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3-D Posers

1. Place 4 at the top, and 7, 1; 2, 6; and 3, 5 down the lateral edges.

2. A = 8, B = 4, C = 5 and D = 8; E = 2, F = 3, followed by 6 and 7 in clockwise rotation.

3 and 4. Many solutions.

5. The magic sum is 63. The layers of one possible scheme are:

Gears

1. In 4 turns because

4 x 13 + 2 = 54 = 6 x 9.

2. In 9 turns because

9 x 14 = 126 = 11 x 11 + 5.

3. In 45 turns because

45 x 11 = 495 = 33 x 15 = 38 x 13 + 1.

Spellathon

1. DEEM, DMEE, DEME, and using symmetry, swap the M and D:

MEED, MDEE, MEDE, and then the E-list

EEMD, EEDM. EMED, EDEM, EDME AND EMDE.

2. DDEE, DEDE, DEED, and by symmetry EEDD, EDED AND EDDE.

3. The numbers of words should be 5! / (3! * 2!) or 10.

They are AAABB, AABAB, AABBA, ABABA, ABAAB, ABBAA.

The B-list is shorter because the other B can be in only four locations:

BBAAA, BABAA, BAABA AND BAAAB.

BONUS: The letters D, E, E, D, E, D should make 6! / (3! * 3!) or 20 distinct words.

DDDEEE, DDEDEE, DDEEDE, DDEEED, DEDDEE, DEDEDE, DEDEED, DEEDDE,

DEEDED and DEEEDD. The other ten words are found by swapping Ds and Es in this list.

Examples 2, 3 and the Bonus can be compared to the Combinations (choice numbers) formula. For example, if we want to choose 3 people from a group of 6, have them line up in front of the room and pass them each a slip of paper labeled with either one of three Ds or one of three Es. Tell them that the Ds will be chosen. When they show the letters they were given, they will "spell" one of the 20 possible words from DEEDED.

Consumers' Dilemma

One solution is 8, 20, and 72 Carrot Patch Kids, Bubble Bunnies and Gummy Hares, respectively. The number of Gummy Hares must be even because of the odd unit price.

Bonus: Dropping 3 Kids and 2 Gummy Hares, we can pick up 5 Bunnies and keep the totals at 100 items and $600. Or we can add 3 Kids and 2 Hares and drop 5 Bunnies. All solutions follow.

Cryptarithms

1. Reversal of Four-tune

2178 * 4 = 8712. There are some interesting extensions of this problem. First, solutions in 5- and 6-digits can be found which point the way to an infinite family of factors and reversed products with 4 as the multiplier. Next, the multiplier 9 leads to another infinite family of reversed products. Does the fact that 1.5 * 4356 = 6534 relate to the situation? Is the number

(10^n + 10^(n-1) – 10 – 1)

a "flipping" candidate?

2. How Far? 7380 + 6873 = 14253.

3. Strange Quintuple. One solution is 1647 x 5 = 8235. Can you find a second?

4. The Angry Heirs. 438984 / 871 = 504

5. A Bit Square, My Father. D is at least 3. The root of 484,416 is 696.

6. Sextuple Switch.
Another way to express the relationship is:

(6000 x ABC) + (6 x DEF) = (1000 x DEF) + ABC

Rearranging and collecting like terms, we get

5999 x ABC = 994 x DEF,

and dividing both sides by 5999 x DEF gives the proportion 994/5999, which reduces to 142/857. The solution is 142857 x 6 = 857142.

7. The Disappearing Leading Digit. Another way to express the relationship is:

(60000 + x) / 25 = x,

and multiplying both sides by 25, we get

60000 + x = 25x,

so 60000 must represent 24x, and that leads to x = 2500. The solution is

62500 / 25 = 2500.

8. They Go Together.
The sums in each identical column end in a different digit, so they must be 19, 20, and 21. F can only be a 1 or 2, so FGHI = 2109. Testing possibilities for the rest yield A = 8 and D + E = B + C = 11 = 5 + 6 = 7 + 4.

9. Special spatial digital problem. One solution is

but are there others?

Quadrisections

Tic Tic Tac Tac No Toe

1. The Game. As long as the first player avoids 3-in-a-row, the second player can as well by playing squares symmetrically. Place a dot in the very center of the grid. When X plays a square, O should play in the square on the opposite side of the dot.

2. The Puzzle. Let’s put an X in the upper left, and we’ll look at the “O” solution in a bit. This X begins three safe words. If the grid were flipped along diagonal D1, the words in the rows and columns are swapped, and the D1 word is unaffected; therefore, the tenth word along D2 cannot change, and that means it is one of the palindromes XOOX or OXXO. If it were the first pal. there would be six words in the grid that begin with X, and that would force a duplicated word, so D2 is OXXO.

If the top row begins with X and ends with O, then columns 2 and 3 must begin with different letters, either XO or OX. Place one of these pairs in and the rest of the spaces will be forced by the rules of the puzzle.

Once you have all ten words in the grid, there are 3 other solutions, but not really. As alluded to, the flip along D1 will exchange rows and columns and will appear to be a second solution. The other two solutions are created from these two by merely swapping Xs for Os and vice versa. Notice also that the solution is unchanged by a flip along D2 – one last instance of symmetry.

Footsie

If the men, namely Mark and Marv lied, then they had both been footsied which is not possible. If the men told the truth, then Flo is implicated. This fact is consistent with the women all telling fibs.

Logical Train

The engineer's name is Sid. The last clue tells us that Sid is not the fireman. Clue d informs that Pete the Passenger is not the one living closest to the conductor because $4000 is not divisible by three. Ivan the passenger is not living closest to the conductor either, since he lives in DC and the conductor lives between NYC and DC. So passenger Sid lives closest to the conductor, and passenger Pete lives in NYC.

Then clue c indicates that Pete is the conductor on the crew. Since Sid is not the fireman nor the conductor, Sid must be the engineer.

Name the 6

The six are 1, 1, 1, 9, 24 and 24. The required members at the outset are 1, 1, a, b, c and 24. Since the median is 5, a+b = 10. The sum of the set is 60 which requires c = 24. Since the mode is one (1), there must be a third occurence of it. Setting a = 1, gives b = 9.

Many similar problems could be constructed to help reinforce the statistical concepts.

Unlucky Digit

Of the 300 locker numbers, there are 138 that contain at least one occurrence of the digit three. The probability of avoiding one of those is 162/300. The odds are slightly in your favor.

Bonus. The probability of avoiding a 4 in any one decimal place is 9/10. To avoid it with 0.5 probability in any of n places:

0.9^n = 0.5 or

n = log(0.5) / log(0.9)

n = 6.57881348

10 ^ n = 3791521.115

so in the interval [3,791,522, inf) you have less than half a chance to avoid your unlucky 4.

5 Degrees of Rationality – one possible set of solutions below.

Pentacubes?

There are 29 distinct pentacubes. There are the 12 pentominos made one unit in thickness. There are six right- and left-handed pairs, and there are five more.

Tetromino Sets

The double set will tile a 5 x 8 rectangle, but the triple set will never tile a 5 x 12. In the double set, the checkerboard coloring indicates that the white and gray squares can be balanced 20 - 20 by the judicious placements of the T-tetrominos.

In a triple set, there must be an unequal number of gray and white squares due to the T, but tiling a 5 x 12 requires 30 gray and 30 white squares.

Beguiling Blockage

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