Code to Zero
Vassil of Lawnswood High School in Leeds sent in a good solution. He worked systematically through the possible values of c using the rearrangement c(c$^2$ -1) = b(10-b) + 99a to establish the range of possible values for a and then testing to find values of b which fitted.
Many thanks to Sue Liu who is in S6 at Madras College in St Andrews for the followingbeautifully neat solution.
To find all 3-digit numbers $abc$ (in base $10$) such that
$$ a + b^2 + c^3 = 100a + 10b + c $$ Rearranging gives $$c^3 - c - 99a = b (10 - b)$$ $$c(c+1)(c-1) - 99a = b (10 - b)$$
For any three consecutive integers one of them is divisible by
$3$. Since $3$ divides $99$ it follows that $3$ divides $b(10-b)$.
Since $3$ is a prime this limits the possible choices of $b$:
Either
| $b=0 10-b=10$ | $b(10-b) = 0$ | |
| $b=3 10-b=7$ | or $b=7 10-b=3$ | $b(10-b) = 21$ |
| $b=6 10-b=4$ | or $b=4 10-b=6$ | $b(10-b) = 24$ |
| $b=9 10-b=1$ | or $b=1 10-b=9$ | $b(10-b) = 9$ |
Hence the possible values of $b(10-b)$ are $0$, $9$, $21$ and $24$.
We now have to find a multiple of $99$ which when subtracted
from a product of $3$ consecutive natural numbers gives $0$, $9$,
$21$ or $24$.
Since $a$ is at least $1$ $c(c+1)(c-1)$ is at least $99$ so $c$ is
at least $5$.
| $c$ | $c(c+1)(c-1)$ | $a$ | $99a$ | $c(c+1)(c-1)-99a$ |
|---|---|---|---|---|
| $5$ | $120$ | $1$ | $99$ | $21$ |
| $6$ | $210$ | $2$ | $198$ |
$12$
|
| $7$ | $336$ | $3$ | $297$ | $39$ |
| $8$ | $504$ | $5$ | $495$ | $9$ |
| $9$ | $720$ | $7$ | $693$ | $27$ |
(Since $0$, $9$, $21$, $24 < 99$ only the multiple of $99$ which is closest to $(c+1)c(c-1)$ needs to be checked.)
From the table we can see that the following are the possibilities for $a$, $b$ and $c$:
-
$a=1$ $b=3$ or $b=7$ $c=5$
-
$a=5$ $b=1$ or $b=9$ $c=8$
giving the solutions $135$, $175$, $518$ and $598$.
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