Answers to Mind Games on the Web

Vol 2, No 5 Answers Presented by Smart Science ™ and Bill Graham as a free service for entertainment and education
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ANSWER TO
THREE GOOD PUZZLES

by Bill Graham

Pigs & Turkeys

Turkeys have two feet and two wings, usually. Pigs have four feet and no wings. Both normally have one head. You may as well assume that all the animals on the farm are normal.

Put the statement into mathematical language. The number of heads and wings equals the number of feet.

heads + wings = feet
1×pigs + 1×turkeys + 2×turkeys = 2×turkeys = 4×pigs A little algebraic manipulation simplifies this mess. Cancel out the 2×turkeys from each side. While you're at it, subtract 1×pigs from each side. 1×turkeys = 3×pigs If farmer Humphrey has one pig, then he'll have three turkeys. Quickly check that this makes sense. Heads = 3 + 1
Wings = 6
Feet = 6 + 4 = 10 It sure looks right to me, and you can simply multiply these numbers (1 pig and 3 turkeys) by any integer. For example, you might find 5 pigs and 15 turkeys on the farm with 50 total feet, 20 heads and 30 wings.

Myrtle and Yertle Get Caught Speeding

First, summarize the information. Myrtle (M) speeds at 3 fpm (feet per minute) and rests one minute every 6 feet. Yertle (Y) speeds at 5 fpm and rests three minutes every 10 feet.

Turtle	Speed	Rest interval	Rest time
M	3 fpm	6 feet	1 min
Y	5 fpm	10 feet	3 min

Next, add to the table a column for time between rests. You calculate this time by dividing the distance by the speed. t = d/s Add another column for the number of running times to reach the total distance and another for the number of resting times. Notice that the resting count will be one less than the running count.

Turtle	Speed	Rest interval	Rest time	Running time	Run count	Rest count
M	3 fpm	6 feet	1 min	2 min	5	4
Y	5 fpm	10 feet	3 min	2 min	3	2

Having constructed this nifty table, you just have to put in one more column that multiplies each rest time by the run count and add to this each rest time times the rest count.

Turtle	Speed	Rest interval	Rest time	Running time	Run count	Rest count	Total time
M	3 fpm	6 feet	1 min	2 min	5	4	2×5 + 1×4 = 14 min
Y	5 fpm	10 feet	3 min	2 min	3	2	2×3 + 3×2 = 12 min

Based on the tables constructed here, Yertle finished two minutes ahead of Myrtle.

A Question of Balance

Twenty-seven cue balls may seem like a lot to test with just three weighings. Settle down and think it over. You have a balance scale with no weights. Therefore, you can compare two weights. Fortunately, you know that one ball weighs more.

Begin by realizing that if you put one ball on each side of the scale, you'll most likely get balance. If you don't, then the heavier of the two balls is the one you seek.

That would solve the problem if you had only two balls. Suppose you have three. Call them A, B, and C. Put A and B on the balance. If you don't have balance, then you know which is the odd ball -- either A or B. If you have balance, then you know that C must be it!

You know have the essential insight into solving this problem. Instead of just getting two possible results from weighing (a binary result), you can have three (a ternary result). The next step up takes groups of balls. For example, if you have six balls, separate them into two groups of three. Put the two sets of three balls on the balance. The odd ball must in the heavier group. You could have another group of three balls off to the side. Then, if the balance shows equal weights, the remaining set of three would contain the odd ball. You already know how to find the odd ball from a group of three.

By now, you probably have the answer. Divide the 27 balls into three groups of nine. Put two on the balance pans to compare them. You know which contains the heavy ball. Divide this group of nine into three sets of three. Apply the balance again. You now have three balls. Use the balance for the third and final time to find out which is the heavy ball.

If you don't know whether the odd ball weighs more or less than the others, then the above strategy fails. You must have at least one more use of the balance. If after the first weighing, you compare a different pair of nine balls, you'll have two results, at least one of which must be unequal and one must be equal. The two equal groups represent the ordinary balls. The unequal weighing will then tell you whether the odd ball is heavy or light because you know which group is the ordinary group. Proceed as before. One additional balance use will suffice.

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