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# Cuboids Poster

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Age 11 to 14

Challenge Level

- Problem
- Student Solutions

The only two solutions are the cuboid with edges $1, 2, 16$ and the cuboid with edges $2, 4, 7$

Use a systematic process to check that there are no more.

Edge lengths $a, b, c$ where $a$ is the shortest edge, $\therefore$ faces have area $ab, bc, ac$

Surface area $2ab + 2bc + 2ac = 100 \Rightarrow ab + bc+ ac = 50$

__Method 1 - rearranging__

If $a=1,$ then $bc + b + c = 50$

Rearrange to get $c:$

$c(b+1) = 50-b$

$c = \dfrac{50-b}{b+1}$

So if $b\leq c$, trying out values of $b$, $c$ could be:

$\frac{49}{2}, \frac{48}{3}, \frac{47}{4}, \frac{46}{5}, \frac{45}{6}, \frac{44}{7}$

$c = \frac{44}{7} = 6\frac27$ when $b = 7$, but $b$ should be $\leq c$ so we have tried all possibilities.

Only $\frac{48}{3} = 16$ is a whole number, $\therefore 1, 2, 16$ is the only possible cuboid when $a=1$

If $a=2,$ then $bc + 2b + 2c = 50$

$c = \dfrac{50-2b}{b+2}$ (or equivalently $b = \dfrac{50-2c}{c+2}$)

$c$ could be:

$\frac{48}{3}, \frac{46}{4}, \frac{44}{5}, \frac{42}{6} = 7, \frac{40}{7}, \frac{38}{8}$

$c = \frac{38}{8} = 4\frac68$ when $b = 6$, but $b$ should be $\leq c$ so we have tried all possibilities.

$\therefore 2, 4, 7$ is the only possible cuboid when $a=2$

If $a=3$, then $bc + 3b + 3c = 50$

$c = \dfrac{50-3b}{b+3}$

$c$ could be $\frac{47}{4}, \frac{44}{5}, \frac{41}{6}, \frac{38}{7}, \frac{35}{8} = 4\frac38\lt 5$ so there are no cuboids with $a=3$

If $a=4$, then $bc + 4b + 4c = 50$

$c = \dfrac{50-4b}{b+4}$

$c$ could be $\frac{46}{5}, \frac{42}{6}, \frac{38}{7}, \frac{34}{8}, \frac{30}{9} = 3\frac39\lt 5$ so there are no cuboids with $a=4$

If $a=5$ then $ac, bc, ac$ are all $\geq25$, so $ac+bc+ac\geq75\gt50$ so there are no cuboids with $a\geq5$

__Method 2 - almost-factorising__

If $a=1$ then $bc + b + c = 50$

$bc + b + c$ "almost-factorises": $(b+1)(c+1) = bc + b + c + 1$

$\begin{align}\therefore bc + b + c &= 50\\

\Rightarrow bc + b + c + 1 &= 51\\

\Rightarrow (b+1)(c+1) &= 51\end{align}$

$b$ and $c$ are whole numbers so use factors of $51$

$b+1$ | $c+1$ | $b$ | $c$ |
---|---|---|---|

1 | 51 | 0 | |

3 | 17 | 2 | 16 |

Only cuboid for $a=1$ is $1, 2, 16$

If $a=2$ then $bc+2b+2c = 50$

$(b+2)(c+2) = bc + 2b + 2c + 4$

$(b+2)(c+2)=54$

$b+2$ | $c+2$ | $b$ | $c$ |
---|---|---|---|

2 | 27 | 0 | |

3 | 18 | 1 | 16 |

6 | 9 | 4 | 7 |

Only new cuboid is $2, 4, 7$

If $a=3$ then $bc+3b+3c = 50$

$(b+3)(c+3)=59$ which is prime

If $a=4$ then $bc+4b+4c = 50$

$(b+4)(c+4)=66$ which is prime

$b+2$ | $c+2$ | $b$ | $c$ |
---|---|---|---|

6 | 11 | 2 | 7 |

No new cuboids

If $a=5$ then $ac, bc, ac$ are all $\gte25$, so $ac+bc+ac\geq75\gt50$ so there are no cuboids with $a\geq5$