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# Opinion | Riddle me this: eight little answers ‘fore I go

Here’s the first part of solutions to my 16 end-of-2019 puzzles, some mathematical and some on logic. Wait until next week for answers to the last eight

The answers to my 16 end-of-2019 puzzles, in two parts. This column has the answers to the first 8 questions; my next column will have the answers to the last 8.

And… here we go!

1) Your friend Kanakadurga thinks of a seven-digit number. She removes one of the digits—without changing the order of the others—leaving a six-digit number. She adds the seven-digit and six-digit numbers, producing 4632159. What was her original seven-digit number?

A) Clue: the answer is odd. This means the six- and seven-digit numbers that we add together cannot end in the same digit (if they did, they would sum to an even number). The only way the two can have different last digits is if the digit removed from the original is the last. Best now to experiment with digits that add to 9. Let’s say you pick 5 and 4 as the last digits of the seven- and six-digit numbers, respectively. Then the penultimate digit of the seven-digit number is also 4, and since the sum in that place is 5, the penultimate digit of the six-digit number must be 1. Continue like this, perhaps run into a dead-end or two and start over … and you will soon find the original seven-digit number: 4211054.

2) Next, Kanakadurga thinks of a four-digit number. It ends with 8. If she moves that 8 to the beginning of the number, she gets a number that’s 1.75 times the first number … and, as a bonus, is the reverse of the first number. What’s the four-digit number Kanakadurga is thinking of?

A) Think of the three digits other than the 8 as a number by itself, call it y. The original four-digit number, then, is 10y + 8. Got that? (e.g. 7238 =10 x 723 + 8). When we move the 8 to the beginning, our new number is 8000+y. (8723=8000+723). So we have to solve this equation:

8000+y=1.75x(10y+8)

This is simply done, to give us x = 484. The original number is thus 4848.

3) Inside a box, we have three purple hats and two yellow hats. Anuradha, Belinda and Catarina each takes a hat from the box and places it on her own head. None of them can see what colour her hat is. We position the three ladies so that Anuradha sees the hats that Belinda and Catarina are wearing, Belinda sees only Catarina’s hat, and Catarina sees no hats.

We ask Belinda. She too says: “No".

We ask Catarina. She says: “Yes!"

What colour is Catarina’s hat and how did she know?

A) If Anuradha sees two yellow hats on Belinda and Catarina, she will know her colour: purple. But she doesn’t know her colour. Thus Belinda and Catarina have either two purples or a purple and a yellow (and both know this because of Anuradha’s response). Now if Belinda sees a yellow on Catarina, she will know hers is purple. But she doesn’t know her colour. Thus Catarina has purple.

4) More about hats and people who wear them. A dictator who hates science has rounded up all 100 mathematicians in his country. (All women, it turns out). The next day, he says, he will line them up so that each one can see everyone in front of her, but nobody behind. He will then put either a black or a white hat on each mathematician’s head, but nobody will know what colour hat she herself is wearing. Starting at the back of the line—the mathematician who can see 99 hats in front of her—he will ask each mathematician in turn what colour hat she is wearing. If she’s right, she lives. If she’s wrong, she’s instantly and painlessly decapitated.

Prospects don’t look good. But being mathematicians, the 100 confer and find a strategy that, at worst, leaves just one of them headless (and at best, saves them all). Who is the mathematician who might die, and what is the strategy?

A)What they decide: All the women count the number of black hats they see. If the last woman’s count is odd, she guesses “black" for herself; if even, “white". It’s a 50-50 guess, so she might die. But her possible sacrifice gives the rest their lives. Let’s say she shouts “black", meaning she sees an odd number of black hats. Now if the penultimate woman also counts an odd number of blacks, her own hat must be white; if she counts even, hers must be black. The same reasoning applies if the last woman calls “white". So the penultimate woman knows her hat’s colour, and each successive woman uses the same logic to learn her colour and save her head.

5) I tie a blindfold over your eyes and take you to my dining table. Scattered on it, I tell you, are many 10 coins, and exactly 87 of them have their heads facing up. The rest show tails. Your task: while still blindfolded, divide the coins into two sets, each with the same number of heads showing. You do it immediately. How?

A) Count out exactly 87 coins: that’s one set, and the rest form the second set. Turn over all 87 coins in the first set. Now both sets have the same number of heads. (Give it a thought).

6) Your new friend Pandharinath tells you he has two children, but forgets to specify their sexes. One day you run into him at the corner store and he introduces you to a little boy, saying: “This is my son Somendranath." What’s the chance he has two boys?

A) Not half. When you know he has two kids, there are four possibilities. In age order, they are: girl-girl, girl-boy, boy-girl, boy-boy. But once you learn that one kid is a son, there are only three possibilities: girl-boy, boy-girl, boy-boy. Thus the chance he has two boys is 1/3.

7) (This may or may not be similar to Pandharinath). I have with me a small box containing two coins. One coin has been minted wrongly, with heads on both sides. The other is normal: heads on one side, tails on the other. You reach in, choose one coin at random and look at one side. If you see heads, what’s the chance that its other side is heads too?

A) Not half. Not 1/3. Call the two coins C1 (the double-header) and C2. There are just three ways you could be looking at heads on a coin you drew from the box: C2’s heads side, or one side of C1, or the other side of C1. In two of these three cases, the other side is heads as well. Thus the chance the other side of your chosen coin is heads too is 2/3.

8) More about boxes: For 35 years, the brothers Tom and Ray Magliozzi hosted a radio show called “Car Talk" in the US. On it, they discussed everything to do with cars — but they also had a weekly feature called “The Puzzler", throwing puzzles hard and easy at their listeners. The hardest of them, the brothers once said, “are so intricate, so convoluted and so obfuscated that they’re sure to start the smoke pouring off your cranium". Here’s one—“Evil King Berman and the Three Boxes"—and do take a selfie showing the smoke.

The fair maiden Rowena wishes to wed. Her father, the Evil King Berman, has other plans for her and has devised a way to drive off suitors. He has prepared a little quiz for them. A very simple quiz, he says. Here it is:

Three boxes sit on a table. The first is made of gold, the second is made of silver, and the third is made of lead. Inside one of these boxes is a picture of the fair Rowena. It is the job of the knights to figure out—without opening them—which one has her picture. Naturally the one who gets that right gets Rowena.

Now, to assist the lovelorn aspirant in this endeavour there is an inscription on each of the boxes. The gold box says, “Rowena’s picture is in this box." The silver box says, “The picture is not in this box." The lead box says, “The picture is not in the gold box."

Only one of those statements is true, says Berman. Many knights are stumped. But one identifies which box holds the picture and wins Rowena. How?

A) If the gold statement is true, the silver one must also be true. But that’s impossible, because only one of the three statements is true. Therefore the gold statement is false; lovely Rowena is not in the gold box. Therefore the lead statement is true. Therefore the silver statement is false and her picture is in that box. Go for silver, young man!

Once a computer scientist, Dilip D’Souza now lives in Mumbai and writes for his dinners. His Twitter handle is @DeathEndsFun