Problem 466

Suppose that Dewa’s four numbers are $w$, $x$, $y$, $z$.

The averages of the four possible groups of three of these are $$\dfrac{w+x+y}{3}, \dfrac{w+x+z}{3}, \dfrac{w+y+z}{3}, \dfrac{x+y+z}{3}$$ These averages are equal to
32, 39, 40, 44, in some order.

The sums of the groups of three are equal to 3 times the averages, so
are 96, 117, 120, 132, in some order.

In other words, $w+x+y$, $w+x+z$, $w+y+z$, $x+y+z$ are equal to 96, 117, 120, 132 in
some order.

Therefore, $$(w+x+y)+(w+x+z)+(w+y+z)+(x+y+z) = 96+117+120+132$$ and so $$3w + 3x + 3y + 3z = 465$$ which gives $$w+x+y+z = 155$$ Since the sum of the four numbers is 155 and the sums of
groups of 3 are 96, 117, 120, 132, then the four numbers are $$155 - 96 = 59 \qquad 155 - 117 = 38 \qquad 155 - 120 = 35 \qquad 155 - 132 = 23$$ and so the largest number is
59.