Asked by Anonymous
                the sum of the digits of three digit number is 14 of hundreds and tens digits are reserved the resulting number is 90 more than the original number if the tens and units digit are resulting number is 27 more than the original number find the original number
            
            
        Answers
                    Answered by
            mathhelper
            
    very sloppy typing, so I have to guess what you mean
Hundreds digit --- x
tens digit --- y
unit digit --- z
So the number is 100x + 10y + z
and we are told that x+y+z = 14
the hundred and tens digit are reversed, so
100y + 10x + z - (100x + 10y + z) = 90
90y - 90x = 90
y - x = 1 ---- x = y-1
the tens digit and the unit are reversed, at least that's what I think you are
trying to say:
(100x + 10z + y) - (100x + 10y + z) = 27
9z - 9y = 27
z - y = 3 ----> z = y+3
sub back into the first
x+y+z = 14
x + y + y-3 = 14
y-1 + y + y+3 = 14
3y = 12
y = 4
then x = 3 and z = 7
the original number was
100x + 10y + z
= 300 + 40 +7
= 347
    
Hundreds digit --- x
tens digit --- y
unit digit --- z
So the number is 100x + 10y + z
and we are told that x+y+z = 14
the hundred and tens digit are reversed, so
100y + 10x + z - (100x + 10y + z) = 90
90y - 90x = 90
y - x = 1 ---- x = y-1
the tens digit and the unit are reversed, at least that's what I think you are
trying to say:
(100x + 10z + y) - (100x + 10y + z) = 27
9z - 9y = 27
z - y = 3 ----> z = y+3
sub back into the first
x+y+z = 14
x + y + y-3 = 14
y-1 + y + y+3 = 14
3y = 12
y = 4
then x = 3 and z = 7
the original number was
100x + 10y + z
= 300 + 40 +7
= 347
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