Question 7795: When the digits of a two-digit number are reversed, the new number is 9 more than the original number, and the sum of the digits of the original number is 11. What is the original number?
Answer by prince_abubu(198)   (Show Source):
You can put this solution on YOUR website! When we have a two digit number, it can be expressed as 10a + b. Actually, let's take the number 82. That's 80 + 2 which really is 10*8 + 2. The a = 8, and the b = 2.
So, let's say that we don't know the original 2-digit number. We have to represent that as 10a + b.
The first thing they say is that when the digits are reversed, the new number is 9 more than the original number. If you reverse the digits of 10a + b, it will be 10b + a. They said that new number is 9 more. This means that the new number is greater, which then means that we must add 9 to the old number so that it will equal the new number. OLD + 9 = NEW. OLD was the 10a + b and NEW was the 10b + a. Let's plug those in:
 <----- Relationship between a two digit number and the number that results in reversing the digits.
They said that the sum of the digits of the original number is 11. So a + b = 11. Let's solve for b: b = 11 - a. Now substitute 11 - a for b in the big equation:
 <---- Simplified
 <---- added 9a to both sides and subtracted 20 from both sides.
 <---- if a + b = 11, and a = 5, then b must be 6. If a = 5 and b = 6, 10a + b would give you 56, the original number. If you reverse 56, you get 65, which indeed is 9 more than the original number.
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