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Question 1109542: The sum of the digits of a three-digit number is 10. When you switch the hundreds place and the tens place digits the original number is decreased by 90. When you switch the ones and hundreds place digits from the original number the original number is increased by 198. What is the original number?
Answer by greenestamps(13214)   (Show Source):
You can put this solution on YOUR website!
First, a solution using formal algebra....
The 3-digit number is ABC; its value is 100A+10B+C.
The number with the hundreds and tens digits switched is BAC; its value is 100B+10A+C.
The digit with the ones and hundreds digits switched is CBA; its value is 100C+10B+A.
The problem tells us
 (BAC is 90 less than ABC)
(1) 
It also tells us
 (CBA is 198 more than ABC)
(2) 
So and 
Then since the sum of the digits is 10,
So B is 2; A is B+1 = 3; and C is B+3 = 5.
The original number ABC is 325.
We can get to the answer with less work using only a little algebra and some logical reasoning.
The derivation of equation (2) above shows that, whenever a 3-digit numbers has its digits reversed (i.e., the ones and hundreds digits are switched), and the value of the number decreases by 198, it means the ones digit is 2 more than the hundreds digit.
We also know that the sum of all three digits is 10.
The numbers with a digit sum of 10 and a ones digit 2 more than the hundreds digit are
163, 244, and 325
Only one of these, 325, satisfies the other condition of the problem -- that switching the hundreds and tens digit decreases the value of the number by 90.
So the original number is 325.
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