(5-1)(10-5) = 9*4 and that most definitely does not equal 9*4

Are you high?

>>9162873
*does not equal 9*5

>>9162726
ab (where a and b are digits) is another way to say 10*(x - 1) + (10 - x)
which equals 10x - 10 + 10 - x = 9x
You fucking brainlet

>>9162726
What you've discovered is ambiguous notation, most likely. A lot of times when I'm tutoring future primary school teachers, we'll give them a problem like this and it just blows their minds. Here's the idea:

Normally, when we write ab, we mean that a and b are numbers which we are then to multiply to get another value, say x, and we'll just denote x by ab.

However, if someone tells you that they have a 3 digit number xyz where x,y,z are digits, they aren't talking about multiplication, they're talking about the number which is x*100+y*10+z. It is just useful notation to get the idea across by writing xyz.

In case you already knew that and were just impressed with the fact that if I wanted to multiply 9*7 i could just reduce 7 by 1 to get 6 then find out 6 plus what equals 9 (it's 3) then put those together to get 63, then congratulations! I like this pattern too. No, you did not find a new mathematical law. We can prove this by examining the equation you wrote, 9x=ab. First, we'll rewrite in a way which we can manipulate algebraically, i.e. 9x=a10+b. we know 9=10-1, so let's write this as (10-1)x=a10+b. Then 0=10a-10x+b+x=(a-x)10+(b+x). This is true only if a-x=-1 and b+x=10. That is, a=x-1 and b=10-x. Hope that clears things up!