Alan

    x = 0.111111.....                 y = 0.01010101......                  z = 0.0001000100010001.....

10x = 1.11111.....             100y = 1.01010101.....         10000z = 1.000100010001.....

  9x = 1                              99y = 1                                9999z = 1

   x = 1/9                                y = 1/99                                   z = 1/9999

   x+y+z = 1213/9999

\(0.0001 = 10^{-4}\\ \frac{10^{46}}{10^{-4}}=10^{50}\\ \sqrt{10^{50}}=10^{25}\\10^{-6}\times 10^{25}=10^{19}\\\text{ Hence : }n=19\)

Here's the picture:

Approach this by first drawing circles centred on point (-5,0) of radii 1, 2, 3 ... 10.

Then draw circles centred on point (5,0) with the same radii.

At the points where these circles intersect the distance from A is a positive integer less than or equal to 10, as is the distance from B.

It is a simple matter to count these intersection points. There are, indeed, 119 of them.

Note that x-y must be larger than x+y if the ratio is to be between 2 and 5. This will be the case if y is negative.

So we have (x-y)/(x+y) > 2,  or. x-y > 2x+2y or x < -3y or x/y > -3 where the < becomes > because we are dividing by a negative number.

We also have (x-y)/(x+y)<5,  or x-y < 5x+5y or 4x > -6y or x > -(3/2)y or x/y < -3/2 where > becomes < because we are dividing by a negative number.

Hence -3 < x/y < -3/2, giving x/y = -2

Max got the right answer with incorrect reasoning! It happens!

Since the numbers in brackets are all greater than 1 for 0 < r < 3, the smallest value of the bracketed term will require the largest value of x.

The smallest is (1 + r/10), so (B) will require the largest x.

Question 2.

Thanks Melody.  This approach still requires some trial and error of course.  In this case that worked out ok at the first guess.  It won't necessarily do so for every problem!

I looked at it this way:

N = 23m + 22        (1)                  m is an integer

N = 21n + 14         (2)                  n is an integer

From (1) and (2) we have:   

21n - 23m = 8

or  21(n-m) -2m = 8 

or 21(n-m)/2  - m = 4    (3)

Now (n-m)/2 must be an integer to satisfy (3).

Try (n - m)/2 = 1    (4)

Put this into (3) to get 21*1 - m = 4  or m = 17    (5)

From (4) and (5)   n = 2 + m = 2 + 17 = 19  (6)

Put (5)  into (1) (or (6) into (2)):  N = 23*17 + 22 = 413.

What if (n - m)/2 = 2 

21*2 - m = 4 so m = 38.  This time n = 4 + m = 42.  Hence N = 23*38 + 22 = 896 which is greater than 500.

Clearly, larger values of (n - m)/2 will produce even larger values of N, so N = 413.

Yes, it is perfectly possible!