## Number Growth

A three-digit number grows by $9$ if we exchange the second and third digits and grows by $90$ if we exchange the first and second digits. By how much will it grow if we exchange the first and third digits?
Source: NCTM Mathematics Teacher, January 2006

SOLUTION
Suppose we start with $134$.
Exchanging the digits $3$ and $4$ yields the new number $143$
$100+30+4=100+40+3-10+1$
$\textrm{-}10+1=\textrm{-}9$ the number grows by $9$
Exchanging the digits $1$ and $3$ yields the new number $314$
$100+30+4=300+10+4-200+20$
$\textrm{-}200+20=\textrm{-}180$ the number grow by $180 >> 90$
It looks like the digits have to be consecutive in order to have a smaller growth.
So let’s try $234$.
Exchanging the digits $3$ and $4$ yields the new number $243$
$200+30+4=200+40+3-10+1$
$\textrm{-}10+1=\textrm{-}9$ the number grows by $9$ as before.
Exchanging the digits $2$ and $3$ yields the new number $324$
$200+30+4=300+20+4-100+10$
$\textrm{-}100+10=\textrm{-}90$ the number grows by $90$ as expected.
Exchanging the digits $2$ and $4$ yields the new number $432$
$200+30+4=400+30+2-200+2$
$\textrm{-}200+2=\textrm{-}198$ the number grows by $198$.
The answer seems to be $198$, but we need to prove it.
Suppose $a,b,c$ represent the three digits of a number.
Step 1: $abc$ becomes $acb$
$(100a+10c+b)-(100a+10b+c)=9$
Simplify
$9c-9b=9$
$c-b=1\qquad\qquad (1)$
Step 2: $abc$ becomes $bac$
$(100b+10a+c)-(100a+10b+c)=90$
Simplify
$90b-90a=90$
$b-a=1\qquad\qquad (2)$
Step 3: $abc$ becomes $cba$
$(100c+10b+a)-(100a+10b+c)=99c-99a$
$=99(c-a)$
Adding Eq. $(1)$ to Eq. $(2)$
$c-b=1$
$b-a=1$
————–
$c-a=2$
The growth equals
$99(c-a)=99(2)=198$

Answer: $198$