# 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:

 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.

# 381654729 : An Interesting Number Happened To Me Today

Image via Wikipedia

You might be thinking why am I writing about an individual number? Actually, in previous year annual exams, my registration number was 381654729. Which is just an ‘ordinary’ 9-digit long number. I never cared about it- and forgot it after exam results were announced. But today morning, when I opened “Mathematics Today” magazine’s October 2010, page 8; I was brilliantly shocked. 381654729 is a nine digit number with each of the digits from 1 to 9 appearing once. The whole number is divisible by 9. If you remove the right-most digit, the remaining eight-digit number is divisible by 8. Again removing the next-right-most digit leaves a seven-digit number that is divisible by 7. Similarly, removing next-rightmost digit leaves a six-digit number that is divisible by 6. This property continues all the way down to one digit.
Further research on this number provided a term for this number as Poly-divisible Number.
And I also noticed that a similar problem has been asked in U S A Mathematical Talent SearchÂ  competition. See the first question in the doc below:

To view this document in appropriate size click on View tab of the doc.

After this beautiful incident, I would like to quote a statement here:

Mathematical Wonders happen with Mathematicians.

Numbers always chase me.

# Derivative of x squared is 2x or x ? Where is the fallacy?

As we know that the derivative of $x^2$ , with respect to $x$ , is $2x$.

i.e., $\dfrac{d}{dx} x^2 = 2x$

However, suppose we write $x^2$ as the sum of $x$ ‘s written up $x$ times..

i.e.,

$x^2 = \displaystyle {\underbrace {x+x+x+ \ldots +x}_{x \ times}}$

Now let

$f(x) = \displaystyle {\underbrace {x+x+x+ \ldots +x}_{x \ times}}$

then,

$f'(x) = \dfrac{d}{dx} \left( \displaystyle {\underbrace {x+x+x+ \ldots +x}_{x \ times}} \right)$

$f'(x)=\displaystyle {\underbrace {\dfrac{d}{dx} x + \dfrac{d}{dx} x + \ldots + \dfrac{d}{dx} x}_{x \ times}}$
$f'(x)=\displaystyle {\underbrace {1 + 1 + \ldots + 1 }_{x \ times}}$
$f'(x) = x$

This argument appears to show that the derivative of $x^2$ , with respect to $x$, is actually x, not 2x..

Where is the error?