Opinion | The final set of eight little answers ‘fore I go

And now for the second set of eight answers to the puzzles in my column here on 27 December 2019.

9) Here’s one more of Tom and Ray’s “tough” puzzles: You have in your possession two pieces of string. Let’s say that each is a couple of feet long, but it doesn’t really matter. And they can both be different lengths, it doesn’t matter either. And they’re burnable, like the fuses used to light dynamite. You could light either end of either string, and it would burn. In fact, if you lit one end of a string, it would burn in exactly an hour.

But here’s the wrinkle: the strings do not burn at a constant rate. For example, the string might burn for two minutes and then go crazy and burn like mad and then slow down. You don’t know what rate the string’s burning at, at any specific time. All you know is that in an hour’s time, the whole string is burned. It’s not linear. And not predictable.

So, the question very simply is, with a lighter and these two strings, how would you measure exactly 15 minutes of time?

A) Light one string from both ends. At the same time, light the second string from one end. The first string will burn out in exactly half an hour. As soon as it does, light the other end of the second string. In exactly 15 minutes, the second string will burn out. There’s your 15 minute measure.

10) Is statement #3 below true or false?

1. There are three numbered statements here.

2. Two of the numbered statements are false.

3. Using only these three statements, you can figure out the answer to this question.

A) Clearly, statement #1 is true. Thus if #2 is true, it claims itself and #3 as false. That is, if #2 is true, it is false: a contradiction, you’ll agree. Thus #2 is false, and since all three statements cannot be false (#1 is true), only one of the statements is false: #2 itself. Therefore #3 is true.

11) How many letters does the correct answer to this puzzle contain?

A) Could it be one? But “one” has three letters. So does “two”. “Seven” has five, etc. We want a number that has its own count of letters. The only such is 4.

12) Which planet is closest to our Earth? Is that a reasonable question? If not, what’s a more reasonable question to ask about planetary distances, and what’s the answer?

A) A truly fascinating inquiry here. The commonest answer is Venus; no planet makes a closer approach to us. But the reality is that given the planets’ respective orbits, Venus spends much of its time further from Earth than … Mercury! So the more reasonable question might actually be: “Which planet is closest to Earth most of the time?” (Or “which planet is, on average, closest to Earth?”), and the answer there is Mercury.

In fact, here’s the amazing thing about our solar system: if you ask the question of any other planet in our solar system, the answer is still Mercury: in the sense of this more reasonable question, it is the closest planet to every other planet. See this terrific video to understand:

13) Your beloved mausi just died and left her entire savings to her favourite niece: you. That’s ₹1 crore coming to you, and while you mourn, you’re also grateful to her.

Going through her belongings, you come across a copy of Alan Paton’s Cry the Beloved Country that she forgot to return to your neighbourhood MCubed Library. It was due on 2 January 1990. The fine print says the fine is ₹1, with interest charged at 1% every week that it is not paid.

Good citizen that you are, you start walking to MCubed to pay the fine. Your buddy Rambehari, who has been busy with a calculator, suddenly shouts: “You crazy? Don’t pay the fine! Just keep the book and forget the library!”

Why’s he shouting and what would you do?

A) He’s shouting because he did some quick calculations. In one week, your fine becomes ₹1.01. In two, ₹1.02. In 25 weeks, ₹1.28. In general, after “n” weeks, the fine will be 1.01^n. Now the book was due 30 years, or 1,560 weeks, ago. The fine, then, will be 1.01^1,560 = ₹55,12,430.17 Over half of your inheritance from your beloved mausi. No wonder Rambehari urges you to keep the book. Best to buy the library a new copy of Cry the Beloved Country, about ₹400 right now on Amazon.

14) This one might or might not be relevant to the world, circa 2019. Well, circa anytime: 100 politicians arrive in Bombay to attend a workshop on parliamentary procedures. You know that each of them is either crooked or honest. The person running the workshop tells you these two things about his attendees:

• At least one of the politicians is honest.

• In any given pair of them, at least one is crooked.

Can you tell the world how many of the 100 politicians are honest and how many crooked?

A) We know there’s at least one honest politician. Are there two? Well, no: because then if you picked that pair of politicians, both would be honest and we know that at least one of every pair of politicians is crooked. Thus there’s one honest politician and 99 crooked ones.

15) You have been sent out to conduct the coin toss for a Test match between Papua New Guinea and Ecuador. Unfortunately, you’ve been given a dud coin, one that doesn’t land heads half the time. The crowd is expectant, the TV coverage is underway … you have no time to get another coin. How can you still carry out a fair (50/50) coin toss and get the match going?

A) Get the captains to call either “heads-tails” and “tails-heads”. Toss the coin twice, repeating tosses pairwise until the first occurrence of either heads followed by tails or vice versa. The captain who correctly called that first occurrence wins.

Why does this work? Say the coin lands heads just 40% of the time. So the probability of landing heads is 0.4, and thus of tails 0.6. Clearly an unfair coin. But the probability of heads-tails (heads followed by tails) is 0.4 x 0.6 = 0.24. The probability of tails-heads is 0.6 x 0.4 = 0.24 again. That is, each of these combinations is equally likely to occur when we flip the coin twice, and that makes for a fair “toss”.

(Question: would getting the captains to call either H-H or T-T be fair?)

16) Below are two number sequences. Tell me what number comes next in the first sequence, and the next few numbers in the second.

a) 2, 12, 1112, 3112, 132112, 1113122112, 311311222112, 13211321322112, 1113122113121113222112, 31131122211311123113322112 …

A) Each new number, when read out, describes the previous one. Try it! You’ll see that the second is “one two”, which describes the first. The third is “one one one two”, which describes the second. The fourth is “three one one two” … and so on. Thus the next number in the sequence must describe the last one above. Thus 132113213221133112132123222112.

(Question: will a 4 ever appear in some number in this sequence?)

b) 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9 …

A) For a given entry in the sequence, note its serial number in the sequence. The entry is the number of times that serial number appears in the sequence. (It took me a minute to think about that and get it clear in my mind; you might want to do the same). Thus the first entry is 1, meaning there’s only one 1. The second is 2, meaning there are two 2s—the second and third entries. The third is 2 again, for two 3s—the fourth and fifth entries. The fourth is 3, for three 4s. And so on.

There are only three 4s, but in the sequence as I’ve given it to you, there are four 6s, four 7s, four 8s and four 9s. Those three 4s refer only to the 6s, 7s and 8s, and thus there must be five 9s, five 10s, and so on. So the sequence must contain another 9, then five 10s, and so on. So next few entries are 9, 10, 10, 10, 10, 10 …

I know you enjoyed these as much as I did! Right?

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