Posted by Haven on Tuesday, May 24, 2011 at 9:08pm.
Let N=number of choir members.
Then for the first night,
N=5m+12
for the second night,
N=6n+1
We look for the smallest number which when divided by 5 gives a remainder of 2, and when divided by 6 gives a remainder of 1.
Possible values:
N=17,22,27,32,37,42,47,52,57,62,67,72,...
N=7,13,19,25,31,37,43,49,55,61,67,72
So the minimum number of choir members is 37. Other possibilities are 67,97,127,...