    Mathematical Puzzles  Puzzle 1 I recently travelled from my home town to a distant music concert, on a pedal tricycle of all things! My wonderful, three wheeled tricycle. I knew that the epic 2,345 mile trip would wreak havoc on the tyres, but luckily I took along 4 spares! Instead of waiting for any single tyre to fail, I decided that I would rotate the tyres evenly, making sure that by the end of the trip all seven tyres had travelled exactly the same distance. What was the distance that each tyre travelled? [Puzzle Code = ZDFX] Copyright © Kevin Stone    Direct Link: www.brainbashers.com?ZDFX Hint: How many tyre miles were travelled? Answer: 1,005 miles. A total of 2,345 miles were travelled and at any one time, three tyres were on the tricycle. Therefore 3 x 2,345 = 7,035 tyre miles were travelled, which was shared equally by the 7 tyres. And 7,035 ÷7 = 1,005. Puzzle 2 Can you find a number such that... ...its double is fourteen more than its quarter? [Puzzle Code = ZZMR] Copyright © Kevin Stone    Direct Link: www.brainbashers.com?ZZMR Hint: The number is less than 20. Answer: 8.We need a number such that: 2 x Number = 14 + 1 x Number                   -                   4 Multiply throughout by 4 to give: 8 x Number = 56 + Number So: 7 x Number = 56 So Number = 8. Puzzle 3 I am compiling the new BrainBashers world almanac and it now contains lots more pages. I know that it takes 333 digits to print the page numbers in sequence. How many numbered pages does the book have and how many times does the number 3 appear? [Puzzle Code = ZNWV] Copyright © Kevin Stone    Direct Link: www.brainbashers.com?ZNWV Hint: Think of the pages from 1-9, and then 10-99. Answer: There are 147 pages and the number 3 appears 35 times. Stage 1 ======= 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. Stage 2 ======= 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. Puzzle 4 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? [Puzzle Code = ZRIL] Copyright © Kevin Stone    Direct Link: www.brainbashers.com?ZRIL 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      and    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.      