Spoiler for New Scientist Enigma 1716:
I commissioned a goldsmith to make some identical solid gold tetrahedrons, in which the original specification gave the length of each of the sides. He calculated what the cost of the gold would be and, as I could afford a certain whole number percentage more, I asked him to increase their size by that percentage. He started making the tetrahedrons, but made the error of increasing each side by that percentage. After he had made one he realised that if he continued in this way he would use much more gold than expected. So he made the second tetrahedron with each side reduced by that same percentage from its original specification. The rest he made to the original specification.
I still got the required number of tetrahedrons and in fact the overall effect was to increase the amount of gold used by the percentage I wanted.
How many tetrahedrons did I order, and what percentage increase did I ask for?
Very straightforward algebra problem. I did think the problem was a little unclear as to whether the tetrahedra had to be regular or not, but decided there wasn’t enough information if the tetrahedra were not regular. Then I had to look up the formula for the volume of a regular tetrahedron: V = (√2/12)a3 where a is the length of a side. For the n tetrahedra as originally ordered, V0 = n(√2/12)a3. For the intended n tetrahedra scaled up in volume by d percent, V1 = n(√2/12)a3(1+d/100). For the order as delivered, V2 = (√2/12)a3(1+d/100)3+(√2/12)a3(1–d/100)3+(n–2)(√2/12)a3(1+d/100)3. These last two quantities are supposed to be equal. Factoring out (√2/12)a3 and expanding (using (1+x)3=1+3x+3x2+x3 and (1–x)3=1–3x+3x2–x3) we end up with n=6d/100 or 50n=3d. d must be a whole number, as is n of course, so one solution is n = 3, d = 50. Of course you can get other formal solutions by multiplying both n and d by a whole number, but any such solution has d≥100 which means the second tetrahedron has zero or negative size! So the only valid solution is: 3 tetrahedra, 50% increase.