Mr and Mrs Wombat live in a house in which the floor of 
every room is a square and covered with identical tiles. 
Mrs Wombat's room contains 101 tiles more than Mr Wombat's 
room. How many tiles does Mrs Wombat's room contain?

Let x = number of tiles in each side of Mr. Wombat's room
Let y = number of tiles in each side of Mrs. Wombat's room

Then there are x² tiles in Mrs. Wombat's room.
And there are y² tiles in Mr. Wombat's room.

>>...Mrs. Wombat's room contains 101 tiles more than Mr. 
Wombat's room...<<

             y² = x² + 101

        y² - x² = 101

 (y - x)(y + x) = 101

The two parentheses on the left must be a factorization
of 101.  Since 101 is a prime number, its only 
factorization is  1×101. Since (y - x) is smaller than
(y + x), we must have

           y - x = 1
and 
           y + x = 101

So we have this system

           y - x = 1
           y + x = 101

Solving that gives x = 51 and y = 50 

So Mrs. Wombat's room has 51 tiles on each side and
Mr. Wombat's room has 50 tiles on each side.

The question is:

>>...How many tiles does Mrs Wombat's room contain?...<<

So her room has 51×51 or 2601 tiles

Checking:
Mrs. Wombat's room has 2601 tiles
Mr. Wombat's room has 50×50 of 2500 tiles
Since 2601 is 101 more than 2500, the solution is correct.

Edwin McCravy
AnlytcPhil@aol.com