word problem- substitution?

[cherica123]

"The sum of the digits of a three-digit number is 9. When the order of the digits is reversed, the newly formed number is 396 greater than the original number. If the leftmost digit is one-third of the middle digit, what is the number?"


I assumed they want me to make a system of equations from this problem and go from there, so I began with:

a+b+c=9
(1/3)a=b

With 'a' being the first digit, etc.....

its the middle part, with the digits being reversed, that i cant figure out an equation for. i'm also not totally convinced im on the right path with my first two equations. if someone could get me going, id really appreciate it!

 

[galactus]

I think I may have something.


Sum of the digits is 9:

\(\displaystyle \L\\a+b+c=9\)

Reverse the digits and they are 396 greater than the original:

\(\displaystyle \L\\100c+10b+a-396=100a+10b+c\)

The leftmost digit is 1/3 the middle digit:

\(\displaystyle \L\\3a=b\)



Now, some algberaic gymnastics:

\(\displaystyle b=\frac{27-3c}{4}\)

\(\displaystyle a=\frac{b}{3}\)


\(\displaystyle \L\\100c+10(\frac{27-3c}{4})+(\frac{\frac{27-3c}{4}}{3})-396=100(\frac{\frac{27-3c}{4}}{3})+10(\frac{27-3c}{4})+c\)

Solving, we find c=5, b=3, and a=1

Let's check it against our criteria:

\(\displaystyle 1+3+5=9\).....check

\(\displaystyle 531-135=396\)....check

and the final, 1 is one-third of 3...check

So, the number is 135

 

[Denis]

Or you can use the "mid digit = 3 times 1st digit" clue to get 3 possibilities only:
13?
26?
39?
Since reversing creates a number higher by 396, then easily seen that
only these are possible, and 135 causes bingo!:
134,135,265,266,396,397