UPDATE (8 July):
The answer to the bonus riddle was found on the Internet by
Li Li, and turns out to have quite a pedigree. Thank you so much for
this, Li Li. It is greatly appreciated.
I'll publish the history of the riddle and its solution, as discovered by Li Li, on this month's solution. Now that this solution has been found, however, I am reinstating the usual restrictions regarding originality of solution. If you want to have your name with an asterisk this month, please solve the problem yourself. Again, thank you very much, Li Li. Dear readers, Thank you all for reading Using your Head is Permitted, and for sending in your riddle suggestions and your solutions. Using your Head is Permitted has published its one hundredth riddle last month, and I wanted to celebrate this "centenary" by mentioning some of what came out of the site over the years. First off, as you know, Using your Head is Permitted is open to all, and requires only a desire to use one's head and tackle interesting mathematical problems. Those of you who have been listed on the site as solvers know that I ask solvers where they are from and what their mathematics background is. Turns out that a love of mathematics is global, and does not depend on what education you may have had. UyHiP solvers come from all over the world and from all walks of life. Some are high school students, others  preeminent professors. Some are professional mathematicians or physicists, others journalists or career soldiers or retirees. The same can be said for the group of people who have sent me riddles during this time, to all of whom I owe a big thank you. In other words, love of working out puzzles of the mind is universal across ages, disciplines, education and origin. (The only place where I see a major skew beyond what is explainable by unrelated factors is in gender bias. Among the site's tripledigit number of distinct solvers, virtually all are male. I can only conclude that female readers tend to not send in their solutions. So, consider this a callout to all you women readers out there: let your voices be heard!) Academically, the site has kept a high standard, and many of the riddles presented here ultimately found their way, in one roundabout way or another, to publication in high ranking journals. Here are some examples:
And, of course, many (if not most) of UyHiP's 100 riddles bring to bear some existing mathematical theory. I did my best to mention this explicitly on the solution pages, so you, dear readers, can follow up on the stateoftheart of the problemspace and its solutions. (And where I didn't have good references, I must thank the readers who were able to track down problems to their origins, who sent me the relevant information and links. It's great to see that you, readers, care as much about the quality of the mathematical material presented here as I do.) Believe me, when I started this journey, back in 2007, I had no idea any of this was going to happen. But enough selfcongratulations. On to the riddle. I had the tough duty to decide on a single riddle, from all the suggestions you have been sending me over the past two months, to be the 100'thanniversary riddle. The winning riddle is one sent to me by RaduAlexandru Todor. Many thanks, Radu, and congratulations! Radu asks: Let a natural n be called a subset sum of a set S if there is a subset of the elements of S that sums up to n. Critically, no element of S can be used more than once. Let S_{p,q}={p^{x}q^{y}  x,y∈ℤ^{≥0}}. Radu's question was: show that any natural except for finitely many is a subset sum of S_{3,4}. Answer this question for credit. However, because this is the centenary riddle, I decided to post, additionally, something unusual for this site: a riddle for which I do not know the solution. This second riddle  both for an asterisk next to your name and for explicit mention on the solution page  is the following: Is it true that for any coprime p,q>1 the same holds, i.e. that any natural except for finitely many is a subset sum of S_{p,q}? For this month's bonus riddle, feel free to search the Internet as much as you want and to collaborate, online or otherwise, as much as you want.

List of solvers:Joseph DeVincentis (1 July 23:10)Dan Dima (2 July 00:31) Guangda Huzhang (2 July 01:42) Uoti Urpala (3 July 05:21) Harald Bögeholz (5 July 01:20) Nis Jørgensen (7 July 23:59) Li Li (*) (8 July 08:31) Andreas Stiller (9 July 03:31) Adrian Neacsu (14 July 01:43) Oded Margalit (14 July 08:44) Lorenzo Gianferrari Pini (20 July 22:01) Daniel Bitin (21 July 19:07) 
Elegant and original solutions can be submitted to the puzzlemaster at riddlesbrand.scso.com. Names of solvers will be posted on this page. Notify if you don't want your name to be mentioned.
The solution will be published at the end of the month.
Enjoy!