Each member of a group of 37 students was sent to one of three rooms.
Later, each was asked "How many other students were in your room?" The average of these responses was 12.
How is this possible?
Let the three groups respectively contain x,y and z students. Then by conditions of the problem:
x+y+z=37, and:
(1/37)*(x(x-1)+y(y-1)+z(z-1)) =12
=> x^2-x+y^2-y+z^2-z=37*12=444
=> x^2+y^2+z^2-(x+y+z)=444
=> x^2+y^2+z^2-37=444
=> x^2+y^2+z^2=481
WLOG, we assume that x<=y<=z and, performing the calculations by a pocket calculator for evaluating prospective solutions for x^2+y^2+z^2= 481, we set up the following table:
x-------y-------z--------x+y+z
2------6------21---------29
6-----11-----18---------35
9-----12-----16---------37
From the above table, we observe that only the triplet (9,12,16) satisfy the equality x+y+z=37
Consequently, the three groups respectively contain 9,12 and 16 students.
Edited on January 4, 2022, 4:16 pm
Puzzle Solution
