Here are answers to some questions I've posed on various other pages:
- Baseball Cards
- 14.7. The first time you roll the dice, you will always get one of the six numbers you're looking for, so the average number of times that you will have to roll the die before one number comes up is 1. The probability of getting a different number the next time you roll the die is 5⁄6, so the number of times you will have to wait for the next number is 6⁄5. Similarly, for the third number you'll have to wait 6⁄4 rolls, and so on. So, the total is
6⁄1 = 14.7
- This problem can be tackled in the same manner as the above problem. To get the answer, you would sum
18⁄16 + ... +
You will have to open approximately 62.9 packs.
- Approximately 35.9 packs.
- If you assume that each pack can contain duplicates, you can treat each card in each pack as an independent random event and we can proceed in the same manner as the first 3 problems. The end result is approximately 382.9 packs.
- The Disadvantages of Being Absent-minded
- The professor had completely forgotten that two of the other students, who always sat next to each other, were identical twins.
- Einstein's Riddle
- The Norwegian drinks water. The Japanese owns the zebra.
- Encircling the Earth
- 5.76". Surprising, eh? Think about it.
- Hole In One
- Based on the information in the problem, and assuming that the odds of making a hole in one on any one hole are independent of the odds of doing so on another hole, the odds of getting two holes in one in a round of golf (18 holes) are
(182) × (1⁄12,500)² = 9.792 × 10−7, or around one in a million.
- Phone Equations
- These were my answers; you may have found another equation. If so, good job!
- −5 + 7 − 6 = 54 − 58
- 3 ÷ 27 = 6 ÷ (9 × (5 + 1))
- Answers may vary.
- The Pigeonhole Principle
- We can group the five selected numbers into three categories based on
the remainder they give when divided by 3.
All numbers will have a remainder of either 0, 1, or 2 when divided by 3.
Two possibilities exist:
- Three of the numbers have the same remainder when divided by 3. In this case, the sum of these three numbers will be a multiple of 3.
- No three numbers have the same remainder when divided by 3. In this case, there will need to be at least one number with a remainder of 0, one number with a remainder of 1, and one number with a remainder of 2. The sum of these three numbers will be divisible by 3.
- Divide the unit square into four smaller squares as shown. Because there are five points and only four subsquares, at least two points will fall in or on the border of the same square. Because the length of the diagonal of the smaller square is
, the distance between at least one set of two points must be no greater than that distance.
- The Hymn Board
- I'm not sure what the best answer is; the best answer that I came up with is $73.50.
- To start with, the church will need at least 11 copies of the numbers 1 through 7 (since, using 7 as an example, hymns 177, 277, 377, 477, and 777 could be selected) and 10 copies of the numbers 8, 9, and 0. So far, we are at 107 plates.
- One savings that could be made is to note that "6" looks like "9" upside-down, so instead of ordering 11 copies of 6 and 10 copies of 9, it will be sufficient to order 14 copies of 6 (14 copies are required because hymns 666, 669, 696, 699, and one containing two copies of 6 or 9 could be used). This brings us to a total of 100 plates.
- Another savings is, instead of printing 10 copies of "0", it is sufficient to print 5 copies of "0" and 3 copies of "00" (since the hymn number never starts with 0). Similarly, instead of printing 10 copies of "8", print 5 copies of "8" and 3 copies of "88". So far, we are at 102 digits on 96 plates.
- Finally, digits can be printed on both sides of the plates, saving half of the required number of plates. The church will still be able to represent all combinations as long as the same number is not printed on both sides of a plate, and as long as there is a variety of numbers on the opposite side of each number (e.g. not all plates with "7" on one side have, say, "5" on the opposite side, etc.). With these savings, 102 digits will be printed on 48 plates, for a cost of 102 × $0.25 + 48 × $1 = $73.50.
- How Many Days 'Til Christmas?
- November 12. In order to display days 0 through 9, all 10 digits have to be on one cube or another; however, since "6" looks like "9" upside-down, we can save a digit by making "6" do double duty. This leaves us with three spare digits on one cube or another. In order to represent the numbers 11, 22, and 33, we will need to put 1, 2, and 3 on both cubes. So, one cube could have the digits 0, 1, 2, 3, 5, and 6, and the other the digits 1, 2, 3, 4, 7, and 8 (we have to put 0 and 4 on different cubes in order to represent 40, but 5, 6, 7, and 8 can be on either cube). This allows us to represent all numbers between 0 and 43. 43 days before Christmas is November 12, so we can start using the decoration on that day.
- Miscellaneous puzzles
- Exactly one of the politicians is honest.
- The only one whose stamp can be deduced is C. Because A answered "no," we know that B and C do not have both Red or Yellow stamps. So, if B saw C with either Red or Yellow, he would know that he does not have the same colour. Therefore, C must have a Green stamp.
- Miscellaneous puzzles 2
- 27. The easiest way to solve this problem is to consider that each player, except for the winner of the tournament, will lose exactly once. Since each match has one loser, there must be 27 matches.
- The amount of wine in the water bottle is the same as the amount of water in the wine bottle. Each bottle currently contains a litre of liquid. So, (amount of water in the water bottle) + (amount of wine in the water bottle) = 1 litre. Since none of the original wine has been lost, (amount of wine in the wine bottle) + (amount of wine in the water bottle) = 1 litre. So, the amount of water in the water bottle equals the amount of wine in the wine bottle, and so the amount of wine in the water bottle equals the amount of water in the wine bottle.
- $1.38. Surprising, eh?