Sunday Afternoon Maths XVI Answers & Extensions

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

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?