The ingenious solution to last week’s candle riddle is to light them both at once, one at one end and the other at both. When the second has been completely consumed, half an hour will have elapsed; then we light the first candle also at the other end and, from that moment, when it is completely consumed, a quarter of an hour will have passed. But Ignacio Alonso raises the reasonable objection that a candle that burns at both ends at the same time cannot be in a vertical position, so its burning time will be altered. That is why the variant with wicks is more plausible instead of candles, since they burn at the same speed in any position, and more evenly than candles. (It is not worth splitting the sails in half, because if you could accurately determine the midpoint, you could also determine a quarter and there would be no problem).
There are different ways to time 9 minutes with an hourglass of 4 and another of 7; One of them can be schematized as follows:
7/0 4/0, 3/4 0/4-4/0, 1/3 0/7-7/0, 0/4 6/1-1/6
That is to say, we start the two clocks at the same time, and when the small one has transferred all its sand, we turn it over, leaving 1 minute left on it when the big one empties (at 7 minutes), at which time that we turn this one around. When the small one transfers the remaining minute, 8 minutes will have passed and there will be a minute of sand in the lower part of the big one: we turn it over and when that minute is transferred, 9 minutes will have passed.
In the problem of the six friends sitting around a circular table, Diana cannot be next to Clara or Eva, and Eva is not to the right of Ana (unless incest is contemplated and Eva is the his own brother’s wife), so it’s on his left. Therefore, the requested sequence is AEBDFC.
numbers by letters
There is a wide range of arithmetic puzzles based on the full or partial substitution of numbers for letters. A well-known classic, and for that very reason a must mention, is the following:
A student asks his parents for money, and he does so by converting his request into a figured sum:
SEND
+MORE
MONEY
How much MONEY does the student ask for? In this problem and in the ones we will see below, each letter corresponds to a different digit, always the same, and vice versa. (Was it necessary to say “and vice versa” or is it redundant?).
These kinds of puzzles are ultimately systems of Diophantine equations with as many unknowns as letters; but they are resolved through ingenious considerations that allow many steps to be skipped from what would be a conventional development. Let’s look at some of the mathematician and philosopher Eric Revell Emmet (1909-1980), professor and author of numerous numerical puzzles (among them those of “crossed numbers”, which we will deal with another time).
1. Solve the sum of two addends in which three different digits are involved:
XD
+HHD
XDH
2. Solve the sum of three addends in which all the digits appear:
MMWXTFGGG
+MMEXWTFGG
+MMYFMMFGG
FTTYMCVFM
3. A multiplication is an addition with the addends repeated, so the following problem belongs to the same “family”:
XYP
xH
PMYX
4. And finally, a variant with tricky digits instead of letters. In the following addition, all the digits are wrong. But the same incorrect digit takes the place of the same correct digit every time it appears, and the same correct digit is always represented by the same incorrect digit.
4751
+9731
46082
You can follow SUBJECT in Facebook, Twitter and instagramor sign up here to receive our weekly newsletter.
Subscribe to continue reading
Read without limits
#Add #continue #ruminating