## a genuine riddle

Posted in Books, Kids, pictures with tags , , , , , on August 5, 2022 by xi'an

A riddle from The Riddler that was pure (if straightforward) logic rather than brute force compulation or mathematical modelling:

Four bags of many marbles are labelled R(ed), B(lue), G(green) and μ (mixed), except that all labels are wrong. Given the possibility to draw two balls, one at a time, from any bag, is it possible to select two monochromatic bags?

Bag μ draw is returning a color, R say, as it is a monochromatic bag. Drawing from another color bag, B say, will produce R or B, in which case it is μ, i.e., mixed (polychromatic), which means the other bags are monochromatic, or G. For this last case, bag B is either polychromatic, in which case bag G is made of blue marbles and bag R of green marbles, or monochromatic, in which case bag G is mixed and bag R is full of blue marbles, but monochromatic for either situation, hence to be chosen on top of bag μ.

## Le Monde puzzle [#1078]

Posted in Books, Kids, R with tags , , , , , , on November 29, 2018 by xi'an

Recalling Le Monde mathematical puzzle  first competition problem

Given yay/nay answers to the three following questions about the integer 13≤n≤1300 (i) is the integer n less than 500? (ii) is n a perfect square? (iii) is n a perfect cube?  n cannot be determined, but it is certain that any answer to the fourth question (iv) are all digits of n distinct? allows to identify n. What is n if the answer provided for (ii) was false.

When looking at perfect squares less than 1300 (33) and perfect cubes less than 1300 (8), there exists one single common integer less than 500 (64) and one single above (729). Hence, it is not possible that answers to (ii) and (iii) are both positive, since the final (iv) would then be unnecessary. If the answer to (ii) is negative and the answer to (iii) is positive, it would mean that the value of n is either 512 or 10³ depending on the answer to (i), excluding numbers below 500 since there is no unicity even after (iv). When switching to a positive answer to (ii), this produces 729 as the puzzle solution.

Incidentally, while Amic, Robin, and I finished among the 25 ex-aequos of the competition, none of us reached the subsidiary maximal number of points to become the overall winner. It may be that I will attend the reward ceremony at Musée des Arts et Métiers next Sunday.

## Le Monde puzzle [#1071]

Posted in Books, Kids with tags , , , , , , , , , , , on October 18, 2018 by xi'an

A “he said she said” Le Monde mathematical puzzle sixth competition problem that reminded me of the “Singapore birthday problem” (nothing to do with the original birthday problem!):

Arwen and Brandwein are privately and respectively given the day and month of Caradoc’s birthday [in the Gregorian calendar] with the information that the month number is at least the day number. Arwen starts by stating she knows Brandwein cannot deduce the birthday, followed by Brandwein who says the same about Arwen. If this “she says he says” goes on for the largest possible number of steps before Arwen says she knows, when is Caradoc’s birthday? Arwen and Brandwein are later given two numbers corresponding to Deirdre’s birthday with no indication of which one is the day and which one is the month. They know both numbers end up with the same digit and that the month number is strictly less than the day number. Arwen states she does not know the date and she knows Brandwein cannot know either. Then Brandwein says he indeed does not the date but he knows whether he got the day or the month. This prompts Arwen to conclude she knows, then Brandwein to do the same. When is Deirdre’s birthday?

Since this was a fairly easy puzzle (and since I had spent too much time debugging the previous R code!), I did not try coding this one but instead drew the possibilities and remove the impossibilities on a blackboard. The first question is quite simple actually since the day numbers stand between 1 and 12 and that each “I cannot know” excludes one of the remaining endpoints, removing first excludes 1 from both lists, then 12, then 2, then …. 8, ending up with 7. And 07/07 as Caradoc’s birthday. The second case sees 13,…,20,23,…,30 eliminated from Arwen’s numbers, then 3,…,10 as well, which eliminates the same numbers from Brandwein’s possibilities. That he knows whether it is a month or a day leaves only 1,2,21,22,31 as possible numbers. Then Arwen’s certainty reduces her numbers to be 2, 21, 22, or 31, and since Brandwein is also sure, the only possible cases are (2,22) and (22,2). Meaning Deirdre’s birthday is on 22/02. I dunno if this symmetry was to be expected! (And I cannot fathom why this puzzle is awarded so many points, when compared with the others.)

## {Monte Carlo}²

Posted in Kids with tags , , , , on May 1, 2012 by xi'an