Hint: Think of the pages from 1-9, and then 10-99.
Answer: There are 147 pages and the number 3 appears 35 times.
For pages 1 - 9 we have a total of 9 digits. For pages 10 - 99 we have a total of 180 digits. This is a total so far of 189, therefore we require another 144 digits, which is another 48 pages. Taking us to 147 pages in total.
From page 1 to 147 we have 15 times where the page number ends in 3. We also have the ten pages in the 30's, and the ten pages in the 130's that have the extra 3 in them. For a total of 35 number 3s.
During a recent BrainBashers thinking contest, the total number of points scored by the first six players was 103 and every score was above zero.
The first player scored half the points of the second player, who in turn scored 6 points fewer than the third player.
The third player in turn scored two thirds the points of the fourth player.
The fifth player managed to score the same number of points as the difference between the first and fourth player's points.
Finally, the sixth player scored 14 fewer than the fifth player.
Can you determine how many points the sixth player managed to score?
Hint: The fourth player is the key to this tricky question.
Answer: 9 points.
Respectively the scores were 7, 14, 20, 30, 23, 9.
If the six player were A, B, C, D, E and F we know that:
A + B + C + D + E + F = 103 
A = B ÷ 2
B = C - 6
C = D x 2 ÷ 3
E = D - A (see note at end)
F = E - 14
Since D is the letter we're missing information for, it's best to find all of the other letters in terms of D. These steps are left as an exercise, but the result is:
A = (2D - 18) ÷ 6
B = (2D - 18) ÷ 3
C = 2D ÷ 3
E = (2D + 9) ÷ 3
F = (2D - 33) ÷ 3
We can then use these all in  to find that D = 30. Which allows us to find F = 9.
Small note: if we chose E = A - D (instead of E = D - A) we'd end up with a negative value for E, which isn't allowed.