Thursday, May 11, 2006

What's so strange about me?

Pick a 3-digit number where the first and last digits differ by 2 or more...

* Consider the "reverse" number, obtained by reading it backwards.
* Subtract the smaller of these two numbers from the larger one.
* Add the result to its own reverse.

What is the peculiarity you observe in the result? Can you explain why it is so? Is there some mathematical logic behind this?

Level: Easy


Anonymous Anonymous said...

Bobach writes: I enjoy your blog. You guys are good. Here's my first response:

Any 3 digit number can be represented as
100x + 10y + z and its reverse would be
100z + 10y - x

Their difference would be
100(x-z) + (z-x) which can be written as 99(x-z).

When 1 < (x-z) < 10, 99(x-z) can be written as
100(x-z-1) + 90 + (10-(x-z)) whose reverse would be
100(10-(x-z)) + 90 + (x-z-1)

Adding these last together we find that all the variables cancel each other so we get
100(10-1) + 180 + (10-1) which always equals 900+180+9=1089

8:04 PM, May 12, 2006  
Blogger Karthik Nagaraj said...

Welcome aboard! Thats a exceelent piece of solution you have there. There is another slight variation of the solution I have come across:

The difference is 99(x-z) and we know that x-z lies between 2 and 9. So, the difference is divisible by 99. Such numbers are 198, 297, on. We notice that such numbers have a 9 in the middle digit and the sum of the other 2 digits is equal to 9. So consider the difference number to be abc, this can be represented as 100a+10b+c and its reverse to be 100c+10b+a. Add them and you get 100(a+c)+10(b)+(a+c) = 100*9+10*9+9=1089 and this is true for any 3 digit number that obeys the conditions given. Good solving Bobach. You will go into the list of the people who solved the puzzles.

This puzzle is SOLVED!

7:41 AM, May 13, 2006  

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