RMO 2001 question paper and solutions

 

Solution source.

Solution:

x = 99p + q where p is a positive integer and 0 <= q <= 98

 x = 101r + s where r is a positive integer and 0 <= s <= 100 

[x/99] = [x/101] => p = r 

=> 99p + q = 101p + s 

=> p = (q-s)/2 

Case 1:

p = 0 => x = q = s 

0 <= q <= 98 so there are 99 such values of (q,s)

But q = 0 and p = 0 => x = 0 which is not allowed.

So 99 - 1 = 98 distinct pairs of (q,s).

Case 2:

p is a  positive integer. 

For q = 98, s = 96,94...0 (49 such values) will give positive p. 

For q = 97, s = 95,93...1 (48 such values) 

For q = 96, s = 94,92...0(48) 

For q = 95, 47 such values ... 

For q = 3, there will be one value of s = 1 

For q = 2 there will be one value of s = 0 

For q = 1, there won't be any s. 

So total distinct pairs of (q,s) = 49 + 2*48 + 2*47 ... +2*1 = 49 + 2*48*49/2 = 2401 

Add 98 which we got for p = 0 => 2401 + 98 = 2499 distinct pairs of (q,s).

Since x = 99p + q = 101p + s, each distinct pair of (q,s) uniquely defines x. 

So answer = 2499.

In the above solution, convince me about the argument that 

"Since x = 99p + q = 101p + s, each distinct pair of (q,s) uniquely defines x. "