How many two-digit integers are increased by exactly nine when the digits are reversed?
K kavitha New member Joined Dec 23, 2010 Messages 17 Jan 2, 2011 #1 How many two-digit integers are increased by exactly nine when the digits are reversed?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Jan 2, 2011 #2 The 'forward' number would be \(\displaystyle 10a+b\) When reversed it is \(\displaystyle 10b+a\) There difference is 9: \(\displaystyle 10a+b-10b-a=9\) \(\displaystyle 10(a-b)-(a-b)=9\) \(\displaystyle 9(a-b)=9\) \(\displaystyle a-b=1\) A difference of 9 occurs when a and b differ by 1. Such as 12 and 21. Count how many up to 99
The 'forward' number would be \(\displaystyle 10a+b\) When reversed it is \(\displaystyle 10b+a\) There difference is 9: \(\displaystyle 10a+b-10b-a=9\) \(\displaystyle 10(a-b)-(a-b)=9\) \(\displaystyle 9(a-b)=9\) \(\displaystyle a-b=1\) A difference of 9 occurs when a and b differ by 1. Such as 12 and 21. Count how many up to 99
