Code to Zero

Age 16 to 18

Challenge Level Yellow star

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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Code to Zero

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