The only days the Lion can say, “I lied yesterday,” are Mondays and Thursdays. The only days the Unicorn can say, “I lied yesterday,” are Thursdays and Sundays. Therefore, the only day they can both say it is Thursday.

On no day of the week is this possible! Only on Mondays and Thursdays could he make the first statement; only on Wednesdays and Sundays could he make the second. So, there is no day he could say both. Poor Alice is stuck in the Land of Forgetfulness forever.

This is a very different situation! It illustrates well the difference between making two statements separately and making one statement that is the conjunction of the two. Indeed, given any two statements X and Y, if the single statement “X and Y” is true, it follows that X and Y are true separately; but if the conjunction “X and Y” is false, at least one of the statements is false.

Now, the only day of the week it could be true that the Lion lied yesterday and will lie again tomorrow is Tuesday (this is the only day that occurs between two of the Lion’s lying days). So, the day the Lion said that couldn’t be Tuesday, for on Tuesdays that statement is true, but the Lion doesn’t make true statements on Tuesdays. Therefore, it is not Tuesday; hence, the Lion’s statement is false, so the Lion is lying. Therefore, the day must be either Monday or Wednesday. Tricky little Lion.

A-4

B-5,8

C-1

D-2, 6

E-3

The hint (Synergy) means the whole is greater than the sum of its parts. This means we must be dealing with a whole number. Therefore, half of “What I’d be” must be a whole number. “What I’d be” must be an even number. “What I am” cannot end in 1. There are four possible arrangements of the three digits:

“What I am” is “Nine less than half what I’d be.” So, (“What I am” + 9) × 2 = “What I’d be.” Examination shows that only arrangement A fits the bill and “What I am” must be 183. No sweat.

* = asterisk (ass to risk).

William Shakespeare.

It was raining.

“...country at the top of the Himalayas” = Nepal = PLANE

“...man from the Far East” = China = CHAIN

“...man from the Middle East” = Iran = RAIN

NECKLACE.

The average is calculated as the total distance (240 miles) divided by the total time (2 hours + 3 hours = 5 hours). Therefore, it’s not 50MPH, but 48MPH.

232 entries. 229 matches were played; therefore, there must have been 229 losers. Add the two who scratched out without playing, and then add the winner of the championship, and you arrive at the total number of entries: 232.

79 years (there was not a year 0).

You win $59.00. You’ll always win the same number of units that heads comes down in the sequence, as long as the final toss is heads.

12365

Yes, in 2.777 minutes.

(*) A reciprocal is a number or quantity that, when multiplied by a given number or quantity, gives the product of 1, that is, 0.8 × 1.25 = 1. To find the reciprocal of 0.8, it is necessary to divide 1 by 0.8, which gives the reciprocal 1.25.

The most common wrong answer is 25. If the problem had been, “What is the smallest number I must pick in order to be sure of getting at least two socks of different colors,” the correct answer would have been 25. But the problem calls for at least two socks of the same color, so the correct answer is three. If I pick three socks, either they are all of the same color (in which case I certainly have at least two of the same color) or else two are of one color and the third is of the other color—so, I have two of the same color.

J.

The bear must be white—it must be a polar bear. The usual answer is that the bear must have been standing at the North Pole. Well, this is indeed one possibility, but it’s not the only one. From the North Pole, all directions are south, so if the bear is standing at the North Pole and the man is 100 yards south of him and walks 100 yards east, when he faces north, he’ll be facing the North Pole again. I’ll buy that.

But, there are many more alternative solutions. It could be, for example, that the man is very close to the South Pole on a spot where the polar circle passing through that spot has a circumference of exactly 100 yards, and the bear is standing 100 yards north of him. Then if the man walks east 100 yards, he would walk right around that circle and be right back at the point he from which he started.

In addition, the man could be a little closer to the South Pole at a point where the polar circle has a circumference of exactly 50 yards, so if he walked east 100 yards, he would walk around that little circle twice and be back where he started. You get the idea.

Of course, in any of these solutions, the bear is sufficiently close to either the North Pole or the South Pole to qualify as a polar bear. There is, of course, the remote possibility that some mischievous human being deliberately transported a brown bear to the North Pole just to spite us—who’s paranoid?

Cerebrum.

Puzzle!

Tulips. She likes only words in which two consecutive letters of the alphabet appear.

If the woman were a Liar, she would have said she was a Truthteller. If she were really a Truthteller, she certainly would have said she was a Truthteller. Therefore, the young man reported truthfully, proving he was a Truthteller. Happy ending.

If his top speed is 100MPH, he’ll never get there. He’ll be going 100MPH and be pushed backward at 100MPH. He’ll just have to wait until the wind dies down. Of course, if he could get the train up to 150MPH, he could do it in four hours, but we’ve just said 100MPH is the top speed. Sorry.

Tom, John, Bill, Pat, Kevin.

None of them can be a Truthteller. The first man’s statement cannot be true. If the second says something to indicate the first man is truthful, he is a Liar. The first man cannot be right. The second man is a Liar, as demonstrated, so the third man, who says the second man is telling the truth, must be a Liar. It’s possible that the second man is a Truthteller, but he can’t really be because the third man says the second is telling the truth, and he’s a proven liar. Therefore, the second man is also lying. Check each one’s statements against the others. Good monitoring! You have a future yet.

24198!!

5.5 – 5 × 55 × (5 – 5/5)

or

555/5 – 55/55

It contains the numbers 1 to 9 in alphabetical order.

“A road map tells you everything except how to refold it.” (Start at A in center.)

36421

He is abstemious. We discovered that he likes only words with all five vowels in them—an odd bird.

That’s easy, man. It’s “wrong.”

ABORT: It is five letters long.

ACT: Last letter cannot be placed first to form another word.

AGT: Not an actual word.

ALP: Does not end with a T.

OPT: Does not start with A.

APT: Not in alphabetical order with rest.

Here’s the mysterious list:

And, of course, the purpose of our journey is to generate as much “MENTAL ENERGY” as possible. How are we doing so far?

Maggie Simpson. That’s what he gets for trying to steal candy from a baby. Mr. Burns certainly learned his lesson the hard way. Kudos to you, for figuring it out.

Note: As you begin to work with the Novell Internet support tools, you’ll undoubtedly discover that they provide multiple menu routes to the same information. Because of this, you might find that the steps you use to locate the information required in this exercise are different from the possible solutions listed in this appendix. Here’s a great start.

[(1.5 + 0.25) × 60/50] = 2 minutes 6 seconds

It helped me remember pi to eight decimal places. The number of letters in each word represents the digits of pi: 3.141559265. I really had a “warped” childhood.

In a pack of 52 cards, there are 32 cards of nine or below. The chance that the first card dealt is one of the 32 is 32/52, for the second card, the chance is 31/51, and so on. The chance of all 13 being favorable is 32/52 × 31/51 ×...× 20/40 = 1/1828. The odds were strongly in Lord Yarborough’s favor.

“This is hard because there are three choices for each number.”

COLD, CORD, WORD, WARD, WARM.

9 + 8 + 7 + 6 / 5 – 4 – 3 + 2 – 1 = 0

Three hours. After x hours,

A had burned x/6, leaving 6–x/6.

B had burned x/4, leaving 4–x/4.

But after x hours, A was twice as long as B. Therefore, 6–x/6 = 2(4–x)/4. Therefore, x = 3.

98 – 76 + 54 + 3 + 21 = 100

28463

It doesn’t matter what x is because at some time during the calculation, you’ll be multiplying by (x–x), which equals 0; therefore, the product will be 0. Tricky!

69382

1, 4, 9, 6, 1, 5, 10, 4, 2

Now change to Roman numerals:

I, IV, IX, VI, I, V, X, IV, II

Very tricky, and I like it for that reason and because the Roman numeral IV appears in my name. Touché.

Actually, the two trains will be at the same distance from Boston when they meet.

The answer is a quarter and a nickel. One of them (namely the quarter) is not a nickel.

Thanks—it’s 3:31 p.m.

The only times that the hands of a watch are at right angles at an exact minute are 3:00 and 9:00. Because it was afternoon, we must be talking about 3:00 p.m. At 3:00 p.m., it was 30 minutes “before the half-hour.” Also, at the next minute, it will be 60 seconds (twice as many seconds) after the same half-hour. Therefore, it’s 3:31 p.m.

DAISY.

Nice try. Roosters don’t lay eggs.

Silence, Wedding Ring, Hole, Debt.

TYPEWRITER.

Fools rush in where angels fear to tread. Which one are you?

Certainty. The sum of the digits 1 to 8 is 36. Any number divides by 9 exactly when the sum of its digits are also divided by 9 exactly. It does not matter in which order the balls are drawn out—the sum will always be 36.

All the names are anagrams:

Zena le Vue = Venezuela (the Americas)

Dr. A. Glebe = Belgrade (Europe)

Rob E. Lumen = Melbourne (Australia)

Ann Ziata = Tanzania (Africa)

That tricky Sherlock Holmes—got me again!

17. All numbers containing only E vowels.

You = 4/7

Me = 2/7

Albert = 1/7

Because you go first, you have twice the chance as I do. And since I go second, I have twice the chance that Albert does—therefore, the chances are 4 to 2 to 1.

Refer to the following figures in Chapter 18 for a detailed description of ACME’s full eDirectory tree:

Refer to the following figures in Chapter 18 for a detailed description of ACME’s full eDirectory tree:

Actually, the two trains will be at the same distance from Boston when they meet.

Linguists. Mutual intelligibility is when two people can understand each other without instruction.

The first answer from Dick cannot be true (if true, it is false). Therefore, Dick is not a Wotta-Wooppa. Because he has made a false statement, Dick can’t be a Pukka either. Therefore, Dick is a Shilli-Shalla and his second answer is true.

Because Dick’s second answer is true, Tom is a Pukka. Therefore, Harry’s second answer is false, as is his first answer, which makes Harry a Wotta-Woppa.

So, Tom is a Pukka, Dick is a Shilli-Shalla, and Harry is a Wotta-Woppa. No sweat.

Mukluks are Eskimo boots made from seal or reindeer hide.

1.   The biologist is not Catherine and not from Canada. Therefore, she belongs to C house (she must have one C).

2.   If Catherine were the doctor, she would have to come from Brazil (we know she’s not from Australia), and if she were the doctor, she couldn’t be from Denmark. Therefore, the doctor is from Brazil—but we are told the doctor is not from Brazil. Therefore, Catherine is not the doctor. So, she must be the author.

3.   The biologist is not from Australia and not from Canada. Therefore, she must be from Denmark.

4.   Because the biologist is from C house and from Denmark, she must be Alice.

5.   Therefore, the doctor is not Alice and the doctor is not Catherine. Therefore, the doctor is Brett and Deirdre must be the cartoonist.

6.   Because the doctor is Brett, she is not from B house. Therefore, the doctor is from A house (the only alternative left). Therefore, the doctor is from Canada.

7.   Because the cartoonist is Deirdre, she cannot be from D house. Therefore, she is from B house. So, the cartoonist is from Australia.

8.   Therefore, Catherine, the author, is from Brazil and was in D house.

Complete solution:

Here’s what was going on inside the cerebrum of C. If anyone were to see two red and two black, he would know that he was white. If anyone were to see two red, one black, and one white, he would know that he could not be black. If he were, the man with the white disc would see two red and two black and would know that he was white. Similarly with red, if anyone were to see one red, two black, and two white.

If anyone were to see one red, one black, and two white, he would know that he could not be black. If he were, either of the men wearing white would see one red, two black, and one white, and would argue as above. If anyone were to see two red and two white, he would argue that he could not be black. If he were, someone would see two red, one black, and one white and would argue as above.

If anyone were to see one red and three white, he would argue that he could not be black. If he were, someone would see one red, one black, and two white and would argue as above; similarly, he would know that he could not be red. Therefore, if anyone sees me wearing red or black, he can deduce his color. Therefore, I must be white.

C really needs to get a life.

Ensign Chekov.

There were 12 friends in the group.

QUEUE—pretty sneaky, huh?

V (vowel) 1 = A, and so on.

C (consonant) 1 = B, and so on.

“Well done! Message successfully decoded.”