Cumbrian Masons | Cumberland and Westmorland Puzzles page

  ANSWER

With two Brethren (A and B), there is one handshake
   (A with B).

With three Brethren (A, B, and C), there are three handshakes
  (A with B and C; B with C).

With four Brethren (A, B, C, and D), there are six handshakes
  (A with B, C, and D; B with C and D; C with D).

In general, with n+1 Brethren, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.

Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66.
This is the quadratic equation n²+ n -132 = 0. Solving for n, we obtain 11 as the answer and deduce that there were 12 Brethren at the Lodge Meeting. Not a good Lodge Meeting but as they say, "It's quality, not quantity that counts."

Since 66 is a relatively small number, you can also solve this problem with a hand calculator. Add 1 + 2 = + 3 = +... etc. until the total is 66. The last number that you entered (11) is n.