## Category Archives: Pre-RMO

### Pre-RMO — training

Problem 1:

If a, b, c, and d satisfy the equations

$a+7b+3c+5d=0$

$8a+4b+6c+2d=-16$

$2a+6b+4c+8d=16$

$5a+3b+7c+d=-16$

then what is the numerical value of $(a+d)(b+c)$?

Problem 2:

Suppose x and y are positive integers with $x>y$ and $3x+2y$ and $2x+3y$ when divided by 5, leave remainders 2 and 3, respectively. It follows that when $x-y$ is divided by 5, the remainder is necessarily equal to

(A) 2

(B) 1

(C) 4

(D) none of the foregoing numbers

Problem 3:

The number of different solutions $(x,y,z)$ of  the equation $x+y+z=10$, where each of x, y, and z is a positive integer is

(A) 36

(B) 121

(C) $10^{3}-10$

(D) $C_{3}^{10}-C_{2}^{10}$, which denote binomial coefficients

Problem 4:

The hands of a clock are observed simultaneously from 12.45 pm onwards. They will be observed to point in the same direction some time between

(A) 1:03 pm and 1:04 pm

(B) 1:04 pm and 1:05pm

(C) 1:05 pm and 1:06 pm

(D) 1:06 pm and 1:07 pm.

More later,

Nalin Pithwa

### Fun with Number Theory — Pre-RMO

Here is an elementary number theory problem which can be looked upon as practice problem for pre-RMO or even RMO or just plain fun with math.

Problem:

Find the least number whose last digit is 7 and which becomes 5 times larger when this last digit is carried to the beginning of the number.

Solution:

This is fun way to learn number theory or some Math. So, go ahead and try it. Your suggestions, answers, comments are welcome 🙂

More later,

Nalin Pithwa

### Percentage play

Alphonse bought two bicycles. He sold one to Bettany for 300 pounds making a loss of 25%, and one to Gemma also for 300 pounds making a profit of 25%. Overall, did he break even? If not, did he make a profit or loss, and by how much?

More later,

Nalin Pithwa

### Pre RMO type practice questions

1. Let $x_{1}, x_{2}, \ldots, x_{100}$ be positive integers such that $x_{i}+x_{i+1}=k$ for all i, where k is a constant. I $x_{10}=1$, find the value of $x_{1}$.
2. If $a_{0}=1, a_{1}=1$ and $a_{n}=a_{n-1}a_{n-2}+1$ for $n > 1$, then find out if $a_{465}$, $a_{466}$ are even or odd.
3. Two trains of equal length L, travelling at speeds $V_{1}$ and $V_{2}$ miles per hour in opposite directions, take T seconds to cross each other. Then, find L in feet (1 mile 1280 feet).
4. A salesman sold two pipes at Rs. 12 each. His profit on one was 20% and the loss on the other was 20%. Then, on the whole, what amount did he gain or lose or did he break even?
5. What is the digit in the units position of the integer $1! +2! +3! + \ldots +99!$?
6. Find the value of the following expression:

$(1+q)(1+q^{2})(1+q^{4})(1+q^{8})(1+q^{16})(1+q^{32})(1+q^{64})$ where $q \neq 1$.

Good luck for the ensuing oct pre RMO 🙂

Nalin Pithwa

### 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

### Rabbits in the Hat

The Great Whodunni, a stage magician, placed his top hat on the table.

‘In this hat are two rabbits,’ he announced. ‘Each of them is either black of white, with equal probability. I am now going to convince you, with the aid of my lovely assistant Grumpelina, that I can deduce their colours without looking inside the hat!’

He turned to his assistant, and extracted a black rabbit from her costume. ‘Please place this rabbit in the hat.’ She did.

Whodunnii now turned to the audience. ‘Before Grumpelina added the third rabbit, there were four equally likely combinations of rabbits. ‘ He wrote a list on a small blackboard:BB, BW, WB and WW. ‘Each combination is equally likely — the probability is 1/4.

But, then I added a black rabbit. So, the possibilities are BBB, BWB, BBW and BWW — again, each with probability 1/4.

‘Suppose —- I won’t do it, this is hypothetical — suppose I were to pull a rabbit from the hat. What is the probability that it is black? If the rabbits are BBB, that probability is 1. If BWB or BBW, it is 2/3. If BWW, it is 1/3. So the overall probability of pulling out a black rabbit is

$\frac{1}{4} \times 1 + \frac{1}{4} \times \frac{2}{3}+\frac{1}{4} \times \frac{2}{3}+\frac{1}{4} \times \frac{1}{3}$

which is exactly 2/3.

‘But. If there are three rabbits in a hat, of which exactly r are black and the rest white, the probability of extracting a black rabbit is r/3. Therefore, $r=2$, so there are two black rabbits in the hat.’ He reached into the hat and pulled out a black rabbit. ‘Since I added this black rabbit, the original pair must have been one black and one white!’

The Great Whodunni bowed to tumultous applause. Then, he pulled out two rabbits fwrom the hat — one pale lilac and the other shocking pink.

It seems evident that you can’t deduce the contents of a hat without finding out what’s inside. Adding the extra rabbit and then removing it again (was it the same black rabbit? Do we care?) is a clever piece of misdirection. But, why is the calculation wrong?

More later,

Nalin Pithwa

### Shaggy Dog Story

Brave Sir Lunchalot was travelling through foreign parts. Suddenly, there was a flash of lightning and a deafening crack of thunder, and the rain started bucketing down. Fearing rust, he headed for the nearest shelter, Duke Ethelfred’s castle. He arrived to find the Duke’s wife, Lady Gingerbere weeping piteously.

Sir Lunchalot liked attractive young ladies, and for a brief moment he noticed a distinct glint through Gingerbere’s tears. Ethelfred was very old and frail, he observed…Only one thing, he vowed would deter him from a secret tryst with the Lady — the one thing in all the world that he could not stand.

Puns.

Having greeted the Duke, Lunchalot enquired why Gingerbere was so sad.

“It is my uncle Elpus,” she explained. “He died yesterday.”

“Permit me to offer my sincerest condolences,” said Lunchalot.

“That is not why I weep so…so piteously, sir knight,” replied Gingerbere. “My cousins Gord, Evan and Liddell are unable to fulfill the terms of uncle’s will.”

“Why ever not?”

“It seems that Lord Elpus invested the entire family fortune in a rare breed of giant-riding dogs. He owned 17 of them.”

Lunchalot had never heard of a riding-dog, but he did not wish to display his ignorance in front of such a lithesome lady. But, this fear, it appeared, could be set to rest, for she said, “Although I have heard much of these animals, I  myself have never set eyes on one.”

“They are no fit sight for a fair lady,” said Ethelfred firmly.

“And, the terms of the will —?” Lunchalot asked, to divert the direction of the conversation.

“Ah, Lord Elpus left everything to the three sons. He decreed that Gord should receive half the dogs, Evan one third, and Liddell one ninth.”

“Mmm. Could be messy.”

“No dog is to be subdivided, good knight.”

Lunchalot stiffened at the phrase good knight, but decided it had been uttered innocently and was not a pathetic attempt at humour.

“Well, —- : Lunchalot began.

“Pah, ’tis a puzzle as ancient as yonder hills!” said Ethelfred scathingly. “All you have to do is take one of your riding dogs over to the castle. Then, there are 18 of the damn things!”

“Yes, my husband, I understand the numerology, but —”

“So, the first son gets half that, which is 9; the second gets one third which is 6; the third son gets one ninth, which is 2. That makes 17 altogether, and our own dog can be taken back here!”

“Yes, my husband, but we have no one here who is manly enough to ride such a dog.”

Sir Lunchalot seized his opportunity. “Sire, I will ride your dog!” The look of admiration in Gingerbere’s eye showed him how shrewd his gallant gesture had been.

“Very well,” said Ethelfred.”I will summon my houndsman and he will bring the animal to the courtyard. Where we shall meet them.”

They waited in an archway as the rain continued to fall.

When the dog was led into the courtyard, Lunchalot’s jaw dropped so far that it was a good job he had his helmet on. The animal was twice the size of an elephant, with thick striped fur, claws like broadswords, blazing red eyes the size of Lunchalot’s shield, huge floppy ears dangling to the ground, and a tail like a pig’s — only with more twists and covered in sharp spines. Rain cascaded off its coat in waterfalls. The smell was indescribable.

Perched improbably on its back was  a saddle.

Gingerbere seemed even more shocked than he by the sight of this terrible monstrosity. However, Sir Lucnhalot was undaunted. Nothing could daunt his confidence. Nothing could prevent a secret tryst with the lady, once he returned astride the giant hound, the will executed in full. Except…

Well, as it happened, Sir Lunchalot did not ride the monstrous dog to Lord Elpus’s castle, and for all he knows the will has still not been executed. Instead, he leaped on his horse and rode off angrily into the stormy darkness, mortally offended, leaving Gingerbere to suffer the pangs of unrequited lust.

It wasn’t Ethelfred’s’ dodgy arithmetic — it was what the Lady had said to her husband in a stage whisper.

What did she say?

🙂 🙂 🙂

More later,

Nalin Pithwa

### Alien Encounter: test your logic skills

Alien Encounter.

The starship Indefensible was in orbit around the planet Noncompostments, and Captain Quirk and Mr. Crock had beamed down to the surface.

“According to the Good Galaxy Guide, there are two species of intelligent aliens on this planet,” said Quirk.

“Correct, Captain —- Veracitous and Gibberish. They all speak Galaxic, and they can be distinguished by how they answer questions. The Veracitors reply truthfully, and the Gibberish always lie.’

‘But physically —‘

‘— they are indistinguishable, Captain.’

Quirk heard a sound, and turned to find three aliens creeping up on them. They looked identical.

‘Welcome to Noncomposimentis,’ said one of the aliens.

‘I thank you. My name is Quirk. Now, you are…’ Quirk paused. ‘No point in asking their names,’ he muttered. ‘For all we know, they will be wrong.’

‘That is logical, Captain,’ said Crock.

‘Because we are poor speakers of Galaxic,’ Quirk improvised, ‘I hope you will not mind if I call you Alfy, Betty and Gemma. As he spoke, he pointed to each of them in turn. Then, he turned to Crock and whispered, ‘Not that we know what sex they are, either.’

‘They are all hermandrofemigynes,’ said Crock.

‘Whatever. Now, Alfy: to which species does Betty belong?’

‘Gibberish.’

‘Ah, Betty: do Alfy and Gemma belong to different species?’

‘No.’

‘Right…Talkative lot, aren’t they? Um…Gemma: to which species does Betty belong?’

‘Veracitor.’

Quirk nodded knowledgeably.’Right, that’s settled it, then!’

‘Settled what, Captain?’

‘Which species each belongs to.’

‘I see. And those species are —?’

‘Haven’t the foggiest idea, Crock. You’re the one who’s supposed to be logical!’

************************************************************************************

So, dear students, have some fun!!

More later,

Nalin Pithwa