anonymous
  • anonymous
The sum of the digits of a two-digit number is 12. If the digits are reversed, the number is decreased by 18. What is the original number?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
75, right?
anonymous
  • anonymous
I'm not sure.....I can't do it......
anonymous
  • anonymous
your number XY. x+y = 12 (10x+y) -(10y+x) = 18 just solve this system

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anonymous
  • anonymous
|x and y| <=9
anonymous
  • anonymous
@bmp is right
anonymous
  • anonymous
How do you get that? I'm really screwed up.
anonymous
  • anonymous
your number XY. x+y = 12 (10x+y) -(10y+x) = 18 just solve this system
anonymous
  • anonymous
the 2 digit number (XY) can be writen like this: 10*x + y = number XY where x y are decimal notation digits form 0...9
anonymous
  • anonymous
GOT IT
anonymous
  • anonymous
Just think that every base 10 number is of the form (10^0)*a + (10^1)*b + (10^2)*c... and so on, where 10^0 is the zeroth decimal number, 10^1 is the second etc. So a two digit number is written as 10^0*y + 10^1*x, or 10x + y, as @myko pointed out.
anonymous
  • anonymous
I understand and I got 75. Thanks!

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