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May 26, 2015

May 26, 2015

Posted by **Haven** on Tuesday, May 24, 2011 at 9:08pm.

- Math -
**MathMate**, Tuesday, May 24, 2011 at 9:26pmLet 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,...