Problem source

Let $k$ be an integer satisfying $0\le k \le 9\,$.
Show that  $0\le 10k-k^2\le 25$ and that $k(k-1)(k+1)$ is divisible by $3\,$.
For each $3$-digit number $N$, where $N\ge100$, let $S$  be the sum of the hundreds digit, the square of the tens digit and the cube of the units digit. Find the numbers $N$ such that $S=N$.
[Hint: write $N=100a+10b+c\,$ where $a\,$, $b\,$ and $c$ are the digits of $N\,$.]

Solution source

First note that $10k - k^2 = 25 - (5-k)^2$ which is clearly bounded above by $25$. The smallest it can be is when $|5-k|$ is as large as possible, ie when $k =0$ and we get a lower bound of $0$.

For $(k-1)k(k+1)$ notice this is the product of $3$ consecutive integers, and therefore must be divisible by $3$. (In fact, it's divisible by six, since $\binom{k+1}{3}$ is the number of ways to choose $3$ objects from $k+1$.

Let $N = 100a + 10b + c$ where $0 \leq a,b,c < 10$ and $1 \leq a$. $S = a + b^2 + c^3$ we want to find

\begin{align*}
&& 100a +10b + c &= a + b^2 + c^3 \\
\Rightarrow && 0 &= \underbrace{99a}_{3 \mid 99 } + 10b - b^2 -\underbrace{c(c+1)(c-1)}_{3 \mid c(c+1)(c-1)} \\
\end{align*}

Therefore $3 \mid 10b - b^2 = b(10-b)$. Therefore $3 \mid b$ or $3 \mid 10-b$ so $b = 0, 3, 6, 1, 4, 7$

We also have $99a \geq 99$ and $10b-b^2 \in [0, 25]$ so we need $c^3-c \geq 99$, so $c \geq 5$

Case $c = 5$, Then $c^3-c = 120$ so $a = 1$ and $10b-b^2 = 21$ so  $b= 3, 7$
$N = 135, 175$ 

Case $c = 6$, so $c^3 - c = 210$ so $a = 2$ and $25-(5-k)^2 = 12$ so no solutions.

Case $c = 7$, so $7^3 - 7 = 336$ so $a = 3$ and $25-(5-k)^2 = 39$ so no solutions.

Case $c = 8$ so $8^3-8 = 504$ so $a = 5$ and $25-(5-k)^2 = 9$, so $b = 1, 9$ and
$N = 518, 598$

Case $c = 9$ so $9^3 - 9 = 720$, so $a = 7$ and $25-(5-k)^2 = 27$ so no solutions.

Therefore all the solutions are $N = 135, 175, 518, 598$