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How Many Fishes in One Year? [A Puzzle in Making]

Sunday, November 13th, 2011 08:49 / 12 Comments

This is a puzzle which I told to my classmates during a talk, a few days before. I did not represent it as a puzzle, instead of a talk suggesting the importance of Math in general life. This is partially solved for me and I hope you will run your brain-horse to help me solve it completely. If you didn’t notice, this puzzle is not a part of A Trip To Mathematics series. Puzzle which I discussed in the talk was something like this:

“Let I have seven fishes in a huge tank of water —four male and three females. Those were allowed to sex independently but under some conditions. One male is allowed to have intercourse with female, unless other has done so. A male can have intercourse with any number of female fishes possible. If we assume that first female fish could give 100 eggs, second female fish could give 110, third 90. We are also known that a female fish might lay eggs in 21 days since the date of sex with male. The children fish are reproductive only after those are 30 days old. In each bunch of children fish of individual female fish, 60% die. In remaining 40% child fishes, the ratio of male and female is 3:2. If two fishes (male and female) can not do intercourse with each other if those are born to same mother fish, one male is allowed to have intercourse with female, unless other has done so, a male can have intercourse with any number of fishes possible, every three female fishes lay eggs in order of 100,110 and 90 eggs out of which only 40% remain alive having a ratio of male and females of 1:1 and the same rule applies to third, fourth and consecutive generations of fishes; then find the number of total fishes in my tank after one year (365 days).”

I have done many proofreads of this puzzle and found it valid. Your comments, your ideas and suggestions might help me working more rigorously on this puzzle. This puzzle is neither too hard nor too easy. I will be updating this post frequently as my work on this puzzle is directed towards a correct way.

The Cattle Problem

Tuesday, October 11th, 2011 17:23 / 1 Comment

This is a famous problem of intermediate analysis, also known as ‘Archimedes’ Cattle Problem Puzzle’, sent by Archimedes to Eratosthenes as a challenge to Alexandrian scholars. In it one is required to find the number of bulls and cows of each of four colors, the eight unknown quantities being connected by nine conditions. These conditions ultimately form a Pell equation which solution is necessary in case of finding the answer of the puzzle.

: Longhorn Cows in the Southwestern Sun, By T.Paden

The Greek puzzle is stated below with a little deviation. I have just tried to make the language simpler than the original, hope you’ll be able to grasp the puzzle easily.

O Stranger! If you are intelligent and wise, find the number of cattle of the Sun, who once upon a time grazed on the fields of an Island, divided into four groups (herds) of different colors, one white, another a black, a third yellow and the last dappled color.In each herd were bulls, mighty in number according to these proportions:

White bulls were equal to a half and a third of the black together with the whole of the yellow.

The black bulls were equal to the fourth part of the dappled and a fifth, together with, once more, the whole of the yellow.

The dappled bulls, were equal to a sixth part of the white and a seventh, together with all of the yellow.

So, these were the proportions of bulls, now the
proportions of the cows were as following:

White cows were equal to the third part and a fourth of the whole herd of the black.

Black cows were equal to the fourth part once more of the
dappled and with it a fifth part, when all cattle, including the bulls, went to pasture together. Now the dappled in four parts were equal in number to a fifth part and a sixth of the yellow herd.

Yellow cows were in number equal to a sixth part and a seventh of the white herd.

Keeping above conditions in focus, find the number of cattle of the Sun, giving separately the number of well-fed bulls and again the number of females according to each color.
But come, this solution is not complete unless you understand all these conditions regarding the cattle of the Sun:

When the white bulls mingled their number with the black, they stood firm, equal in depth and breadth. Number of bulls in a row were equal to the number of columns.

When the yellow and the dappled bulls were gathered into one herd they stood in such a manner that
their number, beginning from one, grew slowly greater till it completed a triangular figure,
there being no bulls of other colors in their midst nor none
of them lacking.

Find the number of cows and bulls of each color separately.
(more…)

A Yes No Puzzle

Friday, October 7th, 2011 00:11 / 19 Comments

This is not just math, but a very good test for linguistic reasoning. If you are serious about this test and think that you’ve a sharp [at least average] brain then read the statement (only) below –summarize it –find the conclusion and then answer that whether summary of the statement is Yes or No.
[And if you're not serious about the test ...then read the whole post to know what the stupid author was trying to tell you.

]
STATEMENT: If the question you answered before you answered the question you answered after you answered the question you answered before you answered this one, was harder than the question you answered after you answered the question you answered before you answered this one, was the question you answered before you answered this one harder than this one? YES or NO?

(more…)

The problem of the Hundred Fowls

Saturday, October 1st, 2011 17:37 / Leave a Comment

This is a popular Chinese problem, on Linear Diophantine equations, which in wording seems as a puzzle or riddle. However, when used algebraic notations, it looks obvious. The problems states :

If a cock is worth 5 coins, a hen 3 coins, and three chickens together 1 coin, how many cocks, hens and chickens, totaling 100 in number, can be bought for 100 coins?

This puzzle in terms of algebraic equations can be written as $5x+3y+\frac{1}{3}z=100$ and $x+y+z=100$
where $x, y, z$ being the number of cocks, hens and chicks respectively.
We find that there are two equations with three unknown quantities. So eliminating one of the unknowns, by putting $z=100-x-y$ from second equation into first one such that $5x+3y+\frac{1}{3} (100-x-y)=100$
or, $15x+9y+100-x-y=300$
or, $14x+8y=200$
or, $7x+4y=100$ .
Which is a linear Diophantine equation (with only two unknown quantities).
The equation $7x+4y=100$ has the general solution [links to WolframAlpha] $x=4 t$ and $y=25-7t$ , so that $z=75+3t$ where $t$ is an arbitrary integer.
Now, since $x, y, z$ are the number of creatures, hence $x, y, z >0$ and thus $4t >0$ , $25-7t >0$ and $75+3t >0$ which imply that $0 < t < 3\frac{4}{7}$ . And because t must have integer values, we have $t=1,2,3$ . Which gives the following three solutions:

Values of $t$	No. Of cocks ( $x=4 t$ )	No. Of hens ( $y=25-7t$ )	No. Of chicks ( $z=75+3t$ )
1	4	18	78
2	8	11	81
3	12	4	84

So there are the three ways to chose the number of cocks, hens and chicken totaling 100 to buy for 100 coins.

Problem Sources:
Elementary Number Theory
David M. Burton, 2006
McGrawHill Publications

Wikipedia article on Diophantine Equations

Image Credit

Three Children, Two Friends and One Mathematical Puzzle

Sunday, September 25th, 2011 00:15 / 3 Comments

Two close friends, Robert and Thomas, met again after a gap of several years.
Robert Said: I am now married and have three children.
Thomas Said: That’s great! How old they are?
Robert: Thomas! Guess it yourself with some clues provided by me. The product of the ages of my children is 36.
Thomas: Hmm… Not so helpful clue. Can you please give one more?
Robert: Yeah! Can you see the number on the house across the street?
Thomas: Yes! I can.
Robert: The sum of their ages equal that number.
Thomas: Sorry! I still could not determine their ages.
Robert: My oldest child has red hair.
Thomas: OH.. Oldest one? Finally I got it. I know age of each of your children.

Question:

What were the ages of Robert’s children and how did Thomas know?

Discussion and probable answer

This is a very good logical problem. To do it, first write down all the real possibilities that the number on that building might have been. Assuming integer ages one get get the following which equal 36 when multiplied:

Age of 1st	Age of 2nd	Age of 3rd	Sum(HouseNo.)
1	1	36	38
1	2	18	21
1	3	12	16
1	4	9	14
1	6	6	13
2	2	9	13
2	3	6	11
3	3	4	10

The biggest clue is that the Thomas DID NOT KNOW after having been told the sum equaled the number on the house. Why didn’t he know? The only reason would be that the number was 13, in which case there are two possible answers. For any other number, the answer is unique and the Thomas would have known after the second clue. So he asked for a third clue. The clue that the oldest had red hair is really just saying that there is an “oldest”, meaning that the older two are not twins. Hence, the answer is that the redhead is 9 years old, and the younger two are both 2 years old.

Source of The Puzzle: This puzzle is a modified form of a puzzle from Science Reporter Magazine, Hindi 1996 and I have changed the names from Ram and Shyam to Robert and Thomas to make this puzzle convenient to read.

Just another way to Multiply (wpgaurav.wordpress.com)

How Genius You Are?

Saturday, August 20th, 2011 19:10 / 13 Comments

Let have a Test:

You need to make a calculation. Please do neither use a calculator nor a paper. Calculate everything “in your brain”.

Take 1000
and add 40.
Now, add another 1000.
Now add 30.
Now, add 1000 again.
Add 20.
And add 1000 again.
And an additional 10.

So, You Got The RESULT! Be Quick! And Click here to check your result.

Quicker you see your result, sharper you are!

Do you think the result is 5000?
Actually, it is not. The correct result is 4100.

A Problem On Several Triangles

Saturday, August 13th, 2011 14:51 / 2 Comments

A triangle $T$ is divided into smaller triangles such that any two of the smaller triangles either have no point in common, or have a vertex in common, or actually have an edge in common. Thus no two smaller triangles touch along part of an edge of them.
For an illustration let me denote the three vertices of T by 1, 2 and 3. Now number each of the vertices of the small triangles by 1, 2, 3. Do this in an arbitrary way, but such that vertices lying on an edge of T must not be numbered by the same number as the vertex of T opposite to that edge.

Show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3.

Solution

To show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3, we show that the number of small triangles whose vertices are labeled with $1,2,3$ is odd and thus actually $>0$ !

We enumerate all small triangles in the picture as $T_1$ , $T_2, \ldots, T_n$ and denote by $a_i$ the number of edges with endpoints $1$ and $2$ in each triangle $T_i$ . Thus, if say the vertices of $T_i$ are labeled by $1,1,2$ , then $a_i=2$ , and so on …

Observe now that obviously we have

$\displaystyle a_1+a_2+a_3+\cdots +a_n= A+2B,$

where $A$ is the number of triangles whose vertices are labeled $1,2,3$ , while $B$ is the number of those triangles labeled by $1,1,2$ or $1,2,2$ . (Actually it is easily seen that $a_i=2$ for such triangles, while $a_i=1$ if the vertices of $T_i$ are $1,2,3$ and $a_i=0$ otherwise.) All we have to show is that $A$ is odd.

Let $C$ denote the number of $12$ -edges lying inside the original triangle $T$ and let $D$ be the number of $12$ -edges lying on the boundary of $T$ . Every interior $12$ -edge lies in two triangles $T_i$ and thus it is counted twice in the sum $a_1+a_2+a_3+\cdots +a_n$ , while every boundary $12$ -edge is counted only once. In conclusion we get

$\displaystyle a_1+a_2+a_3+\cdots +a_n= 2C+D,$

which yields

$\displaystyle A+2B=2C+D.$

Hence $A$ is odd if and only if $D$ is odd. It is therefore enough to show that $D$ is odd.

According to the hypothesis of the problem, edges labeled $12$ or $21$ can occur only on the $12$ -edge of the large triangle $T$ . We start walking along the edge $12$ of the triangle $T$ starting at the vertex $1$ toward the vertex $2$ . Now, only when we first pass an edge labeled $12$ will we arrive at the first vertex labeled $2$ . A number of vertices labeled $2$ may now follow, and only after we have passed a segment $21$ do we reach a label $1$ , and so on. Thus after an odd number of segments $12$ or $21$ we arrive at vertices labeled $2$ , and after an even number of such segments we arrive at vertices labeled $1$ . Since the last vertex we will reach is the vertex $2$ of the big triangle $T$ , it follows that the total number of segments $12$ or $21$ lying on the side $12$ of the big triangle $T$ must be odd! The same reasoning applies for each of the other edges of the big triangle $T$ , so we deduce that $D$ , the total number of $12$ or $21$ -edges lying on the boundary of $T$ , must be odd. Proved

Graphical Proof

It is obvious. As a result of this numbering we get following diagram:

: Problem Image

MY DIGITAL NOTEBOOK

Tag Archives: Puzzles

How Many Fishes in One Year? [A Puzzle in Making]

The Cattle Problem

A Yes No Puzzle

The problem of the Hundred Fowls

Three Children, Two Friends and One Mathematical Puzzle

Question:

Discussion and probable answer

How Genius You Are?

Take 1000
and add 40.
Now, add another 1000.
Now add 30.
Now, add 1000 again.
Add 20.
And add 1000 again.
And an additional 10.

A Problem On Several Triangles

Solution

Graphical Proof

Post navigation

Follow “MY DIGITAL NOTEBOOK”

Tag Archives: Puzzles

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Question:

Discussion and probable answer

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Better You Share!

Like this:

Take 1000 and add 40. Now, add another 1000. Now add 30. Now, add 1000 again. Add 20. And add 1000 again. And an additional 10.

Better You Share!

Like this:

Solution

Graphical Proof

Better You Share!

Like this:

Post navigation

Follow “MY DIGITAL NOTEBOOK”

Take 1000
and add 40.
Now, add another 1000.
Now add 30.
Now, add 1000 again.
Add 20.
And add 1000 again.
And an additional 10.