The ingenious solution to last week’s candle puzzle is to light both at the same time, one on one end and the other on both. When the second is completely consumed, half an hour has passed; Then we also light the first candle at the other end and from the moment when it is completely consumed, fifteen minutes have passed. But Ignacio Alonso rightly raises the objection that a candle that burns at both ends simultaneously cannot be in a vertical position so that its burning time changes. Therefore, the variant with wicks instead of candles is more plausible, since these burn at the same speed and more evenly than candles in any position. (It’s not worth splitting the sails in half because if you could pinpoint the center point you could pinpoint a quarter and there wouldn’t be a problem).
There are different ways to measure 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
More information:
That is, we start the two clocks at the same time, and when the small one has transferred all the sand, we turn it over and leave 1 minute when the big clock is empty (at 7 minutes). time we change that. If the little one transfers the remaining minute, then 8 minutes have passed, and in the lower part of the big one there is a minute of sand: we turn it over and if this minute is transferred, then 9 minutes have passed.
In the problem of six friends sitting at a round table, Diana cannot stand next to Clara or Eva, and Eva cannot stand to the right of Ana (unless incest is contemplated and Eva is his own brother’s wife), so it’s on his left. Therefore, the requested sequence is AEBDFC.
numbers after letters
There is a wide range of arithmetic puzzles based on replacing all or part of letters with numbers. A well-known classic and therefore definitely mentioned:
A student asks his parents for money, and he does so by converting his request into a quantified sum:
SEND
+MORE
MONEY
How much MONEY is the student asking for? In this problem and in those that 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 superfluous?).
Such puzzles are ultimately systems of Diophantine equations with as many unknowns as letters; However, they are solved by sophisticated considerations that allow many steps of traditional development to be skipped. Let’s take a look at the mathematician and philosopher Eric Revell Emmet (1909-1980), professor and author of numerous number puzzles (including the “crossed numbers” puzzle, which we will deal with another time).
1. Solve the sum of two summands involving three different digits:
XD
+HHD
XDH
2. Solve the sum of three summands in which all digits occur:
MMWXTFGGG
+MMEXWTFGG
+MMYFMMFGG
FTTYMCVFM
3. A multiplication is an addition with repeated addends, 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 digits are wrong. But every time it appears, the same wrong digit replaces the same right digit, and the same right digit is always represented by the same wrong digit.
4751
+9731
46082
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