Question 891368
let a = the 100's digit
let b = the 10's
let c = the units
then
100a+10b+c = the original number
:
write an equation for each equation, simplify as much as possible.
:
if the first and third digits of a 3 digit number are interchanged, the difference between the original number and the new one 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 - c = 2
c = a - 2; use for substitution
:
 If the first and second digits of the original are interchanged the new one is 90 less than the original
100a + 10b + c - (100b + 10a + c) = 90
do the same here
100a - 10a + 10b - 100b + c - c = 90
90a - 90b = 90
divide by 90
a - b = 1
b = a - 1; use for substitution
:
The difference between the first and third digits is equal to the second digit.
a - c = b
Replace replace c with (a-2); replace b with (a-1)
a - (a-2) = a - 1
a - a + 2 = a - 1
2 + 1 = a
a = 3
then using our equations replace a with 3
c = 3 -2
c = 1
and
b = 3 - 1
b = 2
:
321 is the original number
:
:
See if that works in the statement," If the first and second digits of the original are interchanged the new one is 90 less than the original
321 - 231 = 90