# Category: Problems

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

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, but 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).”

## The Cattle Problem

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. 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.

Solution:

#### $W$

= number of white bulls
$B$ = number of black bulls
$Y$ = number of yellow bulls
$D$ = number of dappled bulls
$w$ = number of white cows
$b$ = number of black cows
$y$ = number of yellow cows
$d$ = number of dappled cows

The relations come as:

•   $W = (\frac{1}{2} + \frac{1}{3})B + Y$ The white bulls were equal to a half and a third of the black bulls together with the whole of the yellow bulls.
• $B = (\frac{1}{4} + \frac{1}{5})D + Y$ The black [bulls] were equal to the fourth part of the dappled bulls and a fifth, together with, once more, the whole of the yellow bulls
•   $D = (\frac{1}{6} + \frac{1}{7})W + Y$ The remaining bulls, the dappled, were equal to a sixth part of the white bulls and a seventh, together with all of the yellow bulls

## A Yes No Puzzle

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. 🙂 ]

In other words, we could restate the statement as:

If the question you answered before this one was harder than THIS ONE, was the question you answered before this one harder than THIS ONE.

## Three Children, Two Friends and One Mathematical Puzzle

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?

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:

## Do you multiply this way!

Before my college days I used to multiply this way.

But as time passed, I learned new things. In a Hindi magazine named “Bhaskar Lakshya”, I read an article in which a columnist ( I can’t remember his name) suggested how to multiply in single line (row). That was a magic to me.  I found doing multiplications this way, very faster – easier and smarter. There may be many who already know this method, but many others will be seeing it for the first time.

The ‘only’ requirements for using this method is the quick summation. You should be good in your calculations. Smarter your calculations, faster you’re.
I’ll try to illustrate this method below. If you had any problems regarding language (it’s poor off-course) and understandings, please feel free to put that into comments.

Let we try to multiply 498 with 753.
$4 9 8 \ \times 7 5 3$

# Step I

Multiply 8 and 3 and write the unit digit of result carrying other digits for next step. The same is to be done with each step.

# Step 5

The overall work looks like:

I don’t know if there is any algorithm behind it. The pattern of calculation is very simple, which is making crosses and adding numbers.

You can use this method, multiplying larger numbers too. Try this one at your own. Steps are marked for convenience. 🙂