7. 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 write it as: \(1001a = (n - 1)(n + 1)\). Factorize \(1001 = 7 \times 11 \times 13\), so the product \((n - 1)(n + 1)\) must be divisible by 7, 11, and 13. Moreover, 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 \(n = 428, n^2 = 428^2 = 183184\).