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:
 

 

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.
 

 

By elimination, only North can be S. This means that d must sit to the right of North (at West):
 

 

A and a are partners. This is only possible if A is South and a is North:
 

 

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:
$$1=1$$
$$1+2=1+2=3$$
$$1+2+4=1+2+3+43=7$$
$$1+2+4+5=1+2+3+4+53=12$$
$$...$$
$$1+2+4+5+...+(3n+1)=\sum_{i=1}^{3n+1}\sum_{i=1}^{n}3i$$
$$=\frac{1}{2}(3n+1)(3n+2)\frac{3}{2}n(n+1)$$
$$=3n^2+3n+1$$
$$1+2+4+5+...+(3n+2)=3n^2+3n+1+(3n+2)$$
$$=3n^2+6n+3$$
$$...$$
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}?
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