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## Monday, 18 August 2014

### Sunday Afternoon Maths XXV Answers & Extensions

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

#### Whist

Represent each player by a compass point. Let B, D, A and S represent Messrs. Banker, Dentist, Apothecary and Scrivener respectively and b, d, a and s represent the four jobs.
Mr. Banker (B) partners the scrivener (s). Let B sit at West. This means s sits at East. As no other player can also be B or s, the table looks like this:
 North B D A S b d a s
 West B D A S b d a s
 East B D A S b d a s
 South B D A S b d a s
The dentist (d) sits on Mr. Scrivener's (S) right. East cannot be S, so North cannot be d. East cannot be d, so South cannot be S.
 North B D A S b d a s
 West B D A S b d a s
 East B D A S b d a s
 South B D A S b d a s
By elimination, only North can be S. This means that d must sit to the right of North (at West):
 North B D A S b d a s
 West B D A S b d a s
 East B D A S b d a s
 South B D A S b d a s
A and a are partners. This is only possible if A is South and a is North:
 North B D A S b d a s
 West B D A S b d a s
 East B D A S b d a s
 South B D A S b d a s
Therefore, Mr. Banker the dentist sits to the left of the banker.
##### Extension
If each person is partnered with their job namesake, how many possible combinations of names and jobs are possible?

#### Polya Strikes Out

1, 3, 7, 12, 19, ...
1, 7, 19, ...
1=1; 1+7=8; 1+7+19=27; ...
1, 8, 27, ...
The final sequence is the cube numbers. To show why, let $$n$$ be an integer and follow through the process.
Cross out every third number:
1, 2, 3, 4, 5, 6, ..., 3n, $$3n+1$$, $$3n+2$$, ...
1, 2, 4, 5, ..., $$3n+1$$, $$3n+2$$, ...
Find the cumulative sums:
$$1=1$$ $$1+2=1+2=3$$ $$1+2+4=1+2+3+4-3=7$$ $$1+2+4+5=1+2+3+4+5-3=12$$ $$...$$ $$1+2+4+5+...+(3n+1)=\sum_{i=1}^{3n+1}-\sum_{i=1}^{n}3i$$ $$=\frac{1}{2}(3n+1)(3n+2)-\frac{3}{2}n(n+1)$$ $$=3n^2+3n+1$$ $$1+2+4+5+...+(3n+2)=3n^2+3n+1+(3n+2)$$ $$=3n^2+6n+3$$ $$...$$
1, 3, 7, 12, ..., $$3n^2+3n+1$$, $$3n^2+6n+3$$, ...
Cross out every second number, starting with the second:
1, 3, 7, 12, ..., $$3n^2+3n+1$$, 3n2+6n+3, ...
1, 7, ..., $$3n^2+3n+1$$, ...
Find the cumulative sums. The $$m$$th sum is:
$$\sum_{n=0}^{m}3n^2+3n+1$$ $$=3\sum_{n=0}^{m}n^2+3\sum_{n=0}^{m}n+\sum_{n=0}^{m}1$$ $$=\frac{3}{6}m(m+1)(2m+1)+\frac{3}{2}m(m+1)+m+1$$ $$=\frac{1}{2}(m+1)(m(2m+1)+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+m+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+4m+2)$$ $$=(m+1)(m^2+2m+1)$$ $$=(m+1)(m+1)^2$$ $$=(m+1)^3$$
Hence the numbers obtained are the cube numbers.
##### Extension
What happens if you cross out every third number starting at the second? Or every fifth number starting at the fifth? Or every $$n$$th number starting at the $$m$$th?