Here’s the solution:

1st roll: 10
2nd roll: 15
3rd roll: 9
4th roll: 20
5th roll: 12

Answer

Here’s the solution:

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Here’s the solution:

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Here’s how you can draw the figure using one line (start anywhere and follow the arrows):

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You could of course solve this problem by calculating how far Runner B will travel before reaching Runner B, then how far he will travel before returning to Runner A, and so on.  This, however, is extremely time-consuming, and would likely only result in an approximate answer anyway.  The easier way is to simply calculate how long the runners are running; since they start and stop at the same time, they will all run for the same length of time.  To calculate this, start by finding the combined speed with which Runners A and B are moving toward each other: 8 mph (for Runner A) + 10 mph (for Runner B) = 18 mph.  Then, use the distance = rate × time formula to solve for the total time:  18 miles = 18 mph × t hours, so t must equal 1 hour.

Now that we know the total time, we can use the speed of Runner C (11.75 mph) and the same formula to compute his distance traveled: d miles = 11.75 mph × 1 hour = 11.75 miles.  Thus, the answer is that Runner C traveled 11.75 miles.

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Let’s think about this as a simpler problem first.  If Merlin has 4 frogs and 16 potions, how can he determine the spoiled potion?  The key is to give the potions to the frogs so that each potion is given to a unique combination of frogs.  That way, after 10 hours, he can see which frogs have died and know immediately which potion was the deadly one.  Here’s a possible way he could do this:

As an example, if frogs C and D die, but frogs A and B do not, the spoiled potion must be #11; if all four frogs die, the spoiled potion must be #1; and if none of the frogs die, the spoiled potion must be #16.

It may seem impossible to expand this to 500 potions with only 9 frogs, but the key to this solution involves powers of 2.  Since each frog has two possible actions for each potion (to drink it or not to drink it), the number of unique combinations for x frogs is given by 2x.  So with 4 frogs, there are 24, or 16, unique combinations, and with 9 frogs, there are 29, or 512 unique combinations.  As long as Merlin is very careful and keeps track of which frog is which and which potion is which, he’ll have no trouble.

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One interesting fact is that any 3-digit number multiplied by 1,001 is that number repeated.  For instance, 571 times 1,001 equals 571,571.  Try it with a few more numbers if you want.  How does this help me?  Well, 7 times 11 times 13 equals 1,001.  So, when you divide your six-digit number by 7, 11, and 13, you are dividing it by 1,001, and thus ending up with your original 3 digit number (and no remainders!)

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Across:
16 = 4 squared
576 = 24 squared
43264 = 208 squared
961 = 31 squared

Down:
49 = 7 squared
36 = 6 squared
1521 = 39 squared
676 = 26 squared
64 = 8 squared

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Since there is an odd number of face-up cards in the deck right now, Ray clearly cannot just split the deck into two parts.  The key is that Ray can flip cards over; the final number of face-up cards does not have to be (and indeed cannot be) 13.  As it turns out, the only way to be sure that there are the same number of face-up cards in two stacks is to remove 13 cards from the deck (it doesn’t matter which), put them in a stack, and flip them over.  If we call the number of face up cards in the new, 13-card, stack x, then after we flip the new stack over, it will have 13 – x face-up cards in it.  Since the original deck had 13 face-up cards, and we have removed x face-up cards from it, the larger stack will also have 13 – x face-up cards in it.  Try it yourself if you aren’t convinced!

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There are two solutions to this brownie puzzle.  The first is to cut the remaining brownie in half depth-wise, so that one nephew gets the top half and the other gets the bottom half.  This, however, is both impractical and probably unfair, since most people prefer the top part of the brownie to the bottom part.

There is a better solution, however.  The key to this solution is that any cut that passes through both the center of the rectangular pan and the center of the missing piece will cut the remaining brownie into equal pieces.  All we have to do is find the centers of each rectangle (the pan and the missing piece) and cut a straight line through them.  To find the centers, use the property of rectangles that states that the center is located at the intersection of the rectangles’ two diagonals.

The illustration below shows the diagonals and center of the pan in white, and the diagonals and center of the missing piece in black.  The cut is represented by the green line.

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The solution to the riddle is sinecure, which means “a job with minimal duties.”  Moving the s a couple letters to the right gives us insecure, which you probably don’t want yourself or your email to be.  Finally, the last four letters are cure, which is a process for flavoring meat.

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The hint (“think about numbers”) helps you in two ways.  First off, you can convert each letter in the sequence to a number by replacing it with its place in the alphabet.  Each number will then become a number between 1 (A) and 26 (Z).  Thus, our sequence becomes:

C   N   O   I   Z   E   C   E   H   I   ___   ___   ___   ___   ___
3   14  15  9  26  5    3   5    8   9  ...

If those numbers don’t look familiar yet, think of some famous numbers.  Need more help?  Try putting a decimal place after the first “3”.  This gives you the number 3.141592653589..., which is the number pi (π).  To find the next few letters, we need to find the next few digits of pi, which are 793238 (and so on).  If we break these numbers up following the pattern above, and change them to letters, we get:

            7   9   3   23   8
            G   I   C  W   H

The answer, then, is G I C W H.

Answer

For this puzzle, several of the equations are fairly simple to fill in. For instance:

A couple others aren’t too difficult:

Three more require roots (square or cube):

The last one is a bit tricky; it uses the “factorial” operation (represented by “!”).  The factorial of an integer n is equal to n × (n – 1) × (n – 2) ... × 1.  So, in this problem, the 3! is equal to 3 × 2 × 1 = 6.

Answer

Solving this puzzle involves realizing that some letters can be rotated to form other letters.  For instance, if you write it the right way, a capital “E” rotated 90 degrees could be a capital “M”.  Similarly, a lowercase “c” could be a lowercase “n” or “u”, depending on how you rotate it, and a “d” can be rotated to become a “p”. There are several solutions, but here is one.

First cube: J, d (and p), B, O, G, E (and M)
Second cube: u (and c and n), L, R, Y, S, and F
Third cube: A, u (and c and n), E (and M), V, T, and any other letter

For example, “DEC” (for December) can be spelled using “d” from the first cube, “E” from the third cube, and “c” (same as u and n) from the second cube.

Answer

Here is a solution (there are many other possible solutions):

The trick is to make sure that either 0 or 9 goes in the center square.  If 0 goes in the center, 1 must go in the outer square; if 9 goes in the center, 8 must go in the outer square.

Answer

The nine-letter word (in all caps of course) is CHECKBOOK:

Some other long words that can be written in this manner are: CODEBOOK, COOKBOOK, EXCEEDED, BEDECKED, DECIDED, and DEICIDE.  Indeed, any word that contains only the letters B, C, D, E, H, I, K, O, and X can be written in this manner.  These are the letters with vertical symmetry.

Alternatively, you could try writing the word vertically and relying on the horizontal symmetry of certain letters.  Such letters include A, H, I, M, O, T, U, V, W, and X; depending on how you write your capital Y, it may be included as well.  Here are some long words (I couldn’t find any longer than seven letters) that have this horizontal symmetry: AUTOMAT, TOMATO, WHAMMY, and MOUTHY.

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The longest word you are likely to know that can be typed with one hand is stewardesses, with 12 letters. Technically, there are a few longer left-handed words: aftercataracts, tesseradecades, and tetrastearates all have 14 letters, though I doubt anyone actually uses any of them in everyday language.
 
The longest common right-handed word is lollipop (8 letters).  The 11-letter word hypolimnion is the longest right-handed word of any kind, unless a hyphen is allowed (it is, after all, on the right half of a QWERTY keyboard); in that case, the longest right-handed word is johnny-jump-up (12 letters).

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Your first instinct (like mine) may have been to split the coins into two piles and weigh them to see which is the heaviest, then split the heavy pile into two piles, and so on until you have identified the heaviest coin.  This is indeed a good way, but it doesn’t work with a pile of 25 coins, since 25 does not divide evenly by 2, and dividing the coins into a pile of 12 and a pile of 13 won’t work either (the pile of 13 would presumably always be heavier). Besides, there is an even faster way of finding the answer.

Instead of dividing the coins into two groups, divide them into three groups, two of which have the same number of coins.  Then, put the two equal-sized groups on the scale.  If one of them is heavier, the forged coin is in that group; if they weigh the same, then the forged coin is in the third group.  So, we start by dividing the 25 coins into groups of 9, 8, and 8.  We balance the two piles of 8; let’s assume the worst-case scenario, that they weigh the same.  This means the pile of 9 coins contains the forgery.  Next, split the coins into three piles of 3 coins each.  Whichever pile weighs the most must contain the forged coin.  Balance two of those three coins; if one weighs more than the other, it is the forgery, but if they weigh the same, the third coin is the forgery.  You’ve just found the fake coin in only 3 weighings!

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Instead of picking a royal flush from the deck, my friend picked four 10s and a Jack of spades.  This meant that the best hand that I could pick was a straight-flush from 5 to 9, since all the tens were taken already.  I wasn’t worried, though, since a straight-flush of any type still beats four of a kind.  Then, my friend discarded the 10s of diamonds, hearts, and clubs, leaving him with the 10 of spades and the Jack of spades.  He then selected the Queen, King, and Ace of spades from the deck, giving him a royal flush that beat my straight flush.  At this point, there were no cards that I could discard and draw to make a royal flush, since all the 10s were either in his hand or in the discard pile.  I graciously conceded defeat, and decided never to only play face-down poker from then on.

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Here is the solution. The two keys are to (1) not restrict your lines to within the imaginary "box" formed by the dots in the grid, and (2) realize that you can draw a line that is not at a 90 or 45 degree angle to the grid.

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The question tells us how many students got an A, and how many got an F (8 and 9, respectively).  That leaves 33 (50 – 8 – 9) students who got a B, C, or D.  We know that 30 got a C or worse—in other words, a C, D, or F.  We also know that 35 got a C or better—an A, B, or C.  Since 9 got an F, we know that 21 got a C or a D (30 – 9 = 21).  Since 8 got an A, we know that 27 got a B or a C (35 – 8 = 27).  Since 33 got a B, C, or D, and 27 got a B or a C, that means 6 got just a D.  Similarly, since 33 got a B, C, or D, and 21 got a C or a D, 12 got just a B.  Which leaves 15 people who got a C.  So our final breakdown is 8 A’s, 12 B’s, 15 C’s, 6 D’s, and 9 F’s.

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The sentence “James while John had had had had had had had had had had had a better effect on the English teacher.” can be punctuated as follows:

James, while John had had “had had,” had had “had;” “had had” had had a better effect on the English teacher.

Or, in other words:

While John had written the words “had had,” James had written the word “had;” the English teacher preferred “had had.”

Yes, this sentence with crazy grammar is about grammar!

Answer

To figure out how many times “3” will appear on the odometer, it helps to restrict your scope to just one of the six digit places on the odometer.  It’s easiest to just think about the left-most (hundred-thousands) place, since all the mileages that have a 3 in that place happen consecutively—300,000 to 399,999.  Thus, the digit “3” appears 100,000 times in that place.  Multiply that by six places and we get a total of 600,000 3s.

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Assuming that by “one day”, the judge means to sentence you to 24 hours in jail, the best day to pick is the one day a year that (in most locations in the U.S. and Canada) you can serve 24 hours while only experiencing 23 hours.  How is this possible?  If you choose to serve your sentence on the day that Daylight Savings Time goes into effect, you will get credit for the hour (between 2 a.m. and 3 a.m.) that is skipped when the clocks are moved ahead.  Daylight Savings Time currently goes into effect on the second Sunday in March, which in 2009 will be March 8.  That hour may not sound like much, but any hour spent outside of jail is a good hour!

If, on the other hand, the judge meant “from sunrise to sunset” when he said “one day” (which seems less likely, but who knows?), you would want to choose the shortest day of the year (or Winter Solstice in the northern hemisphere). Choosing the start of Daylight Savings Time wouldn’t help you in this case, since the clocks are set forward at 2 a.m.  In 2008 and 2009, the winter solstice occurs on December 21. To give you an idea of how much jail time you’ll save this way, in 2008, if you served in New York City on December 21, you would be in jail from 7:16 a.m. (sunrise) to 4:31 p.m. (sunset).  That’s 9 hours and 15 minutes.  If you served on June 22, on the other hand, you’d be in jail from 5:24 a.m. to 8:31 p.m.  That’s 15 hours and 7 minutes, or almost 6 hours more than you’d serve on the Winter Solstice!

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I should have envelopes with the following amounts: $1, $2, $4, $8, $16, $32, $64, $128, and $256 (a total of 9 envelopes).  These envelopes total $511, leaving you with $489 in dollar bills outside of the envelopes.  With some combination of $489 and the 9 dollar values in the envelopes, I can come up with any dollar value between 1 and 1000.  How?  Well, by using powers of 2 for the envelopes, we can easily get any value between 1 and 511:

511 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256;
510 = 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256;
509 = 1 + 4 + 8 + 16 + 32 + 64 + 128 + 256;
508 = 4 + 8 + 16 + 32 + 64 + 128 + 256;
507 = 1 + 2 + 8 + 16 + 32 + 64 + 128 + 256;
506 = 2 + 8 + 16 + 32 + 64 + 128 + 256;
505 = 1 + 8 + 16 + 32 + 64 + 128 + 256;
504 = 8 + 16 + 32 + 64 + 128 + 256;
503 = 1 + 2 + 4 + 16 + 32 + 64 + 128 + 256;
and so on…

For any value higher than $511, we start with the $489 that are outside of the envelopes, then add from envelopes.  Since we already know that we can get any value less than or equal to $511 from the envelopes, we will have little trouble getting any value less than $1000.  For instance, if you asked me for $777, I would give you the $489 loose dollars, plus $288 in envelopes: $256 + $32.

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The rescue team can reach 20 kilometers away from the cabin if only 6, 7, or 8 of them are on the team.  Here’s how a team of six people can make it 20 kilometers:

Before leaving on the first day, the six people have 42 days’ worth of food and water (7 days’ worth each).  After the first day, they have 36 days of supplies, and are 5 km from the cabin.  On the second day, one of the six people returns to the cabin, taking just enough food and water to reach it safely (1 day’s worth).  He gives the rest of his supplies (5 days’ worth) to the other rescuers, who split it among themselves.  Now, there are 5 rescuers, with a total of 35 days of supplies. 

After the second day, the 5 people are left with 30 days of supplies (6 days’ worth each), and are 10 km from the cabin.  Before the third day starts, another person returns to the cabin, taking with him two days of supplies, and giving his remaining four days’ worth of supplies to the remaining four rescuers.  We now have 4 people, and 28 days of supplies. 

After the third day, the 4 people have 24 days of supplies, and are 15 km from the cabin.  The process repeats again on the fourth day, after which there are 3 people left in the group, with 18 days of supplies.  They are 20 km from the cabin.  If they were to find the missing person then, they would need 16 days of supplies to return to the cabin (4 people * 4 days’ travel).  If they traveled for another day, even if one person returned to the cabin as on the previous days, they’d be left with 3 people (2 rescuers and the rescuee) and 5 days’ travel, but only 12 days of supplies.

A similar solution works with seven people; after the fourth day, there would be 4 rescuers and 21 days of supplies, of which they would require 20 to return to the cabin (5 total people * 4 days’ travel = 20).  If 8 people start out, 5 rescuers with 24 days of supplies would remain after the fourth day; they’d need all 24 to get back (6 total people * 4 days’ travel = 24).  If there are fewer than 5, or more than 8, rescuers, they can go no farther than 15 km from the cabin.  Try it yourself to find out why.

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The brothers’ hamburger stand should be fine.  We can compute the probabilities easily if we convert the percentages in the problem to decimals (30% = 0.3, etc.).  The probability of each combination of brothers being sick can be calculated by multiplying probabilities.  Thus, if we wanted to find the probability that Al and Bob are well, but Carl is sick, we multiply the chances of Al not being sick (100% – 30% = 70% = 0.7) by the chances of Bob not being sick (100% – 40% = 60% = 0.6) and the chances of Carl being sick (20% = 0.2).  The result, 0.7 × 0.6 × 0.2 = 0.084 = 8.4%, is one part of our answer.  The table below shows the data for all the possible combinations of sick and well brothers:

If we add up the top four rows of the table, we can find the total percentage for all the combinations in which at least two brothers are at the stand: 33.6% + 8.4% + 22.4% + 14.4% = 78.8%.  Since this is greater than 75%, the brothers’ hamburger stand will (barely) be able to stay afloat.

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To fill in the remaining letters in the word “L121212UE”, you can take either of two approaches.  The “hard” approach is to think about the clue, namely that the word is about “someone who uses words”.  If you think of some words about words that start with “L”, you may notice the root log, which is often used at the end of a word (often with a “ue” on the end): monologue, dialogue, epilog, etc.  This root, which means “speech,” gives us all we need to fill in the rest of the word.  If it starts with “L-O-G”, then the digit 1 must be “O” and the digit 2 must be “G”.  This means that the resulting word is “logogogue”.  The suffix –gogue means “one who leads”, as seen in the vocab words demagogue (one who leads people; i.e., a ruler) and pedagogue (one who leads children; i.e., a teacher).  Thus, logogogue means “one who leads others in the use of speech or words”.  Hopefully, by writing C2’s Words of the Week, I am acting as a logogogue!

The easier way to solve this puzzle (which some may consider cheating) is to utilize an online dictionary with wildcard capability, such as the OneLook dictionary (http://www.onelook.com/).  Wildcards are symbols (such as * and ?) to search for words when you don’t know all the letters of the word.  This is especially useful when trying to find an answer for a crossword puzzle.  On these sites, the ? symbol is used to represent any letter, so if you searched for “b?d”, you would get the following results: bad, bed, bid, and bud.  So, if you wanted to search for the word from this puzzle, you could enter in “l??????ue” (six ?’s).  If you do this, only one result will come up: logogogue.

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The key to figuring out the missing numbers is to add all the numbers together as they are read.  Then, you just subtract the sum of those 99 numbers from the sum of all the numbers from 1 to 100, which is relatively easy to calculate (more on this below).  So, let’s say you add up all the numbers and get 4992.

Since the sum of any set of consecutive integers is equal to the median of the set times the number of integers in the set, we just need to figure out the median of the 100 integers to find their sum. Since there are an even number of integers in the set, the median is the average of the middle two, in this case 50 and 51.  Thus, the median is 50.5.  Now, we just multiply 50.5 by the number of integers in the set (100) to get our sum: 5050.

Now, we just subtract 4992 from 5050 and we have our missing number (in this case, 58).  As a shortcut, you don’t have to remember the first two digits of your sum—they will always be “49” if the last two digits are 50 or higher, and “50” if the last two digits are less than 50.  In other words, the range of possible sums is from 4950 to 5049.  Using this shortcut, we may get a sum of, say “78”.  This represents 4978, which, subtracted from 5050, would give us a missing number of 72.

Answer

Since 4 gumdrops are added and 3 are taken away each minute, it would be tempting to say that it takes 50 minutes for the bowl to be filled. There is one trick, though. After 45 minutes, there are 45 gumdrops in the bowl. After 46 minutes, there are 46 gumdrops. Then, in minute 47, Willy adds 4 gumdrops. At that point, the bowl of candy is full. It doesn't matter that Veruca will soon take 3 gumdrops from the bowl, since the question doesn't specify that it has to be full at the end of a minute. Thus, the answer is 47 minutes.

Answer:

The shortest word in the English language that has all 5 main vowels is the rarely used medical term eunoia, which means (roughly) “well mind”.  This word comes from the Greek prefix eu- (“good”) and the Greek root –noia (“mind)”.  So, the next time you feel better after a time of sadness, you could say that you have returned to a state of eunoia.  The shortest common word that has all five vowels is sequoia, the type of tree.  Sequoia trees, which grow in California, are the tallest in the world, with many measuring over 300 feet in height.  The world’s tallest tree, named “Hyperion,” is a 379-foot sequoia!

Answer:

The two words are facetiously and abstemiously.  Both have the vowels a, e, i, o, u, and y—in order—and no other vowels.  So you know, facetious means “not literal or serious; humorous” and abstemious means “sparing or moderate, particularly related to eating and drinking.”  So, I could lighten the mood of a dreary gathering by talking facetiously, and I could try to lose weight by eating abstemiously.  And in both cases, I’d be using all the vowels!

Answer:

There are a variety of 6-letter words that can be formed using the letters A-E: decade, accede, deeded, bedded, beaded, and a few obscure words.  The only 7-letter word that I have found is acceded.  What was the longest word you found?  Can you think of one that is 8 or more letters?  If so, let us know!

Answer:

The next three numbers in the sequence are: 3, 6, and 6. The solution has nothing to do with the numbers themselves, so trying to find a mathematical pattern will only prove fruitless. The key is connecting each number’s position in the sequence to the number itself:

How does that help?  Well, all you have to do now is spell out the “Position” numbers:

See the pattern yet?  Each number refers to the number of letters in its position.  Since the next three positions will be “ten”, “eleven”, and “twelve”, our next three numbers must be 3, 6, and 6, or the number of letters in each of those positions.

Answer:

Why didn’t the cooling fans on the electrical equipment help keep the equipment cool?  Well, fans cool by blowing air… and in space, there’s no air!  Though today’s space shuttles are pressurized and have air in the cabins, this was not the case in the early days of the space program. Because of this, early astronauts had to wear their suits at all times, of course.

Answer:

Carla shook 4 hands.  How do you figure this out?  Well, a good place to start is with the person who shook 8 hands (let’s call him James).  Since there are 10 people at the party, and James obviously didn’t shake his own hand or his wife’s hand, he must have shaken hands with every other person at the dinner party.  That means that his wife (Anita) must be the person who shook 0 hands, since she is the only person who didn’t shake hands with her husband.

The person who shook 7 hands (Ivan) thus shook everyone’s hand but his own, his wife’s, and Anita’s.  This means that his wife (Betty) must be the person who shook only 1 hand, since she is the only one who shook James’ hand but not Ivan’s.  Next, we look at the person who shook 6 hands (call him Gary).  Gary shook hands with everyone but himself, his wife, Anita, and Betty.  Following the same process, we can see that his wife (Debby) must have shaken 2 hands (only James’ and Ivan’s).  Similarly, the person who shook 5 hands, Frank, must be married to the person who shook 3 hands, Eva.

Who’s left over?  Only Carla and her husband (we’ll call him Harry).  We know that Harry must be the person who answered that he shook 4 hands (the only answer that we have not yet accounted for).  He shook hands with the four other men: James, Ivan, Gary, and Frank.  Looking back at our analysis, we see that none of the women shook hands with Carla; only James, Ivan, Gary, and Frank did.  Thus, Carla (like Harry) shook hands with 4 people.

To help you see this, we’ve included below a graphical version of this solution and a table of all the handshakes.  A box indicates a marriage; a line indicates a handshake.  Carla’s handshakes are in orange:

Answer:

The problem with Joseph’s friend’s calculator?  It was in hexadecimal mode.  In hexadecimal, there are 16 digits instead of 10 (0 through 9, then A through F).  Thus, the number 15 in the normal decimal system would be expressed as F in hexadecimal mode.  Instead of having a ones place, a tens place, and a hundreds place (etc.) like a decimal number, a hexadecimal number has a ones place, a sixteens place, and a two-hundred-fifty-sixes place (256 = 16 * 16).

So, when Joseph punched in the addition “17 + 19” and got the result “30” in hexadecimal, he was punching in the equivalent of the decimal addition “23” (16 + 7) + “25” (16 + 9) = “48” (16 * 3 + 0).  “15 + 14 = 29” is, in decimal, “21 + 20 = 41”.  And the final addition he punched in, “15 + 19” was the hexadecimal equivalent of “21 + 25,” and its answer, which is “46” in decimal, was “2E” in hexadecimal.  When Joseph saw the E, he knew that he was in hexadecimal mode, and changed back to the standard decimal mode.

Answer:

What this question is really asking (in math terms) is this: What is the largest number that cannot be expressed as the sum of multiples of 6, 9, or 20?  Let’s start by listing some multiples of those numbers:

6, 12, 18, 24, 30, 36, 42, 48, 54, 60, …
9, 18, 27, 36, 45, 54, 63, …
20, 40, 60, …

Now, let’s make a chart of some numbers, and whether they can be expressed as a sum of these numbers:

So far, 43 is the largest number that cannot be expressed in this way.  After 43, something interesting happens:

We could keep going, but we don’t have to.  Why?  Because now that we have found six consecutive numbers that are sums of the multiples, we can add six to each of those numbers to prove that the next six (50 through 55) do not fit, like so:

…and so on for each set of six numbers after that.  Thus, we know that 43 is the last number that we cannot express in this way.  In other words, 43 is the largest number of nuggets that you cannot buy.

Answer:

Much like with the Green Glass Door riddle, the key to this riddle is in the letters.  The rule for the White Stone Hoop, however, doesn’t rely on double letters (as you have no doubt guessed).  I’ve bolded the key letters in the examples below:

cabbage will go through the hoop, but lettuce will not
ghosts can go through the hoop, but not spirits
cars can go through, but not trucks
snow can go through, but not rain

Have you figured it out?  In the letter pairs ab, gh, rs, and no, the second letter always follows the first alphabetically.  Thus, any word that contains two consecutive letters from the alphabet will go through the White Stone Hoop, which incidentally contains a hint to the solution (white, stone, and hoop all have consecutive-letter pairs).  Can you think of any other examples?

Answer:

This key to this classic riddle is to consider, not the objects themselves, but the letters in the words.  What do the words “sheep,” “trees,” and “boots” have that the words “cows,” “bushes,” and “shoes” do not?  The answer is a double-letter combination: sheep, trees, boots. Any word with the same letter twice in a row will “go through” the Green Glass Door.  You can form your own examples by thinking of a word with a double letter (like, say “letter”) and then thinking of a synonym or similar word that does not have a double letter (like “vowel” or “consonant”).  Thus, you could get the example “Letters can go through it, but not vowels or consonants.”

By the way, there is a clue to the solution in the title of the riddle: every word in Green Glass Door has a double-letter combination in it.

Answer:

The letters that complete all three blanks are “notable.” Here’s how it works:

The notable doctor was not able to operate because she had no table.

There you have it!

Answer:

The trick to this puzzle is that the two lines do not have to intersect. If the lines intersect, then the clock will be divided up into four areas. Since the numbers on a clock total 78, and 78 does not divide evenly by 4, there is no way for four regions to all have the same value!

Instead, we will divide the clock into three regions. The first line starts between the 10 and the 11 and ends between the 2 and the 3. The second line starts between 8 and 9 and ends between 4 and 5. The three regions are as follows:

1. 11, 12, 1, and 2 at the top (11 + 12 + 1 + 2 = 26)
2. 10, 9, 3, and 4 in the middle (10 + 9 + 3 + 4 = 26)
3. 8, 7, 6, and 5 at the bottom (8 + 7 + 6 + 5 = 26)

Here is a picture version of the solution. Pretty neat, huh?

Answer:

Let’s start by examining a random light bulb in the hall, say, the 14th one. How many times will its switch be flipped? Obviously, the first person, who flips every switch, will flip it. The 2nd person (who flips every 2nd switch) and the 7th person will also flip it; finally, the 14th person will flip it. That’s a total of 4 flips, which, since it started in the “OFF” position, means that the bulb will be off once all is said and done. The same process can be repeated for every switch, but that would be time consuming.

The trick here is to realize that in order for a bulb to be on, its switch must be flipped an odd number of times (ON-OFF-ON, ON-OFF-ON-OFF-ON, etc.). As you may have noticed from our example, a bulb’s switch is flipped by the people representing the factors of the number of the bulb (i.e., the 14th switch is flipped by the 1st, 2nd, 7th, and 14th people; 1, 2, 7, and 14 are the factors of 14). Put these two facts together, and you realize that the bulb numbers that end up in the “ON” position must have an odd number of factors.

What numbers have an odd number of factors? It turns out that this is only true of perfect squares. Since all factors have a pair that they are multiplied by to equal the number (i.e., 2 and 7 are a pair, since 2 * 7 = 14), most numbers will have an even number of factors. But with perfect squares, one of the factors (the square root of the number) is paired with itself. As an example, 9 has the factors 1, 3, and 9 (1 and 9 are pairs, and 3 is paired with itself).

To solve the problem, then, we just count up all the perfect squares less than or equal to 1000. They are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, and 961. Thus, 31 light bulbs are on after the process is complete (a shortcut would be to just take the square root of 1000, ~31.6228, and round down to 31; this gives you the last integer that, when squared, is less than 1000).

Answer:

Two things are clear from Mr. Smith’s wishes:

1. He wants his daughter to receive twice as much as his wife (two-thirds to one-third) and
2. He wants his wife to receive twice as much as his son (by the same ratio).

So, if we assume that the son receives one part of the inheritance, the wife must thus receive two parts, and the daughter four parts. Adding these together, we get a total of seven parts. Thus, the daughter should receive four-sevenths of the estate, the wife two-sevenths, and the son one-seventh.

Answer:

First, fill the 9-ounce glass. Then, fill the 4-ounce tumbler using the water in the glass. This should leave you with 5 ounces in the glass and 4 ounces in the tumbler.

Next, dump the water in the tumbler down the drain, and fill the tumbler again using the water in the glass. Now there is 1 ounce in the glass and 4 ounces in the tumbler. Again, dump the water in the tumbler down the drain.

Next, transfer the 1 ounce of water from the glass to the tumbler. Then, fill the entire glass with water (9 ounces worth). There are now 9 ounces in the glass and 1 ounce in the (4-ounce) tumbler.

Finally, fill the tumbler using the water from the glass. Since this requires 3 ounces of water, the glass will now have 6 ounces in it. Now I can add the drink mix and enjoy!

Answer:

Here are some nine-letter words that you can build using the method described (one letter at a time, with every step along the way forming a valid English word):

• I » in » sin » sing » sting » siting » sitting » spitting » splitting
• I » in » sin » sing » singe » singer » stinger » stringer » stringier
• a » at » ate » sate » state » estate » restate » restated » restarted
• I » in » pin » ping » aping » taping » tapping » trapping » strapping

There is also at least one 10-letter word (assuming you think that “estating” is a valid word):

• I » in » sin » sing » sting » sating » stating » estating » restating » restarting

Finally, here is an obscure 11-letter word that works (at least according to the Scrabble dictionary):

• I » pi » pin » ping » oping » coping » comping » compting » competing » completing » complecting

So you know, complecting means “joining by weaving,” compting is an archaic form of “counting,” comping means “providing something for free, ” and oping is an archaic form of “opening.”

Answer:
The most important digit for us to consider for this brainteaser is the last one, since we know it changes with every mile that is driven. Since the initial mileage is a 5-digit palindrome, our first (ten-thousands) digit must be the same as our last (ones) digit. After a mile has been driven, the last four digits are a palindrome, so the second (thousands) digit must be the same as the new last (ones) digit. Similarly, after the second mile has been driven, the last three digits are a palindrome, so the third (hundreds) digit must be the same as the new last (ones) digit.

If we call the initial first digit n, we can express our initial mileage as (the one caveat here is that we are dealing with digits here, not integers, so if n is 9, then n + 1 = 0, not 10):

n    n+1    n+2    n+1    n

Similarly, after one mile, our mileage must read:

n    n+1    n+2    n+1    n+1

You may have noticed that the last four digits here are not a palindrome; in order for them to be a palindrome, the fourth (tens) digit must be n+2, not n+1. What this means is that adding 1 to n requires us to carry a 1 into the next column, thus increasing it by 1 as well. Therefore, the digit n must be 9 (since 9 + 1 = 10). Plugging this into our formula above, our three mileages are:

90109
90110
90111

Answer:
The last number (20,865,620,175,800,955,129) is the perfect square (of 4,567,890,123).  The key is in the final digit: perfect squares cannot end in the digits 2, 3, 7, or 8. We can determine this by squaring all 10 digits, and seeing what the last digit of each square is:

02 =   0
12 =   1
22 =   4
32 =   9
42 = 16
52 = 25
62 = 36
72 = 49
82 = 64
92 = 81

Thus, regardless of how many digits come before the last digit, the last digit of a perfect square must be 0, 1, 4, 9, 6, or 5.  This eliminates the first four numbers, leaving the fifth as the only possible perfect square.

Answer:
Well, the first and third numbers end in an even number, so they are divisible by 2, and thus are not prime. That leaves the three odd numbers. If we can eliminate two of them, we’ll know the answer. So, let’s check to see if any of the three odd numbers is divisible by 3. To do this, let’s add up their digits. A fun fact related to multiples of 3 is that any number whose digits add up to 3 is also a multiple of 3. So, let’s add some digits:

1 + 2 + 7 + 6 + 4 + 7 + 8 + 7 + 8 + 4 + 6 + 3 + 5 + 8 + 4 + 4 + 1 + 4 + 7 + 1 = 97

1 + 4 + 0 + 0 + 9 + 0 + 5 + 7 + 6 + 8 + 0 + 8 + 4 + 6 + 5 + 8 + 7 + 6 + 5 + 7 = 96

1 + 5 + 1 + 8 + 9 + 3 + 1 + 6 + 5 + 4 + 9 + 8 + 7 + 3 + 1 + 3 + 6 + 7 + 9 + 3 = 99

Since 96 and 99 are both multiples of 3, the last two numbers are multiples of 3 and cannot be prime. Thus, the number 12,764,787,846,358,441,471 must be prime.

Answer:
To maximize your chances of winning on “Wild Cards,” you need to consider two sets of probabilities: the chance that you will pick a given stack, and the chance that you will select an ace from the chosen stack. As it turns out, your best chance is to split the deck into five stacks. Four of the stacks will consist of just one card (an ace), while the other stack contains the other 48 cards. Thus, your odds of selecting one of the ace “stacks” is 4/5; if you select an ace stack, you are guaranteed to win (since the chances are 1/1 that you will then pick the ace), whereas if you select the other stack, you are guaranteed to lose (since the chances are 0/48 that you will get an ace).

If you divide the deck into fewer stacks (say, 4), the best you can do is to have 3 stacks containing just an ace, and a fourth stack with one ace and the other 48 cards. The odds of winning in this case are 3/4 + 1/49, or approximately 0.77 (less than 4/5, or 0.8). If you divide the deck into more stacks (say, 6), the best you can do is to have 4 stacks with just an ace, and two stacks without (it doesn’t matter how you divide the last two). The odds then are 4/6 + 0, or about 0.67.

Answer:
The trick to this brainteaser is to make the 4 an exponent (this doesn’t constitute “adding a sign” since there is no operator for exponents):

3^4 – 62 = 19 (3^4 = 81; 81 – 62 = 19)

Answer:
First, turn on the first switch. Leave it on for a few minutes. Then, turn it off, and turn on the second switch. Now, go into the room with the light bulbs. The bulb that is on obviously corresponds with the second switch. Feel the other two bulbs—one of them will be warm. The warm bulb is the one that had been on for several minutes, and thus the one that corresponds with the first switch. The remaining bulb is paired with the third switch.

Answer:
Once each of the four prisoners has stolen a Jeep and filled its tank completely (for a total of 4 tanks of gas), there are at least two ways for them to escape:

1. Each prisoner drives his stolen Jeep one-sixth of the way to the gas station, using up one-third of the gas in each tank. There are now eight-thirds of a tank of gas left (4 tanks – 4/3 tank = 8/3 tank). One of the prisoners abandons his Jeep, dividing its remaining gas equally among the other three vehicles. Each Jeep now has 8/9 of a tank (8/3 of a tank divided by 3 vehicles).  The four prisoners now drive in the three Jeeps, traveling another 1/6 of the distance to the gas station. Each of the three vehicles uses up another 1/3 of a tank, leaving each with 5/9 of a tank (8/9 – 1/3 = 5/9). There is a total of 5/3 of a tank of gas left. The prisoners abandon another Jeep, dividing its gas between the two remaining Jeeps, each of which now has 5/6 of a tank of gas. The four prisoners are now in two Jeeps; they drive another 1/6 of the way to the gas station, so that they are halfway there. Each of the two Jeeps uses another third of a tank, leaving each with half a tank (5/6 – 1/3 = 1/2). The four prisoners then put all the gas into one Jeep (giving it a full tank) and pile into it. They drive the remaining distance to the gas station, arriving just as the gas runs out.
2. Same as #1, except the prisoners drive four cars 1/4 of the distance, transfer the remaining gas to just two cars (filling them completely), drive another 1/4 of the distance, and transfer the gas to just one car. As before, they are now halfway to the station, and have one car with a full tank, just enough to reach the station.

Answer:
The two babies are two-thirds of a set of triplets (or two of four quadruplets, etc.).

Answer:
Let’s say there are B mathematicians. We know that B divided by 7 has no remainder, since when the mathematicians divide into groups of 7, there is no one left over. Thus, B is a multiple of 7. Similarly, we know that B has a remainder of 1 when divided by 2, 3, 4, 5, or 6. This means that B is 1 more than a multiple of 2 (and thus odd), 1 more than a multiple of 3, etc. Thus, B – 1 is a multiple of 2, 3, 4, 5, and 6. If we find the least common multiple (LCM) of 2, 3, 4, 5, and 6, we will have a good starting point in searching for B. It turns out that the LCM of those numbers is 60.

What this means is that B – 1 must be a multiple of 60. To find B, we must add 1 to all the multiples of 60 until we find a B that is a multiple of 7:

Thus, there are 301 mathematicians at the meeting.

Answer:
(11)^3 = 1331

There are 9 possibilities for the number AA–11, 22, 33, 44, 55, 66, 77, 88, 99. B cannot be 1, since any of those numbers to the first power is only two digits, not four. Next, we try a B of 2, which means we square each of the possible values of AA. The results of these nine calculations are 121, 484, 1089, 1936, 3025, 4356, 5929, 7744, and 9801, none of which fits the ABBA pattern that we need. Next, we try a B of 3. The first possible value of AA is 11: 113 = 1331. Voila! We have an answer. A = 1, and B = 3.

Answer:
In order to win the largest prize possible, Amy must quit now, even though she only gets $10 for doing so. This is because she has already eaten too many jelly beans to win the grand prize or second prize. The hint says that she must average 4 beans per minute to win the grand prize, and 6 beans per minute for the second prize. Unfortunately, the show is half over when she gets this hint—and she has already eaten 13 jelly beans per minute for the first half of the show! Let’s say the show is 60 minutes long (though it could be any length):

• 13 jelly beans per minute * 30 minutes (half of 60 minutes) = 390 jelly beans
• To win the grand prize: 4 jelly beans per minute * 60 minutes = 240 jelly beans
• To win the second prize: 6 jelly beans per minute * 60 minutes = 360 jelly beans

Thus, Amy has already eaten 30 too many jelly beans to win one of the bigger prizes. If she realizes this, however, she can at least quit now and get the $10 consolation prize.

Answer:
First, four of the children take one apple each. Then, the fifth child takes the last apple, puts it in the basket, and holds the basket. Thus, an apple is in the basket, and each child has exactly one apple.

Answer:
The master's hat was black. He was able to deduce this from the fact that each master had an equal chance to win. In order for this to be true, all three hats had to be black. If only one of the masters had a black hat, then that master would see two white hats and thus know that his must be black; the other two masters, however, would not be certain of their hat color, making the contest unfair.

Similarly, if two of the masters had black hats and the third had a white hat, then the two with black hats would have had an advantage. They would be able to see one black and one white hat each; each one would suppose that if he had been given a white hat, then the master with the black hat would see two white hats, and thus the contest would be unfair (as described above). Eliminating this possibility leaves only the possibility that all three hats are black. Thus, the winner was the master who realized this first, and proclaimed that his hat was black.

Answer:
The wisest one must have thought like this:
I see all hands up and 2 red dots, so I can have either a blue or a red dot. If I had a blue one, the other 2 guys would see all hands up and one red and one blue dot. So they would have to think that if the second one of them (the other with red dot) sees the same blue dot, then he must see a red dot on the first one with red dot. However, they were both silent (and they are wise), so I have a red dot on my forehead.

Here is another way to explain it:
All three of us (A, B, and C (me)) see everyone's hand up, which means that everyone can see at least one red dot on someone's head. If C has a blue dot on his head then both A and B see three hands up, one red dot (the only way they can raise their hands), and one blue dot (on C's, my, head). Therefore, A and B would both think this way: if the other guys' hands are up, and I see one blue dot and one red dot, then the guy with the red dot must raise his hand because he sees a red dot somewhere, and that can only mean that he sees it on my head, which would mean that I have a red dot on my head. But neither A nor B say anything, which means that they cannot be so sure, as they would be if they saw a blue dot on my head. If they do not see a blue dot on my head, then they see a red dot. So I have a red dot on my forehead.

Answer:
You can hang the iron rods on a string and watch which one turns to the north (or hang just one rod).

Gardner gives one more solution: take one rod and touch with its end the middle of the second rod. If they get closer, then you have a magnet in your hand.

The real magnet will have a magnetic field at its poles, but not at its center. So as previously mentioned, if you take the iron bar and touch its tip to the magnet's center, the iron bar will not be attracted. This is assuming that the magnet's poles are at its ends. If the poles run through the length of the magnet, then it would be much harder to use this method.

In that case, rotate one rod around its axis while holding an end of the other to its middle. If the rotating rod is the magnet, the force will fluctuate as the rod rotates. If the rotating rod is not magnetic, the force is constant (provided you can keep their positions steady).

Answer: 76

Where do we start to find the answer?! Let's try a random guess: is it 10? No. Why? As 10 isn't a multiple of 4 it would have to be between 60 and 69 (by #1). Therefore, if it isn't a multiple of 4, it must lie between 60 and 69 (and not be a multiple of 4), i.e. 61, 62, 63, 65, 66, 67 or 69. It can't be a multiple of 3 (by #2) therefore only 61, 62, 65, 67 remain. However, by #3 none of these numbers is allowed (as none is a multiple of 6). Therefore we've reached a contradiction, and the number MUST be a multiple of 4. Let's try another random number now, 8 say, but by #3 the number must be between 70 and 79, so another contradiction. So, either the number is a multiple of 6, or between 70 and 79 (and not a multiple of 6). Every multiple of 6 is also a multiple of 3, therefore by #2 we're looking for a number between 50 and 59, which is a multiple of 3, i.e. 51, 54, 57, BUT by #1 none of these is allowed. Therefore by #3, the number lies between 70 and 79. We know it's a multiple of 4 (by #1) which leaves only 72 and 76, but by #2 (and #3) we can rule out 72. Which only leaves 76!