This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### 2009

\(2009=7\times 7\times 41\), so there are four possible sets of dimensions of the cuboid:

$$1\times 1\times 2009$$
$$1\times 7\times 287$$
$$1\times 41\times 49$$
$$7\times 7\times 41$$
In the first three cuboids, there is a face with an area of 2009 units (\(2009\times 1\), \(7\times 287\) and \(41\times 49\) respectively) and so 2009 stickers will not be enough. Therefore the cuboid has dimensions \(7\times 7\times 41\) and a surface area of 1246, leaving

**764**stickers left over##### Extension

For which numbers \(n\) can a cuboid be made with \(n\) unit cube such that \(n\) unit square stickers can cover the faces of the cuboid.

#### 3\(n\)+1

**(i)**Let \(a,b\in S\). Then \(\exists \alpha,\beta\in \mathbb{N}\) such that \(a=3\alpha+1\) and \(b=3\beta+1\).

(This says that if \(a\) and \(b\) are in \(S\) then they can be written as a multiple of three plus one.)

$$a\times b=(3\alpha+1)\times (3\beta+1)$$
$$=9\alpha\beta+3\alpha+3\beta+1$$
$$=3(3\alpha\beta+\alpha+\beta)+1$$
This is a multiple of three plus one, so \(a\times b\in S\).

**(ii)**No, as \(36\times 22=4\times 253\) and 36,22,4 and 253 are all irreducible.

##### Extension

Try the task again with \(S=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}\).

#### The Ace of Spades

The problem can be expressed using the following probability tree:

The probability that the card turned over from C is an Ace of Spades is:

$$\frac{1\times 2\times 2+1\times 50\times 1+50\times 1\times 2+50\times 51\times 1}{51\times 52\times 53}$$
$$=\frac{52}{51\times 53}$$
Surely $2009 \neq 7^4$...

ReplyDeleteYes, I mixed up 49 and 41. I've corrected it

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