### Exhaustion

Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2

# Code to Zero

##### Age 16 to 18Challenge Level

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 $598$.