#### Always a Multiple?

Let the two digit number chosen by \(10a+b\), with \(a\) and \(b\) one digit integers. Reversed this will be \(10b+a\). The sum of these will be \(11a+11b\) which is divisible by 11.

This will work for any integer with an even number of digits. Let our number have \(2n\) digits. It can be written as:

$$\sum_{i=1}^{2n}10^{i-1} a_i $$

Adding it to its reverse, we get:

$$\sum_{i=1}^{2n}10^{i-1} a_i + \sum_{i=1}^{2n}10^{2n-i} a_i = \sum_{i=1}^{2n}(10^{i-1}+10^{2n-i}) a_i $$

\(10^{i-1}+10^{2n-i}\) is divisible by 11 (for \(n \in \mathbb{N}\), \(i \in \mathbb{N}\), \(i\leq 2n\)). This can be shown by induction on \(n\):

If \(n=1\): \(10^{i-1}+10^{2n-i} = 10^{i-1}+10^{2-i}=11\), which is clearly divisible by 11.

Suppose result is true for \(n-1\). Now consider \(10^{i-1}+10^{2n-i}\).

If \(i>1\), then \(10^{i-1}+10^{2n-i}= 10(10^{(i-1)-1}+10^{(2n-2)-(i-1)}\) which is divisible by 11 by the inductive hypothesis.

If \(i=1\), then:

$$10^{i-1}+10^{2n-i} = 1+10^{2n-1} = \sum_{j=1}^{j=2n-2}9\times 10^j +11$$
$$=\sum_{j=1}^{j=n-1}99\times 10^{2j} +11$$
$$=11\left(\sum_{j=1}^{j=n-1}9\times 10^{2j} +1\right)$$

##### Extension

Which numbers with an odd number of digits will be divisible by 11 when added to their reverse?

#### Pocket Money

If Amy gets \(n\)p for pocket money then brother 1 gets \(\frac{n}{2}\)p, brother 2 gets \(\frac{n}{3}\)p, brother 3 gets \(\frac{n}{4}\)p and brother 4 gets \(\frac{n}{5}\)p.

If Tom is brother 1 and Peter is brother 2, then:

$$\frac{n}{2}-\frac{n}{3}=30$$
$$\frac{n}{6}=30$$
$$n=180$$

If Tom is brother 1 and Peter is brother 3, then:

$$\frac{n}{2}-\frac{n}{4}=30$$
$$\frac{n}{4}=30$$
$$n=120$$

If Tom is brother 1 and Peter is brother 4, then:

$$\frac{n}{2}-\frac{n}{5}=30$$
$$\frac{3n}{10}=30$$
$$n=100$$

If Tom is brother 2 and Peter is brother 3, then:

$$\frac{n}{3}-\frac{n}{4}=30$$
$$\frac{n}{12}=30$$
$$n=360$$

If Tom is brother 2 and Peter is brother 4, then:

$$\frac{n}{3}-\frac{n}{5}=30$$
$$\frac{2n}{15}=30$$
$$n=225$$

If Tom is brother 3 and Peter is brother 4, then:

$$\frac{n}{4}-\frac{n}{5}=30$$
$$\frac{n}{20}=30$$
$$n=600$$

So the possible amounts of money Amy could have received are £1.80, £1.20, £1, £3.60, £2.25 and £6.

##### Extension

Which values could the 30p be replaced with and still give a whole number of pence for all the possible answers?