UPDATE (3 July): I received several solutions already that
retain the original month+year order.
Also: Several people noted in their solutions that Martin Gardner's age should be rounded up, because he was slightly over the half-mark to the next year. This Using your Head is Permitted riddle, however, will stick to the more common practice of rounding down. My apologies to those whose solutions I did not accept for this reason. Do try again.
UPDATE (11 July): One of the solutions to be published at the end of the month will show how to do this using the digits of the year only (and preserving digit order).
This month, "Using your Head is Permitted" will digress slightly from its standard format in order to honour Martin Gardner, one of the world's foremost mathematical riddlers, who, sadly, passed away a few weeks ago.
In order to commemorate Gardner's work, this month's riddle attempts to be a riddle in his distinctive style. Here goes:
Take the date in which Gardner died and, by using the digits of the month and year only, write a computation that results in the age in which he died.
Allowed operations are anything that can be done in pure ASCII text, without resorting to adding any more alphanumeric characters. (So, you should use all digits in the month and year, each digit exactly in the number of times that it appears, no more no less, and you cannot use operations like "cos()", "floor()", "sqrt()" and the like, that require the addition of letters.) Adding symbols, such as "+" or "-", is allowed, of course, as long as they are ASCII symbols (so, no "√", for example).
Also not allowed is digit concatenation. Each digit should be taken as a number between 0 and 9.
I am not aware of solutions that leave the month+year digit order intact. However, solvers looking for an extra challenge will discover a solution that retains the original digit order when the year is written before the month.
In keeping with Martin Gardner's tradition (as well as with "Using your Head is Permitted"'s regular style), I am also presenting this month a question with more mathematical depth. It is a not-for-credit bonus question, and is an invention of Gardner himself. His wording was better.
The bonus question is this: suppose that a family has n siblings, and suppose that exactly m of these siblings have blue eyes. The numbers m and n in this particular family happen to satisfy the following property: if two siblings are chosen at random (uniformly, from the set of all possible pairs) then the chance of both of them having blue eyes is exactly 50%.
If the family is a human family then one can be reasonably sure what both m and n are. If we are to take this as an abstract mathematical problem, however, and assume that families can be arbitrarily large, then there is an infinite number of solutions.
The challenge is to find what the ratio of two consecutive solutions for n converges to. Solvers interested in a slightly more difficult question can seek an explicit formula to calculate the i'th solution.
A hint for the bonus question is available here.
List of solvers:Albert Stadler (1 July 16:28)
Bojan Bašić (1 July 20:43)
Robert DiMarco (2 July 04:32)
Tai-Jin Lee (2 July 04:39)
Nick McGrath (2 July 04:55)
Anurag Anshu (2 July 15:20)
Oded Margalit (2 July 21:19)
Itsik Horovitz (3 July 10:13)
Yoav Yaari (3 July 16:40)
Zilin Jiang (6 July 01:19)
Jan Herout (7 July 03:38)
Daniel Bitin (8 July 01:22)
Joseph DeVincentis (8 July 03:53)
Erick Wong (8 July 08:13)
Jens Voss (8 July 21:23)
Dan Dima (9 July 08:25)
Naftali Peles (11 July 17:15)
Mauro Bampo (28 July 07:14)
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.
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