## Monthly Archives: August 2015

### More Clock Problems

Example.

At what time between 4 and 5 o’clock will the minute-hand of a watch be 13 minutes in advance of the hour hand?

Solution.

Let x denote the required number of minutes after 4 o’clock; then, as the minute hand travels twelve times as fast as the hour hand, the hout hand will move over x/12 minute divisions in x minutes. At 4 o’clock, the minute hand is 20 divisions behind the hour hand, and the finally minute hand is 13 divisions in advance; therefore the minute hand moves over $20+13$, that is,, 33 divisions more than the hour hand.

Hence, $x=\frac{x}{12}+33$ which implies $\frac{11x}{12}=33$ and hence, $x=36$.

Thus, the time is 36 minutes past 4.

If the question be asked as follows: “At what times between 4 and 5 o’clock will there be 13 minutes between the two hands, then we must also take into consideration, the case when the minute hand is 13 divisions behind the hour hand. In this case, the minute hand gains $20-13$ or 7 divisions.

Hence,, $x=\frac{x}{12}+7$ which gives $x=7 \frac{7}{11}$

Therefore, the times are $7\frac{7}{11}$ past 4, and $36^{'}$ past 4.

Homework for fun:

1. At what time between one and two o’clock are the hands of a watch first at right angles?
2. At what time between 3 and 4 o’clock is the minute hand one minute ahead of the hour hand?
3. When are the hands of a clock together between the hours of 6 and 7?
4. It is between 2 and 3 o’clock, and in 10 minutes the minute hand will be as much before the hour hand as it is not behind it; what is the time?
5. At what times between 7 and 8 o’clock will the hands of a watch be at right angles to each other? When will they be in the same straight line?

Hope you had enough fun! 🙂

More fun later,

Nalin Pithwa

### Interchanging the Hands of a Clock

Problem.

The biographer and friend of the immortal physicist Albert Einstein, A. Moszkowski, wished to distract his friend during an illness and suggested the following problem:

The problem he posed was this: “Take the position of the hands of a clock at 12 noon. If the hour hand and the minute hand were interchanged in this position the time would would still be correct. But, at other times (say at 6 o”clock) the interchange would be absurd, giving a position that never occurs in ordinary clocks: the minute cannot be on 6 when the hour hand points to 12. The question that arises is when and how often do the hands of a clock occupy positions in which interchanging the hands yields a new position that is correct for an ordinary clock?

“Yes,” replied Einstein, “this is just the type of problem for a person kept to his bed by illness; it is interesting enough and not so very easy. I am afraid thought that the amusement won’t last long because I already have my fingers on a solution.”

“Getting up in bed, he took a piece of paper and sketched the hypothesis of the problem. And, he solved it in no more time than it took me to state it.”

How is the problem tackled?

Solution:

We measure the distance of the hands around the dial from the point 12 in sixtieths of a circle.

Suppose one of the required positions of the hands was observed when the hour hand moved x divisions, from 12, and the minute hand moved q divisions. Since the hour hand passes over 60 divisions in 12 hours, or 5 divisions every hour, it covered the x divisions in x/5 hours. In other words, x/5 hours passed after the clock indicated 12 o’clock. The minute hand, covered y divisions in y minutes, that is, in y/60 hours. In other words, the minute hand passed the figure 12 a total of y/60 hours ago, or $\frac{x}{5}-\frac{y}{60}$

hours after both hands stood at twelve. This number is whole(from 0 to 11) since it shows how many whole hours have passed since twelve.

When the hands are interchanged, we similarly find that $\frac{y}{5}-\frac{x}{60}$

whole hours have passed from 12 o’clock to the time indicated by the hands. This is a whole number from 0 to 11.

And, so we have the following system of equations: $\frac{x}{5}-\frac{y}{60}=m$ $\frac{y}{5}-\frac{x}{60}=n$

where m and n are integers (whole numbers) that can vary between 0 and 11. From this system, we find $x=\frac{60(12m+n)}{143}$ $y=\frac{60(12n+m)}{143}$

By assigning m and n the values from 0 to 11, we can determine all the required positions of the hands. Since each of the 12 values of m can be correlated with each of the 12 values of n, it would appear that the total number of solutions is equal to 12 times 12, that is, 144. Actually, however, it is 143 because when $m=0, n=0$ and also when $m=11, n=11$ we obtain the same position of the hands.

Put $m=11, n=11$, we have $x=60, y=60$

and the clock shows 12, as in the case of $m=0, n=0$.

We will not discuss all possible positions, but only two.

First example: $m=1, n=1$ $x=\frac{60.13}{143}=5\frac{5}{11}$ and $y=5\frac{5}{11}$

and the clock reads 1 hour $5 \frac{5}{11}$ minutes; the hands, have merged by this time and they can of course be interchanged (as in all other cases of coincidence of the hands).

Second Example. $m=8, n=5$. $x=\frac{60(5+12.8)}{143} \approx 42.38$ and $y=\frac{90(8+12.5)}{143} \approx 28.53$.

The respective times are 8 hours 28.53 minutes and 5 hours 42.38 minutes.

More clock problems later,

Nalin Pithwa

### Complex Numbers — a typical IITJEE Main problem

Example 1.

If $\omega$ is the imaginary cube root of unity, then value of the expression $1(2-\omega)(2-\omega^{2}) + 2(3-\omega)(3-\omega^{2})+\ldots+(n-1)(n-\omega)(n-\omega^{2})$ is

(a) $\frac{1}{4}{n^{2}}(n+1)^{2}-n$

(b) $\frac{1}{4}{n^{2}}(n-1)^{2}+n$

(c) $\frac{1}{4}{n^{2}}(n+1)-n$

(d) $\frac{1}{4}n(n+1)^{2}-n$

Solution:

rth term of the given expression is $r(r+1-\omega)(r+1-\omega^{2})=(r+1-1)(r+1-\omega)(r+1-\omega^{2})=(r+1)^{3}-1$

because $x^{3}-1=(x-1)(x-\omega)(x-\omega^{2})$.

Thus, the value of the expression is given by $\sum_{r=1}^{n-1}r(r+1-\omega)(r+1-\omega^{2})=\sum_{r=1}^{n-1}[(r+1)^{3}-1]$ $=2^{3}+3^{3}+ \ldots + n^{3}-(n-1)$ $=1^{3}+2^{3}+\ldots + n^{3}-n$ $=\frac{1}{4}n^{2}{(n+1)}^{2}-n$

More later,

Nalin Pithwa

### It’s complex feeling!

Example.

If $z_{1}, z_{2}, z_{3}$ are complex numbers such that $|z_{1}|=|z_{2}|=|z_{3}|=|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}|=1$, then $|z_{1}+z_{2}+z_{3}|$ is

(a)  equal to 1

(b) less than 1

(c) greater than 3

(d) equal to 3

Solution:

Since $|z_{1}|=|z_{2}|=|z_{3}|=1$, we get $z_{1} \overline{z_{1}}=z_{2} \overline{z_{2}}=z_{3} \overline{z_{3}}=1$

Now, $1=|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}|=|\overline{z_{1}}+\overline{z_{2}}+\overline{z_{3}}|=|\overline{z_{1}+z_{2}+z_{3}}|$ $\Longrightarrow 1 = |z_{1}+z_{2}+z_{3}|$

### Heron Suit

No cat that wears a heron suit is unsociable.

No cat without a tail will play with a gorilla.

Cats with whiskers always wear heron suits.

No sociable cat has blunt claws.

No cats have tails unless they have whiskers.

Therefore.

No cat with blunt claws will play with a gorilla.

Is the deduction logically correct?

Nalin Pithwa

### Puzzling poem

One of my students Anubhav C Singh shared the following puzzling poem he had found from the internet:

Ten weary, footsore travellers,
All in a woeful plight,
Sought shelter at a wayside inn
One dark and stormy night.

‘Nine rooms, no more,’ the landlord said
‘Have I to offer you.
To each of eight a single bed,
But the ninth must serve for two.’

A din arose. The troubled host
For of those tired men not two
Would occupy one bed.

The puzzled host was soon at ease –
He was a clever man –
And so to please his guests devised
This most ingeneous plan.

In a room marked A two men were placed,
The third was lodged in B,
The fourth to C was then assigned,
The fifth retired to D.

In E the sixth he tucked away,
In F the seventh man.
The eighth and ninth in G and H,
And then to A he ran,

Wherein the host, as I have said,
Then taking one – the tenth and last –
He logged him safe in I.

Nine singe rooms – a room for each –
Were made to serve for ten;
And this it is that puzzles me
And many wiser men.

What is wrong in this logic?
More later,
Nalin Pithwa

### Deceptive Dice

The Terrible Twins, Innumeratus and Mathophila, were bored.

“I know,” said Mathophila brightly. “Let’s play dice!”

“Don’t like dice.”

“Ah, but these are special dice,” said Mathophila, digging them out of an old-chocolate box. One was red, one yellow and one blue..

Innumeratus picked up the red dice. “There’s something funny about this one,” he said. “It’s got two 3’s, two-4’s and two 8’s.”

“They are all like that,” said Mathophila carelessly. The yellow one has two 1’s, two 5’s and two 9’s — and the blue  one has two 2’s, two 6’s and two 7’s.

“They look rigged to me,” said Innumeratus, deeply suspicious.

“No, they are perfectly fair. Each face has an equal chance of turning up.”

“How do we play, anyway?”

We each choose a different one. We roll them simultaneously, and the highest number wins. We can play for pocket money. Innumeratus looked skeptical, so his sister quickly added: “Just to be fair, I will let you choose first! Then you can choose the best dice!”

“Welllll…”, said Innumeratus, hesitating.

“Should he play? If not, why not?”

More puzzles to keep you entertained later,

Nalin Pithwa

### An Age-Old Old-Age Problem

The Emperor Scrumptius was born in 35 BC, and died on his birthday in AD35. What was his age when he died?

More later,

Nalin Pithwa

### To Find Fake Coin

In February 2003, Harold Hopwood of Gravesend wrote a short letter to the Daily Telegraph, saying that he had solved the newspaper’s crossword every day since 1937, but one conundrum had been nagging away at the back of his mind since his schooldays, and at the age of 82 he had finally decided to enlist some help.

The puzzle was this. You are given 12 coins. They all have the same weight, except for  one, which may be either lighter or heavier than the rest. You have to find out which coin is different, and whether it is light or heavy, using at most three weightings on a pair of scales. The scales have no graduations for weights; they just have two pans, and you can tell whether they are in balance, or the heavier one has gone down and the lighter one has gone up.

Within days, the paper’s letters desk had received 362 letters and calls about the puzzle, nearly all asking for the answer, and they phoned Ian Stewart. He recognized the problem as one of the classic puzzles, typical of the weights and scales genre, but he had forgotten the answer. But his friend Marty, who happened to be in the room when he answered the phone, also recognized the problem. The same puzzle had inspired Marty as a teenager, and his successful solution had led him to become a mathematician.

Of course, he had forgotten how the solution went, but they came up  with a method in which they weighed various sets of coins against various others, and faxed it to the newspaper.

In fact, there are many answers, including a very clever one which Ian Stewart finally remembered on the day that the Telegraph printed out less elegant method. Professor Stewart had seen it twenty years earlier in New Scientist magazine, and it had been reproduced in Thomas H. O’Beirne’s Puzzles and Paradoxes, which Professor Stewart had on his bookshelf.

Puzzles like this seem to come round every twenty years or so, presumably when a new generation is exposed to them, a bit like an epidemic that gets a new lease of life when the population loses all immunity. O’Beirne traced it back to Howard Grossman in 1945, but it is almost certainly much older, going back to the seventeenth century. It wouldn’t surprise me if one day we find it on a Babylonian cuneiform tablet.

O’Beirne offered a ‘decision tree’ solution, along the lines that Marty and Professor Stewart had concocted. He also recalled the elegant 1950 solution published by “Blanche Descartes” in Eureka, the journal of Archimedeans, Cambridge University’s undergraduate mathematics society. Ms Descartes was in actuality Cedric A. B. Smith, and his solution is a masterpiece of ingenuity. It is presented as a poem about a certain Professor Felix Fiddlesticks, and the main idea goes like this:

F set the coins out in a row

And chalked on each a letter so,

To form the words “F AM NOT LICKED”

(An idea in his brain had clicked.)

And now his mother he’ll enjoin:

MA DO LIKE
ME TO FIND
FAKE COIN

This cryptic list of three weightings, one set of four against another solves the problem, as Eureka explains, also in verse. To convince you, Professor Stewart listed all the outcomes of the weightings, according to which coin is heavy or light. Here, L means that the left pan goes down, R that the right pan goes down, and — means they stay balanced. $\begin{tabular}{|| l | l | l | l ||} \hline False coin & 1st weighing & 2nd weighing & 3rd weighing \\ F heavy & --- & R & L \\ F light & --- & L & R \\ A heavy & L & --- & L \\ A light & R & --- & R \\ M heavy & L & --- & L \\ M light & R & R & --- \\ N heavy & --- & R & R \\ N light & --- & L & L \\ O heavy & L & L & R \\ O light & R & R & L \\ T heavy & --- & L & --- \\ T light & --- & R & --- \\ L heavy & R & R & R \\ L light & L & L & L \\ C heavy & --- & --- & R \\ C light & --- & --- & L \\ K heavy & R & --- & L \\ K light & L & --- & R \\ E heavy & R & L & L \\ E light & L & R & R \\ D heavy & L & R & --- \\ D light & R & L & --- \\ \hline \end{tabular}$

You can check that no two possibilities give the same results.

The Telegraph’s publication of a valid solution did not end the matter. Readers wrote in to object to Prof Stewart and his colleagues answer, on spurious grounds. They wrote to improve it, not always by valid methods. They e-mailed to point out Ms Descartes’s solutions or similar ones. They told them about other weighing puzzles. The readers thanked the duo for setting their minds at rest. The readers cursed the duo for reopening old wounds. It was as if some vast secret reservoir of folk wisdom had suddenly been breached. One correspondent remembered that the puzzle had featured on BBC television in the 1960s, with the solution being given the following night. Ominously, the letter continued, “I do not recall why it was raised in the first place, or whether that was my first acquaintance with it; I have a feeling that it was not.”

More puzzles for you later,

Nalin Pithwa

### White-Tailed Cats

“I see you’ve got a cat.” said Ms Jones to Ms Smith. “I do like its cute white tail! How many cats do you have?”

“Not a lot,” said  Ms Smith. “Ms Brown next door has twenty, which is a lot more than I’ve got.”

“You still haven’t told me how many cats you have?”

“Well…let me put it like this. If you chose two distinct cats of mine at random, the probability that both of them have white tails is exactly half.”

“That doesn’t tell me how many you’ve got!”

“Oh yes it does.”

How many cats does Ms Smith have — and how many have white tails?

More later,

Nalin Pithwa