Question 316464
{{{n1=100a+10b+c}}}
{{{n2=100b+10a+c}}}
.
.
{{{n2=1.2n1}}}
{{{10n2=12n1}}}
{{{1000a+100b+10c=1200b+120a+12c}}}
.
.
{{{880a-1100b-2c=0}}}
{{{440a-550b-c=0}}}
{{{c=440a-550b}}}
.
.
a,b,c must all be integers.
a,b must be between 1 and 9.
c must be between 0 and 9.
Since c must be positive or zero, 
{{{440a>=550b}}}
{{{a>=(5/4)b}}}
Since a must take on integer values, then
{{{a>= b+1}}}
Let {{{a=b+1}}}
{{{c=440(b+1)-550(b)}}}
{{{c=440b+440-550b}}}}
{{{c=440-110b}}}
{{{c=0}}} when {{{b=4}}} and then {{{a=5}}}
Let's look at other values,
Let {{{a=b+2}}}
{{{c=440(b+2)-550b}}}
{{{c=440b+880-550b}}}
{{{c=880-110b}}}
{{{c=0}}} when {{{b=8}}} but then {{{a=10}}}
So that solution is not valid.
No other solutions exist because 
{{{c=440(b+3)-550b}}}
would not have a solution in the valid ranges for a,b, and c.
.
.
.

The original number is 450 and the new number is 540.