Project Euler Problem 932
For the year 2025 Given positive integers a and b, the concatenation ab we call a 2025-number if ab = (a+b)^2.
Solution
Answer: 72673459417881349
Let the concatenation $ab$ mean
$$N = a\cdot 10^k + b$$
where $k$ is the number of digits of $b$. A $2025$-number satisfies
$$N=(a+b)^2.$$
Define $x=a+b$. Then
$$x^2 = a\cdot 10^k + b$$
and since $b=x-a$,
$$x^2 = a(10^k-1)+x.$$
So
$$a=\frac{x^2-x}{10^k-1}.$$
Thus for each possible $k$, we need:
- $10^k-1 \mid x(x-1)$,
- $a>0,; b=x-a>0$,
- $b$ has exactly $k$ digits,
- $x^2$ has at most 16 digits.
This structure makes the search finite and efficient because $x(x-1)\equiv 0\pmod{10^k-1}$, so $x$ must be congruent to either $0$ or $1$ modulo each prime-power factor of $10^k-1$. Using CRT to enumerate all such residues and summing all valid $2025$-numbers up to 16 digits gives:
$$T(4)=5131$$
matching the problem statement, confirming correctness.
Therefore,
Answer: 72673459417881349