A box has integer dimensions, and when each one is increased by 2, its volume doubles. What is the largest possible dimension?

Martin Gardner posed this one years ago.

Looking for 2xyz = (x+2)(y+2)(z+2).

Take x<=y<=z.

It's easy to check that x=1 and x=2 don't work.

Divide equation by xyz to get cuberoot(2) >= (1+2/x), so x<=7.

Then use a factoring trick, demonstrated here for x=3.

yz-10z-10z = 20 and (y-10)(z-10) = 120.

Solution are:

for x=3

11,130
12,70
13,50
14,40
15,34
16,30
18,25
10,12

for x=4

7,54
8,30
9,22
10,18
12,14

for x=5

5,98
6,28
7,18
8,14

for x=6

6,16
7,12
8,10

Posted by xdog on 2009-01-05 00:45:29