Smallest Number n such that 9 Divides n, 16 Divides n+1, 25 Divides n+2

Find  
\[n\]
  such that 9 divides  
\[n\]
, 16 divides  
\[n+1\]
  and 25 divides  
\[n+2\]
. If 9 divides  
\[n\]
  then  
\[n=9a\]

If 16 divides  
\[n\]
  then  
\[n+1=16b \rightarrow n=16b-1\]

From these two equations we can write  
\[16b-9a=1\]
.
We can find the general solution using General Solution to gcd(a,b)=ax+by and get  
\[a=7+16j, b=4+9j\]
.
Using  
\[n=9a=63+144j\]
.
\[n= 63, \; 207, \; 351, \; 495, \; 639, \; 783, \; 927, \; 1071, \; 1215, \; 1359, 1503, \; 1647, 1791, \; 1935, \; 2079, \; 2223 \]
.
Of these numbers 2223 is the smallest such that when we add 2 it is divisible by 25, so  
\[n=2223\]
.