**Puzzle 21**

Al, Bob, Carl, Dan and Ed share the money they have with each other in the following manner: Al gives half the money he has to Bob. Bob then gives half the money he then has to Carl. Carl then does the same for Dan and Dan does the same for Ed. At that point, they all have the same amount of money. A) How much money did Ed have before all the sharing occurred? B) What fraction of the total amount of money did Al have at that time?

Show SolutionAlternatively, for many people, the easiest way to solve this puzzle is to use actual amounts and work backward. Assume that each man has $1 at the end. The fourth (and last) step of sharing must have been Dan giving Ed $1 of the $2 that he had. The third step must have been Carl giving Dan $1 of the $2 that he had. The second step was Bob giving Carl $1 of the $2 that he had and the first step was Al giving Bob $1 of the $2 that he had, At the beginning, then, Al had $2, Ed had nothing and each of the other three men had $1, for a total of $5, so Al had 2/5 of the total amount.

**Puzzle 22**

A tunnel is lit by a thousand lights, numbered from 1 to 1ooo, currently off (no pun intended… or maybe it was). The first of 1000 people to go through the tunnel turns all the lights on. The second person to go through it turns off every second light, that is, every light whose number is a multiple of 2. Likewise, the third person changes the status of every third light, that is, the person goes to every light whose number is a multiple of three and turns on the lights that are off and turns off the lights that are on. The fourth person does the same for every fourth light, and so on. After the thousandth person goes through, which numbered lights are on? And why does it work out that way? (The second question is the important one.)

Show Solution………..1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

1 N N N N N N N N N N N N N N N N N N N N N N N N N

2 F F F F F F F F F F F F

3 F N F N F N F N

4 N N F N N F

5 F N N F F

6 F N F N

7 F N N

8 F F F

9 N N

10 F N

11 F N

12 F N

13 F

14 F

15 F

16 N

17 F

18 F

19 F

20 F

21 F

22 F

23 F

24 F

25 N

The light numbers of the lights that are on after the 25th person goes through the tunnel are 1, 4, 9, 16, 25, namely, the perfect squares. And this pattern would continue, no matter how many lights there were. Each light’s on/off status is changed a number of times equal to the number of factors of the light’s number. Numbers that are not perfect squares have an even number of factors, because each factor can be paired up with a distinctly different factor. Examples: 5 = 1 • 5 (two factors), 12 = 1 • 12 = 2 • 6 = 3 • 4 (six factors), 48 = 1 • 48 = 2 • 24 = 3 • 16 = 4 • 12 = 6 • 8 (ten factors). Perfect squares have an odd number of factors, because the square root of the number is paired with itself. Examples: 4 = 1 • 4 = 2 • 2 (three factors), 100 = 1 • 100 = 2 • 50 = 4 • 25 = 5 • 20 = 10 • 10 (nine factors). Since the first change in status is from off to on, a light whose status is changed an odd number of times is on at the end and all the other lights are off. The reasoning that leads to this conclusion can be done without making the table, but most people, myself included, are more likely to come up with this reasoning *after* they have found in the table which lights are on.

**Puzzle 23**

The sum of the ages of the members of a family of four is 90. The father is five years older than the mother and the son is three years older than the daughter. Seven years ago, the sum of all the ages was 63. How old is the father now?

Show Solution*unless*seven cannot be subtracted from one or more of the current ages. Since we have to account for only one year, if the daughter (the youngest member of the family) is six years old now, then seven years ago the sum of their ages would be 90 – 27 = 63, which is consistent with the fourth statement. The son is therefore 6 + 3 = 9 years old now. Substituting these two ages (S = 9, D = 6) and M + 5 for F into the first equation, we get (M + 5) + M + 9 + 6 = 90, so M = 35 and F = 40. The father is 40 years old now.

**Puzzle 24**

If each domino of a collection is exactly the size of two adjacent squares of a checkerboard, then 32 of these dominoes could cover the entire 64=square board with no dominoes overlapping or sticking out over the edge of the board. (I leave it to the reader to verify this.) If one square in each of two consecutive corners of the board are cut out of the board, the remaining figure, containing 62 squares, could be covered, with the same proviso, using 31 dominoes. (You can confirm that, too.) The puzzle is this: If one square in each of two opposite corners of the checkerboard are cut out of the board, could the remaining figure, also containing 62 squares, also be covered, with the same proviso? As usual, if your answer is “yes”, you must show how to do it and if your answer is “no”, you must prove that it can’t be done.

Show Solution*the same number of squares of each color are covered*. When one square in each of two opposite corners are cut out of the board, two squares of the same color are removed leaving 30 squares of that color and 32 squares of the other color, which is impossible to cover in the manner prescribed.

**Puzzle 25**

Find the value of

Show Solution