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The bowls, in order, contain 18, 30, 4, 26 and 22 nuts. There are several ways to solve. A guess and check strategy with a clever revision will yield the solution. For example, guessinng 24 and 24 for the first two bowls, we must have 10, 20 and 28 for the others in order. The total 106 must be adjusted down.

A very clever student might see that bowls 1, 2, 4 and 5 contain all but 4 of the nuts. The other amounts follow quickly.

1 = 9 – 87 + 65 – 4 – 3 + 21

1 = 98 – 76 – 54 + 32 + 1, and there might be others.

0 = 9 – 87 + 6 + 54 – 3 + 21

100 = 9 + 8 + 76 + 5 + 4 – 3 + 2 – 1

[and you fill in the rest.]

1/2 = 7269 / 14538, and there is a second solution!

1/3 = 5823 / 17469

1/4 = 3942 / 15768

1/5 = 3297 / 16485, and there is a second solution!

1/6 = 5697 / 34182, and ... you guessed it!

1/7 = 7614 / 53298 as given in the problem, but ... can you find a second solution?

1/8 = 7123 / 56984

1/9 = 8361 / 75249

Recently I found third solutions for one-half and one-fifth. I believe that one-ninth has only one solution. Can it be proved?

1. Switcheroo

89 + 9 = 98

2. Double Duty

D = 1, since 1.1 + 11 = 1.1 * 11. Also D = 0 is ruled out by the instructions.

Bonus: 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.

1. Turn D and F for 180 / 18 = 10.

2. Turn K, G and H for 186 – 11 x 8 = 98.

3. Turn N then P, and M and L for (1689 + 66) / (196 – 1) = 9.

A. 63 – 29 = 8 x 2 x 2 + 2 B. 6 x (15 – 3) = 100 – 7 x 2 x 2

C. 38 x 1 + 63 = 37 x 3 – 10 D. 46 – 8 x 5 = 96 / 16

E. 510 / (2 x 3 x 5) = 28 – 11 F. 225 – 5 x 5 x 5 = 66 / 6 + 5 + 84

Zero (0) = 4 + 1 – 3 – 2. Answers may vary. Possible solutions for the negative integers are in the table below. Negative 17 was the toughest for me.

1 + 3 – 24 | -20 | 2 – 4 * 3 * 1 | -10 |

4 * 1 – 23 | -19 | 32 – 41 | -9 |

23 – 41 | -18 | 1 – 2 – 3 – 4 | -8 |

(3 – 4 / .2) / 1 | -17 | 24 – 31 | -7 |

4 * (1 – 3 – 2) | -16 | 3 – 1 – 2 * 4 | -6 |

2 – 4 – 13 | -15 | 4 + 3 – 12 | -5 |

3 + 4 – 21 | -14 | 1 + 2 – 3 – 4 | -4 |

3 – 4 – 12 | -13 | 3 – 2 * (4 – 1) | -3 |

2 * (1 – 4 – 3) | -12 | 4 – 1 – 3 – 2 | -2 |

31 – 42 | -11 | 4 * 1 – 3 – 2 | -1 |

The row totals are 61, -11, -55 and 5. The column totals are -69, 11, 43 and 15.

The sum of all entries is zero.

The (unseen) addends that created the addition table are 10, 30, 38 and 31 going across and -12, -30, -41 and -26 down the side, but this solution is not unique.

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.

Answers will vary. One student observed that the median is a better cushion for the single "disastrous" score; however, for the same reason, another student remarked that for the student satisfied with his/her grades, there would be little motivation to take the last test.

The park (P) is 25,200 and the Town Hall site is 16,800 square feet. Since rectangles in one row or one column share either a length or a width, the ratio of their areas must equal the ratio of the unshared dimension. Unknown areas can be solved with proportions. For example, (Lot B4 / 4000) = (11,200 / 3200). B4 = 14000 square feet. Lots D2, D3 and A2 can be used to solve A3. Lots A3, A4 and B4 will solve P. Lots A2, A3 and P can be used to solve T.

1. 699 pounds 2. The cylinder is 3000 / 62.5 or 48 cubic feet. It's diameter is 2.6 feet. The Big Dude would be wider because bone is less dense than water.

For 3, 4 and 5 most answers will vary except that the blue whale mentioned needs at most 39,000 pounds of bones.

The optimum solutions seem to fix two cache locations and go no further than 205 miles for the first cache. The second cache should be located by working backwards as will be shown. All distances are miles, and also figs!

First solution: Begin with 1000 figs, travel 205 miles, cache 590 figs, return to start to pick up 1000 more. Travel 205 miles to the cache, pick up 205 from cache (leaving 385), and travel forward to ... where?

In order to have 530 figs at market, we need to leave from 470 miles away from market with 1000 figs. So we proceed to the second cache 530 miles from the start. The camel eats 325 figs going from the first to the second cache, and we have 675 when we arrive. We can afford to leave 350 figs and still get back to the first cache where we pick up 205 for the return to the starting point – where we pick up the last 1000. At the first cache we still have 180 figs. We travel to the first cache to pick those up and we have 1000 – 205 + 180 or 975 for the final push forward.

With 975 figs we proceed to the second cache using 325 of them, where we pick up 350 more and proceed the final 470 miles to deliver 530 figs to market.

Bonus. Changing only the location of the first cache to 204 miles from start, we again get 530 figs to market, having to leave two figs at a cache. Can the 530 figs be bettered?

Three coins and two coins was not a fair split. Since 3 people shared 5 loaves equally, each ate 5/3 loaves. One of the bedouins had 6/3 loaves and contirbuted 1/3 to the stranger. The other had 9/3 and contributed 4/3 loaves to the stranger. A fair split would be 4 coins to one.

A sphere 6 feet in diameter has a volume of 113.1 cubic feet. Other answers will vary. My informal lab test showed that a sack 2" in diameter holds roughly 100 quarter dollars.

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