Thursday, June 23, 2011

Ones and nines

Show that all the divisors of any number of the form 19...9 (with an odd number of nines) end in 1 or 9.  For example, the numbers 19, 1999, 199999, and 19999999 are prime (so clearly the property holds), and the (positive) divisors of 1999999999 are 1, 31, 64516129 and 1999999999 itself.
Show further that this property continues to hold if we insert an equal number of zeroes before the nines.  For example, the numbers 109, 1000999, 10000099999, 100000009999999, and 1000000000999999999 are prime, and the (positive)divisors of 10000000000099999999999 are 1, 19, 62655709, 1190458471, 8400125030569, 159602375580811, 526315789478947368421, and 10000000000099999999999 itself.

Ant in a field

An ant, located in a square field, is 13 meters from one of the corner posts of the field, 17 meters from the corner post diagonally opposite that one, and 20 meters from a third corner post.  Find the area of the field.  Assume the land is flat.
Square field, containing an ant, as described above.

Missing digits

Given that 37! = 13763753091226345046315979581abcdefgh0000000, determine, with a minimum of arithmetical effort, the digits a, b, c, d, e, f, g, and h.  No calculators or computers allowed!

Five card trick

Two information theoreticians, A and B, perform a trick with a shuffled deck of cards, jokers removed.  A asks a member of the audience to select five cards at random from the deck.  The audience member passes the five cards to A, who examines them, and hands one back.  A then arranges the remaining four cards in some way and places them face down, in a neat pile.
B, who has not witnessed these proceedings, then enters the room, looks at the four cards, and determines the missing fifth card, held by the audience member.  How is this trick done?
Note: The only communication between A and B is via the arrangement of the four cards.  There is no encoded speech or hand signals or ESP, no bent or marked cards, no clue in the orientation of the pile of four cards...

Three children

On the first day of a new job, a colleague invites you around for a barbecue.  As the two of you arrive at his home, a young boy throws open the door to welcome his father.  “My other two kids will be home soon!” remarks your colleague.
Waiting in the kitchen while your colleague gets some drinks from the basement, you notice a letter from the principal of the local school tacked to the noticeboard.  “Dear Parents,” it begins, “This is the time of year when I write to all parents, such as yourselves, who have a girl or girls in the school, asking you to volunteer your time to help the girls' soccer team.”  “Hmmm,” you think to yourself, “clearly they have at least one of each!”
This, of course, leaves two possibilities: two boys and a girl, or two girls and a boy.  Are these two possibilities equally likely, or is one more likely than the other?
Note:  This is not a trick puzzle.  You should assume all things that it seems you're meant to assume, and not assume things that you aren't told to assume.  If things can easily be imagined in either of two ways, you should assume that they are equally likely.  For example, you may be able to imagine a reason that a colleague with two boys and a girl would be more likely to have invited you to dinner than one with two girls and a boy.  If so, this would affect the probabilities of the two possibilities.  But if your imagination is that good, you can probably imagine the opposite as well.  You should assume that any such extra information not mentioned in the story is not available.

Dice game

Two students play a game based on the total roll of two standard dice.  Student A says that a 12 will be rolled first.  Student B says that two consecutive 7s will be rolled first.  The students keep rolling until one of them wins.  What is the probability that A will win?


Let p be the probability that student A wins.  We consider the possible outcomes of the first two rolls.  (Recall that each roll consists of the throw of two dice.)  Consider the following mutually exclusive cases, which encompass all possibilities.
  • If the first roll is a 12 (probability 1/36), A wins immediately.
  • If the first roll is a 7 and the second roll is a 12 (probability 1/6 · 1/36 = 1/216), A wins immediately.
  • If the first and second rolls are both 7 (probability 1/6 · 1/6 = 1/36), A cannot win.  (That is, B wins immediately.)
  • If the first roll is a 7 and the second roll is neither a 7 nor a 12 (probability 1/6 · 29/36 = 29/216), A wins with probability p.
  • If the first roll is neither a 7 nor a 12 (probability 29/36), A wins with probability p.
Note that in the last two cases we are effectively back at square one; hence the probability that A subsequently wins is p.
Probability p is the weighted mean of all of the above possibilities.
Hence p = 1/36 + 1/216 + (29/216)p + (29/36)p.
Therefore p = 7/13.

Five men, a monkey, and some coconuts

Five men crash-land their airplane on a deserted island in the South Pacific.  On their first day they gather as many coconuts as they can find into one big pile.  They decide that, since it is getting dark, they will wait until the next day to divide the coconuts.
That night each man took a turn watching for rescue searchers while the others slept.  The first watcher got bored so he decided to divide the coconuts into five equal piles.  When he did this, he found he had one remaining coconut.  He gave this coconut to a monkey, took one of the piles, and hid it for himself.  Then he jumbled up the four other piles into one big pile again.
To cut a long story short, each of the five men ended up doing exactly the same thing.  They each divided the coconuts into five equal piles and had one extra coconut left over, which they gave to the monkey.  They each took one of the five piles and hid those coconuts.  They each came back and jumbled up the remaining four piles into one big pile.
What is the smallest number of coconuts there could have been in the original pile?

Cryptogram Puzzle

Cryptograms are word puzzles in which your primary task is to decode a block of cipher-text. You are faced with a sequence of characters that looks like gibberish. The characters can be letters, numbers, or any other symbols you like. The idea is that each character represents a letter of the alphabet, and when all the characters are replaced with their corresponding letters, the gibberish turns into something meaningful and (hopefully!) recognizable.Here's an example of a cryptogram that has been embellished with a theme and an author of the encrypted phrase - just to spice things up a notch...
Topic - Realism
Word Puzzle Example - Cryptogram
In this example, I've given you a toehold into the grid by telling you that the number 22 represents the letter T. The cryptograms you meet won't necessarily have hints, or themes for that matter (though the themeless cryptograms seem a bit dry and academic for me).As you work your way through the puzzle you make educated guesses at the letters being represented by certain characters by focusing on key word patterns. I'll let you have a go at the one above to get a feel for what I mean by 'key word patterns' here.

Download the pdf file from :

Lets not ruin the interest by giving the solution, I will verify if you post the answer.

Sunday, June 19, 2011

Nature's Numbers

If you've ever looked for a four-leafed clover, you know that nature rarely delivers such a curiosity. Nearly every clover plant you check has the usual three leaves. If you study the flowers in your garden or in the countryside, you'll discover the most common number of petals is five. Buttercups, geraniums, pansies, primroses, rhododendrons, tomato blossoms, and many more all have five petals.

Five also shows up in arrangements of seeds. Cut an apple in half across its core (rather than the usual way down the core from the stem), and you'll see the seeds arranged in a beautiful five-pointed star. What numbers do cucumbers, tomatoes, pears, and lemons feature?

Pineapples have eight rows of scales, seen as roughly diamond-shaped markings, sloping in one direction and 13 sloping in the other. Pine cones show the same sort of feature.

The head of a sunflower highlights other numbers. In a perfect head, the tiny flowers, or florets, that will become seeds are arranged in two spirals, one winding clockwise and the other counterclockwise. Depending on the species of sunflower, you might find 34 and 55, 55 and 89, or even 89 and 144 florets along a spiral.

Similarly, floret spirals at the center of certain types of daisies feature the numbers 21 and 34. You can look for similar patterns on brocolli or cauliflower.

Take a look at the numbers that nature seems to like (at least in plants): 3, 5, 8, 13, 21, 34, 55, 89, and 144. Can you find a pattern?

Here's a clue: Start with 1 + 1 = 2, and then add the two numbers on each side of the equal sign. Keep on doing this with each new equation that you get.

1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34, and so on.

The sums you get are all members of a famous sequence of numbers named for the mathematician Leonardo of Pisa, also known as Fibonacci, who studied them about 800 years ago. Scientists have long wondered why these number come up in plants. The answer may have something to do with the way plants grow, especially the way petals or buds space themselves to gather the most sunlight and nutrients.

Wherever you look, nature certainly has a way with numbers.

Monday, June 13, 2011

Find Faulty Coin

There are 12 coins. One of them is false; it weights differently. It is not known, if the false coin is heavier or lighter than the right coins. How to find the false coin by three weighs on a simple scale?


Two Boys Cross a River Puzzle

The Puzzle:

Two boys wish to cross a river. The only way to get to the other side is by boat, but that boat can only take one boy at a time. The boat cannot return on its own, there are no ropes or similar tricks, yet both boys manage to cross using the boat.


Our Solution:

The boys start on opposite sides of the river.

Bridge crossing

  This problem was published in MAA on line: Crossing a Rickety Bridge at Night By Flashlight.
    A group of four people has to cross a bridge. It is dark, and they have to light the path with a flashlight. No more than two people can cross the bridge simultaneously, and the group has only one flashlight. It takes different time for the people in the group to cross the bridge:
      Annie crosses the bridge in 1 minute, 
      Bob crosses the bridge in 2 minutes, 
      Volodia Mitlin crosses the bridge in 5 minutes,
      Dorothy crosses the bridge in 10 minutes.
How can the group cross the bridge in 17 minutes?


Really Interesting

Somebody thought of two natural numbers greater than 1 and calculated their sum and product (both turned out to be less than 100). He told the sum to S. and the product – to P. The following is their discussion:
P.: I don’t know the numbers.
S.: Me neither.
P.: And I knew from the beginning that you would not know.
S.: And I knew from the beginning that you would know from the beginning.
P.: But I still don’t know them!
S.: But now I do.

What are the numbers?
Try it and post in the comments.....
Its interesting one try yourself......
Easy But hard.....

"Light-bulb" Math Riddles

  1. How many mathematicians does it take to change a light bulb? None. The answer is intuitively obvious or. It's left to the reader as an exercise.
  2. How many numerical analysts does it take to change a light bulb? 3.9967 (after six iterations).
  3. How many mathematical logicians does it take to change a light bulb? None. They can't do it, but they can easily prove that it can be done.
  4. How many classical geometers does it take to change a light bulb? None. You can't do it with a straight edge and a compass.
  5. How many math analysts does it take to change a light bulb? Three. One to prove existence, one to prove uniqueness and one to derive a nonconstructive algorithm to do it.
  6. How many number theorists does it take to change a light bulb? I don't know the exact number, but I am sure it must be some rather elegant prime.
  7. How many statisticians does it take to screw in a light bulb? We really don't know yet. Our entire sample was skewed to the left!
  8. How many math students does it take to change a light bulb? Ten. One to do it and nine to watch.
  9. How many topologists does it take to change a light bulb? Just one. But what will you do with the doughnut?
  10. How many professors does it take change a light bulb? One. With eight research students, two programmers, three post-docs and a secretary to help him.
  11. How many university lecturers does it take to change a light bulb? Four. One to do it and three to co-author the paper.
  12. How many graduate students does it take to change a light bulb? Only one. But it takes nine years.
  13. How many administrators does it take to change a light bulb? None. What was wrong with the old one?

A cat has nine tails.

Theorem. A cat has nine tails.

Proof. No cat has eight tails. Since one cat has one more tail than no cat, it must have nine tails.

Farmer's Sheep


Farmer Longmore had a curious aptitude for arithmetic, and was known in his district as the "mathematical farmer."
The new vicar was not aware of this fact when, meeting his worthy parishioner one day in the lane, he asked him in the course of a short conversation, "Now, how many sheep have you altogether?"
He was therefore rather surprised at Longmore's answer, which was as follows:
"You can divide my sheep into two different parts, so that the difference between the two numbers is the same as the difference between their squares.
Maybe, Mr. Parson, you will like to work out the little sum for yourself."

Can the reader say just how many sheep the farmer had?

Supposing he had possessed only twenty sheep, and he divided them into the two parts 12 and 8.
Now, the difference between their squares, 144 and 64, is 80.
So that will not do, for 4 and 80 are certainly not the same.
If you can find numbers that work out correctly, you will know exactly how many sheep Farmer Longmore owned.


If he divided this sheep (which is best done by weight) into two parts, making one part two-thirds and the other part one-third, then the difference between these two numbers is the same as the difference between their squares—that is, one-third.

Any two fractions will do if the denominator equals the sum of the two numerators..

The farmer had one sheep only!

Wizard's Cats


A wizard placed ten cats inside a magic circle as shown in our illustration, and hypnotized them so that they should remain stationary during his pleasure.
He then proposed to draw three circles inside the large one, so that no cat could approach another cat without crossing a magic circle.


The illustration requires no explanation.
It shows clearly how the three circles may be drawn so that every cat has a separate enclosure, and cannot approach another cat without crossing a line.

The Wizard's Arithmetic

Once upon a time a knight went to consult a certain famous wizard.

"And art thou learned also in the magic of numbers?" asked the knight.
"Show me but one sample of thy wit in these matters."

The old wizard took five blocks bearing numbers, and placed them on a shelf, apparently at random, so that they stood in the order 41096, as shown in our illustration.
He then took in his hands an 8 and a 3, and held them together to form the number 83.

"Sir Knight, tell me," said the wizard, "canst thou multiply one number into the other in thy mind?"

"Nay, of a truth," the good knight replied.
"I should need to set out upon the task with pen and scrip."

"Yet mark ye how right easy a thing it is to a man learned in the lore of far Araby, who knoweth all the magic that is hid in the philosophy of numbers!"

The wizard simply placed the 3 next to the 4 on the shelf, and the 8 at the other end. It will be found that this gives the answer quite correctly—3410968.
Very curious, is it not? How many other two-figure multipliers can you find that will produce the same effect?
You may place just as many blocks as you like on the shelf, bearing any figures you choose.

Sunday, June 12, 2011

Julius Caesar's Last Breath

What's the chance that the breath you just inhaled contains at least one air molecule that was in Julius Caesar's last breath--the one in which he said (according to Shakespeare) "Et tu Brute? Then die Caesar"?

Assume that the more than two thousand years that have passed have been enough time for all the molecules in Caesar's last breath to mix evenly in the atmosphere, and that only a trivial amount of the molecules have leaked out into the oceans or the ground. Assume further that there are about 1044 molecules of air, and about 2 x 1022 molecules in each breath--yours or Caesar's.

That gives a chance of 2 x 1022/1044 = 2x 10-22 that any one particular molecule you breathe in came from Caesar's last breath. This means that the probability that any one particular molecule did not come from Caesar's last breath is [1-2x10-22]. But we want the probability that the first molecule did not come from Caesar's last breath and that the second molecule and that the third molecule and so forth. To determine the probability of not just one thing but of a whole bunch of things that are causally unconnected happening together, we multiply the individual probabilities. Since there are 2x10-22 different molecules, and since each has the same [1-2x10-22] chance of not coming from Caesar's last breath, we need to multiply the probability of any single event--[1-2x10-22]--by itself 2x1022 times. That gives us:


as the probability that none of the molecules in the breath you just inhaled (assuming you are still holding out) came from Julius Caesar's last breath.

How to evaluate this? Recall that if a number x is small, then (1-x) is approximately equal to e-x, where e=2.718281828... is the so-called base of the natural logarithms. So we can rewrite the equation above as:


And remember that when we raise numbers with exponents to further exponents, we simply multiply the exponents together. In this case, one exponent (the chance that a molecule came from Caesar) is very small, and the other (the number of molecules in a breath) is very large. When we multiply them together, we get: [-2x10(-22)] x [2x10(22)] = -4. e-4 is approximately 1/(2.72 x 2.72 x 2.72 x 2.72) = 1/54.7 = 0.018.

Thus there is a 1.8% chance that none of the molecules you are (still) holding in your lungs came from Caesar's last breath. And there is a 98.2% chance that at least one of the molecules in your lungs came from Caesar's last breath.

From John Allen Paulos's Innumeracy.

Amazing Mathmatics Number

Sequential Inputs of numbers with 8
1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321

Sequential 1's with 9
1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 + 10 = 1111111111

Sequential 8's with 9
9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888

Numeric Palindrome with 1's
1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111 = 12345678987654321

Without 8
12345679 x 9 = 111111111
12345679 x 18 = 222222222
12345679 x 27 = 333333333
12345679 x 36 = 444444444
12345679 x 45 = 555555555
12345679 x 54 = 666666666
12345679 x 63 = 777777777
12345679 x 72 = 888888888
12345679 x 81 = 999999999

Sequential Inputs of 9
9 x 9 = 81
99 x 99 = 9801
999 x 999 = 998001
9999 x 9999 = 99980001
99999 x 99999 = 9999800001
999999 x 999999 = 999998000001
9999999 x 9999999 = 99999980000001
99999999 x 99999999 = 9999999800000001
999999999 x 999999999 = 999999998000000001

Sequential Inputs of 6
6 x 7 = 42
66 x 67 = 4422
666 x 667 = 444222
6666 x 6667 = 44442222
66666 x 66667 = 4444422222
666666 x 666667 = 444444222222
6666666 x 6666667 = 44444442222222
66666666 x 66666667 = 4444444422222222
666666666 x 666666667 = 444444444222222222

Day of The Date

Day of the Week:
January has 31 days. It means that every date in February will be 3 days later than the same date in January(28 is 4 weeks exactly). The below table is calculated in such a way. Remember this table which will help you to calculate.
January 0
February 3
March 3
April 6
May 1
June 4
July 6
August 2
September 5
October 0
November 3
December 5

Step1: Ask for the Date. Ex: 23rd June 1986
Step2: Number of the month on the list, June is 4.
Step3: Take the date of the month, that is 23
Step4: Take the last 2 digits of the year, that is 86.
Step5: Find out the number of leap years. Divide the last 2 digits of the year by 4, 86 divide by 4 is 21.
Step6: Now add all the 4 numbers: 4 + 23 + 86 + 21 = 134.
Step7: Divide 134 by 7 = 19 remainder 1.
The reminder tells you the day.
Sunday 0
Monday 1
Tuesday 2
Wednesday 3
Thursday 4
Friday 5
Saturday 6
Answer: Monday

Monty Hall Problem

The Monty Hall Dilemma was discussed in the popular "Ask Marylin" question-and-answer column of the Parade magazine. Details can also be found in the "Power of Logical Thinking" by Marylin vos Savant, St. Martin's Press, 1996.

Marylin received the following question:

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?
Craig. F. Whitaker
Columbia, MD


Assume that you always start by picking Door #1, and the host then always shows you some other door which does not contain the car, and you then always switch to the remaining door.

If the car is behind Door #1, then after you pick Door #1, the host will open another door (either #2 or #3), and you will then switch to the remaining door (either #3 or #2), thus LOSING.

If the car is behind Door #2, then after you pick Door #1, the host will be forced to open Door #3, and you will then switch to Door #2, thus WINNING.

If the car is behind Door #3, then after you pick Door #1, the host will be forced to open Door #2, and you will then switch to Door #3, thus WINNING.

Hence, in 2 of the 3 (equally-likely) possibilities, you will win. Ergo, the probability of winning by switching is 2/3.

Lewis Carroll's Pillow Problem

This problem is cited by M. Gardner in his Mathematical Circus and also Gardner's Workout.

A bag contains a counter, known to be either white or black. A white counter is put in, the bag is shaken, and a counter is drawn out, which proves to be white. What is now the chance of drawing a white counter?
Lewis Carroll offers two solutions - one short and wrong (Proof 1 below), another right but long. (He of course new which is which.) Martin Gardner [Circus, p. 189] calls the second of Carroll's proofs "long-winged" and affers instead a shorter one (Proof 2 below) by one of his readers, Howard Ellis from Chicago.

Solution #1

As the state of the bag, after the operation, is necessarily identical with its state before it, the chance is just what it was, viz. 1/2.

Solution #2

Let B and W1 stand for the black or white counter that may be in the bag at the start and W2 for the added white counter. After removing white counter there are three equally likely states:

Inside bagOutside bag

In two of these states a white counter remains in the bag, and so the chance of drawing a white counter the second time is 2/3.


  1. M. Gardner, Gardner's Workout, A K Peters, 2001, pp. 129-132
  2. M. Gardner, Mathematical Circus, Vintage, 1981