5. Find the smallest three-digit number with the property that if a number, which is 1 greater, is appended to it on the right, then the result (a six-digit number) will be a perfect square.
Answer: 183

Solution: Let the required number be a, then $1000a + a + 1 = n^2$. We can rewrite this as: $1001a = (n - 1)(n + 1)$. Factorize $1001 = 7 \times 11 \times 13$, meaning the product (n-1)(n+1) must be divisible by 7, 11, and 13. Additionally, for the square to be a six-digit number, $n$ must be in the interval $[317; 999]$.

Consider the following cases:

a) $n-1$ is divisible by $143, n+1$ is divisible by 7, then we find $n=573$;

b) $n-1$ is divisible by $7, n+1$ is divisible by 143, then $n=428$;

c) $n-1$ is divisible by $77, n+1$ is divisible by 13, then $n=155$ - does not fit;

d) $n-1$ is divisible by $13, n+1$ is divisible by 77, then $n=846$;

e) $n-1$ is divisible by 91, $n+1$ is divisible by 11, then $n=274$ - does not fit;

f) $n-1$ is divisible by $11, n+1$ is divisible by 91, then $n=727$.

The smallest $n=428, n^2=428^2=183184$.