\[n\]
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
\[n=2223\]
.