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Monday, 9 June 2014

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?