A candy merchant receives 3 opaque boxes. One box contains mint candies, another contains anise candies, and the last box contains a mixture of mint and anise. The boxes are labeled Mint, Anise, and Mixed. All of the boxes are labeled incorrectly. What is the minimum number of candies the merchant will have to sample to correctly label each box?
The fifth enigma is stated as follows: “A student asks his teacher, ‘How old are your three daughters?’ The teacher replies, ‘If you multiply their ages you get 36. If you add their ages you get my house number.’ ‘I am missing a detail,’ protests the student. ‘Oh yes,’ says the teacher, ‘the older one plays piano.’ How old are the 3 daughters?”
The seventh enigma is an elementary algebra problem with a twist. A mother is 21 years older than her son. In 6 years, the son will be one-fifth his mother’s age. The enigma asks what the father is currently doing.
The minimum is three - he takes two from one box and gets one mint, one anise. Now he knows that one is mixed, and so he needs to take just one from one of the other boxes.
Of course, that’s the minimum - he might not get so lucky.
The first one is an old chestnut I’ve seen many a time.
Only one draw is necessary – one draw from the box labelled “mixed” will tell you what solid flavor is in there (ie, if a mint one is drawn, that clearly has to be the “all mint” box).
Once that one is properly labelled, the other two are forced (to continue my example, the one labelled ‘anise’ can’t be anise, and can’t be mint, so that one is in reality the “mixed” one).
The other two I’m not sure about – why couldn’t the ages by 3,3, and 4, or 2,2,9?
The minimum is one. Taste a candy from the box labeled “Mixed.” It will be either mint or anise. You know that this box is labeled incorrectly, so put either the “Mint” or “Anise” label on that box. Whichever one of “Mint” and “Anise” is left is the “Mixed” box. The last box gets the final label, whichever flavor the “Mixed” box was not.
Second one:
Yes, 9, 2, and 2 is correct. But another possible answer for the factors of is 6, 6, and 1, which is why the student needed to know that there is only one “older” daughter.
Third one:
I did the algebra, and the son is currently aged minus 9 months. What the father is doing right now is left as an exercise to the reader.
>>Yes, 9, 2, and 2 is correct. But another possible answer for the factors of is 6, 6, and 1, which is why the student needed to know that there is only one “older” daughter.
As someone upthread suggested, couldn’t you have “3, 3, and 4”?
This still leaves toy with just one “older” daughter, and while it’s unlikely that a four year old can play the piano, it’s not outside the realm of possibility.
Here’s all the puzzles:
Each of the enigmas relayed through the PDA have a one-minute deadline and a simple solution—that is, they do not rely upon any advanced mathematical knowledge. After the candy puzzle, the second enigma is a series of 169 ones and zeros which, when placed in a 13-by-13 square and interpreted as pixels, form an image of a skull.
The third enigma describes a sealed room containing one light bulb. Outside of the room there are three switches, only one of which operates the bulb. The puzzle solver begins outside the room, able to operate the switches in any way he sees fit, but when the door is opened for the first time, he must determine which switch operates the light. The enigma is to determine how this is to be done.
The fourth enigma is to describe how one can use a four-minute hourglass and a seven-minute hourglass to measure a period of nine minutes.
The fifth enigma is stated as follows: “A student asks his teacher, ‘How old are your three daughters?’ The teacher replies, ‘If you multiply their ages you get 36. If you add their ages you get my house number.’ ‘I am missing a detail,’ protests the student. ‘Oh yes,’ says the teacher, ‘the older one plays piano.’ How old are the 3 daughters?”
The sixth enigma is a Knights and Knaves puzzle. Two doors are guarded by two men, one of whom always lies and one of whom always tells the truth; however, the puzzle solver does not know which man is which. One of the doors leads to freedom and one to captivity. The enigma is to determine a single question that, if asked of one of the guards, would reveal the door to freedom with certainty.
The seventh enigma is an elementary algebra problem with a twist. A mother is 21 years older than her son. In 6 years, the son will be one-fifth his mother’s age. The enigma asks what the father is currently doing.
