It’s tempting to think that our bear from last week is at the south pole: you walk 10 kilometers north along one meridian, then another 10 kilometers east along a parallel, and finally another 10 kilometers south along another meridian to return to point starting point after describing an elegant equilateral geodesic triangle. In fact, this is the solution that is often given in posts where this riddle appears; but there is a small flaw: in Antarctica there are no bears and, therefore, we are in the Arctic, 10 kilometers south of the parallel, close to the north pole, whose length is 10 kilometers: we reached it by walking 10 kilometers towards the north, we cross it completely when going 10 kilometers towards the east and we return to the starting point when going 10 kilometers towards the south. Can you calculate the latitude of the meeting point with the bear? Is the solution unique?
The number of cannonballs in the second problem is easy to calculate: in the base there are 15², in the second level 14², in the third 13² and so on, so in total there will be:
15² + 14² + 13² + 12² + 11² + 10² + 9² + 8² + 7² + 6² + 5² + 3² + 2² + 1 = 1240
It is clear that, for any number n of bullets on the bottom side of the pyramid, the total number of bullets is the sum of the squares of the first n natural numbers; but if n is large the operations are long and cumbersome, so can you find a simple formula that gives the sum of the squares of the first n numbers?
And, going from arithmetic to geometry, can you calculate the height of the pyramid of bullets, assuming that they are spheres with a diameter of 20 cm? And moving from geometry to physics, why don’t the bullets/balls roll out from the perimeter of the base?
As for unequal-armed scales, the double-weighing rule greatly improves the situation for customers, but still allows cheating merchants to pilfer a bit. Indeed, suppose that the arms of the balance measure 9 and 10 inches (or any other units) respectively and that we weigh 1 kilo of merchandise; If we place it on the pan of the short arm and call x the weight necessary to balance the scale, we will have:
x/9 = 1/10, whereby x = 9/10 = 0.9
If we place the merchandise on the other pan and call y the weight needed to balance it now:
y/10 = 1/9, from where y = 10/9 = 1,111…
The average of both weighings is 2.011/2 = 1.0055, so the cheating merchant charges 5.5 grams more for each kilo of merchandise he sells.
heterogeneous banquets
And since we have revisited several classic brain teasers, let’s end with one that deserved the attention of mathematicians as illustrious as Luca Pacioli and Niccolò Fontana, better known as Tartaglia, and another very similar one proposed by Euler:
A banquet is attended by 41 people, including men, women and children, who spend a total of 40 drachmas. Since men eat more than women and adults more than children, each man pays 4 drachmas, each woman pays 3 drachmas, and each child pays 4 denarii. Knowing that a drachma is equal to 12 denarii, how many men, women, and children attend the banquet?
And, now without children, the Euler variant:
A group of men, some of them accompanied by their wives, spent 100 drachmas on a banquet. Each man paid 19 drachmas and each woman 13. How many men and how many women were there?
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