Let 10^2017 = n (say)
also, 200n+1= P (say)
and, n^2+2n+3=Q (say)
Then, we have:
P+1 P
A-B = -------------- - ----
Q+2n+3 Q
so, the denominator is Q(Q+2n+3) which is obviously POSITIVE.
The numerator is equal to:
(P+1)Q - P(Q+2n+3)
= Q - P(2n+3)
But, P(2n+3)
= (200n+1)(2n+3)
=400 * n^2 + 602n+3
> n^2 + 2n +3
= Q
Accordingly,
Q- P(n+3) < 0
Since the denominator is positive, it then follows that A < B
Consequently, B is the greater number.
Puzzle Solution
