Question 622057
The three digits are a, b, c
then
100a + 10b + c = "the number"
:
Write an equation for each statement:
:
"The tens digit is greater than the ones digit,"
b > c
:
"the sum of the digits is 9,"
a + b + c = 9
:
" and if the digits are reversed and the resulting three-digit number is subtracted from the original number, the difference is 198."
100a + 10b + c -(100c + 10b + a) = 198
Removing the brackets changes the signs
100a + 10b + c - 100c - 10b - a = 198
Combine like terms
100a - a + 10b - 10b + c - 100c = 198
99a - 99c = 198
simplify, divide by 99
a - b = 2
a = b+2
:
Replace a with (b+2) in the sum equation
(b+2) + b + c = 9
2b + c = 9-2
2b + c = 7
From this assume b=3, then c=1 and a=5
531 is the number