SOLUTION: find a three digit number such that if the digits at the tens place and the hundred place are reversed then the number obtained is twenty percent greater than the original number

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Question 316464: find a three digit number such that if the digits at the tens place and the hundred place are reversed then the number obtained is twenty percent greater than the original number
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
n1=100a%2B10b%2Bc
n2=100b%2B10a%2Bc
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n2=1.2n1
10n2=12n1
1000a%2B100b%2B10c=1200b%2B120a%2B12c
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880a-1100b-2c=0
440a-550b-c=0
c=440a-550b
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a,b,c must all be integers.
a,b must be between 1 and 9.
c must be between 0 and 9.
Since c must be positive or zero,
440a%3E=550b
a%3E=%285%2F4%29b
Since a must take on integer values, then
a%3E=+b%2B1
Let a=b%2B1
c=440%28b%2B1%29-550%28b%29
c=440b%2B440-550b}
c=440-110b
c=0 when b=4 and then a=5
Let's look at other values,
Let a=b%2B2
c=440%28b%2B2%29-550b
c=440b%2B880-550b
c=880-110b
c=0 when b=8 but then a=10
So that solution is not valid.
No other solutions exist because
c=440%28b%2B3%29-550b
would not have a solution in the valid ranges for a,b, and c.
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The original number is 450 and the new number is 540.