**Solution #87: Dinner for Three**
The shepherd who had three loaves should get one coin and the shepherd who had five loaves should get seven coins.
If there were eight loaves and three men, each man ate two and two-thirds loaves. So the first shepherd gave the hunter one third of a loaf and the second shepherd gave the hunter two and one-third loaves. The shepherd who gave one-third of a loaf should get one coin and the one who gave seven-thirds of a loaf should get seven coins.
**Problem #88: One Clock**
In the days before watches were invented, clocks were valuable items. There was a man who had one clock in his house. It kept good time, but one day he found that it had stopped. He had no idea what the correct time was. He walked to the next valley to visit his friend who had a clock showing the right time. He spent a little while chatting with his friend then he walked home. He did not know the exact length of the journey before he started. How did he manage to set his one clock correctly on his return?
**Solution #88: One Clock**
He wound his clock and set it at some particular time before he left. He noted the exact time of his arrival at and departure from his friend's house. He noted the time showing on his clock when he returned. He walked at the same pace on the two journeys. The elapsed time on his clock is the duration of the two journeys plus the length of his visit to his friend. Knowing the time he spent with his friend, he subtracts this from the elapsed time on his clock and divides the result by two in order to calculate the duration of the journey. He adds this to the exact time he left his friend's house in order to set his clock at the correct time. For example, assume he set his clock at 12, arrived at his friend's at 6:30, and left at 7:30. When he returned, his clock showed 4:00. Then his journey took one and one half hours and the correct time is 9:00.
**Problem #89: The Mongolian Postal Service**
The Mongolian Postal Service has a strict rule stating that items sent through the post must not be more than 1 meter long. Longer items must be sent by private carriers, and they are notorious for their expense, inefficiency, and high rate of loss of goods.
Boris was desperate to send his valuable and ancient flute safely through the post. Unfortunately, it was 1.4 meters long and could not be disassembled as it was one long hollow piece of ebony. Eventually he hit on a way to send it through the Mongolian Postal Service. What did Boris do?
**Solution #89: The Mongolian Postal Service**
Boris placed the flute diagonally in a suitcase that measured 1 meter by 1 meter. This suitcase was quite acceptable to the postal officials its sides measured 1 meter. From corner to corner, it measures 1.414 meters -- the square root of two.
Incidentally, if his flute had measured 1.7 meters, he could have fitted it across the diagonal of a box whose sides were 1 meter long. The diagonal of a cube is the square root of the three times its side -- 1.73 meters.
From a theoretical mathematical viewpoint, there is no reason why this process cannot be extended indefinitely. If Boris could construct a four-dimensional box with 1-meter sides, then he could get a 2-meter flute in it (square root of 4) and a 25-dimensional construction could contain a 5-meter flute while still meeting the rules of having no side longer than 1 meter.
**Problem #90: The Amorous Commuter**
John Jones lives in Maidenhead. He has one girlfriend in Reading and another in Slough. He has no car and therefore takes a train whenever he goes to see them.
Trains stopping in Maidenhead can go either east or west. If they are westbound, they will go to Reading. If they are eastbound, they will go to Slough. There are an equal number of trains going in each direction.
John likes his two girlfriends equally. Because he finds it hard to choose between them, he decides that when he goes to the station, he will take the first arriving train, regardless of whether it is going east or west. After he has done this for a month, he finds that he has visited the girlfriend in Slough 11 times as often as he visited the girl in Reading. Assuming that he arrived an the Maidenhead station at random times, why should the poor girl in Reading have received so little attention?
**Solution #90: The Amorous Commuter**
The Slough trains depart from Maiden head at five past every hour. The Reading Trains depart at ten past the hour.
If John arrives at any time between ten past the hour and five past the next hour, then the first train to arrive will be bound for Slough. He will catch the Reading train only if he happens to arrive between five past and ten past the hour. It is therefore 11 times more likely that he will catch the Slough train than the Reading train.
**Problem #91: Short Roads**
There are four main towns in Lateralia. We will call them A, B, C and D. They lie at the corners of a ten-mile square. In order to improve communications between the towns, the Lateralian Department of Transport decided to build a new road linking all four towns together. Because they had very little money, it was decided that the new road system should be as short as possible and still allow access from any one town to any other. The engineers came up with three designs shown below.
Number one uses 40 miles of road, number two uses 30 miles of road, and number three uses 28.3 miles of road. The designers naturally recommend plan number three because it employed the smallest road area and, therefore, cost the least. However, when they submitted their plan to the Minister of Finance, he accused them of extravagance and quickly pointed out a better design that required even less total road surface. What was his superior solution?
**Solution #91: Short Roads**
The shortest solution is shown. It represents some 27.3 miles of road in total and therefore saves a precious mile in road-building expense compared to the two diagonals. Someone starting from A would have a shorted journey to D but a longer one to B or C.
**Problem #92: A Weighty Problem I**
A shopkeeper wants to be able to dispense sugar in whole pounds ranging from one pound up to 40 pounds. He has a standard, equal-arm balance weigh scale. Being of an extremely economical outlook, he wants to use the least possible number of weights to enable him to weigh any number of pounds between 1 and 40. How many weights does he need and what are they?
**Solution #92: A Weighty Problem I**
If the weights can be placed in either of the scale pans, then you can solve the problem with weights of 1, 3, 9, and 27 pounds only. With that combination any weight from 1 to 40 pounds can be measured. If the weights can be placed in one scale pan only, then you need the weights 1, 2, 4, 8, 16, and 32 pounds, which enables you to measure any weight up to 63 pounds. This latter solution is really an example of counting with a binary number system where any number can be expressed as the sum of 2 raised to various powers. For example, 63 expressed in binary form is 111111.
**Problem #93: A Weighty Problem II**
You have an equal-arm balance scale and twelve solid balls. You are told that one of the balls has a different weight from all the others, but you do not know whether it is lighter or heavier. You can weigh the balls against each other in the scale balance. Can you find the odd ball and tell if it is lighter or heavier in only three weighings?
**Solution #93: A Weighty Problem II**
Let us call the balls A, B, C, D, ..., L. Start by weighing four against four. If they balance, then weigh any of the remaining three against any three of the good balls. If they balance then we know the odd one is the remaining ball and we can identify whether it is heavier or lighter in the final weighing. If the three against three do not balance then we take the three containing the odd ball and weigh an one against another.
If the first weighing of four against four does not produce a balance, then the second weighing involves three against three with balls switched between the two pans and a good ball introduced. So:
If A + B + C + D > E + F + G + H
We try A + B + E against C + F + J
If A + B + E = C + F + J,
then we know that either D is heaver or G or H is lighter, so we weight G against H.
If A + B + E > C + F + J,
then we know that either F is lighter or A or B is heavier, so we weight A against B.
If A + B + E < C + F + J,
then we know that either E is lighter or C is heavier, so we weight either against a good ball (e.g. K against E).
An alternative second weighing is A + B + E against C + D + F, which follows similar lines to the above.
**Problem #94: Eight Years Old**
A girl was eight years old on her first birthday. How could that be?
**Solution #94: Eight Years Old**
She was born on February 29, 1896. The year 1900 was not a leap year (only centuries divisible by 400 are leap years), so the next February 29 fell in 1904 when she was eight. She was twelve on her second birthday.
**Problem #95: Cover That Hole**
A manhole is a hole which allows someone to gain access to the sewers or other pipes which are below ground. Our local town council recently decided that all the town's manhole covers should be changed from square to round ones. We are used to the town council making silly decisions, but this time they were absolutely right. Why?
**Solution #95: Cover That Hole**
A square or rectangular manhole cover can fall down the hole, while a round manhole cover cannot. The square cover will fit down the diagonal of the hole (unless the rim it sits on is very large) but no matter how you turn a circle it never measures less than its diameter. So for safety and practicality all manhole covers should be round.
Alex Freuman also notes: "In addition to the advantage that a circular manhole cover cannot fall into a manhole, one can roll a circular manhole cover rather than lifting it."
**Problem #96: Pond Problem**
A man wishes to reach the island in the middle of an ornamental lake without getting wet. The island is 20 feet from each edge of the pond (see diagram) and he has two planks each 19 feet long. How does he get across?
**Solution #96: Pond Problem**
Hey lays the planks as shown in this diagram:
**Problem #97: River Problem I**
A man came to a river carrying a fox, a duck, and a bag of corn. There was a boat in which he could ferry of the three items across the river at any one time. He could not leave the fox alone with the duck, nor the duck alone with the corn, so how did he get all three across?
**Solution #97: River Problem I**
First the man took the duck across, then he came back and took the fox over. He left the fox on the far side of the river and returned with the duck. He then left the duck on the near side and took the corn over. Then he returned and took the duck across. Pretty straightforward, eh?
**Problem #98: River Problem II**
This time the man reached the river with a fox, a duck, and a bag of corn, but this fox ate corn as well as ducks! There was the same boat as before in which he could take only one of the three with him. He could not leave the fox with either the corn or the duck, and of course, the duck would gladly eat the corn if they were left together. How did he get all three across?
**Solution #98: River Problem II**
The man tied the duck to the back of the boat wit a rope. The duck swam along behind the boat as the man ferried the fox and corn over in turn.
**Problem #99: The Bicycles and the Fly**
Two boys on bicycles 20 miles apart, began racing directly toward each other. The instance they started, a fly on the handle of one bicycle started flying straight toward the other cyclist. As soon as it reached the other handle bar it turned and started back. The fly went back and forth this way, from handle bar to handle bar, until the two bicycles met.
If each bicycle had a constant speed of 10 miles per hour, and the fly flew at a constant speed of 15 miles per hour, how far did the fly fly?
**Solution #99: The Bicycles and the Fly**
Each bicycle travels at 10 miles per hour, so they will meet at the center of the 20-mile distance in exactly one hour. The fly travels at 15 miles per hour, so at the end of the hour it will have gone 15 miles.
**Problem #100: Where does the square go?**
Paste a sheet of graph paper on a piece of cardboard. Draw the square shown in figure 1, then cut along the lines to make five pieces. When you rearrange these same five pieces, in the manner shown in Figure 2, a hole will appear in the center of the square!
The square in Figure 1 is made up of 49 smaller square. The square in Figure 2 has only 48 small squares. Which small square has vanished and where did it go?
**Solution #100: Where does the square go?**
When the two largest pieces are stitched, each small square that is cut by the diagonal line becomes a trifle higher than it is wide. This means that the large square is no longer a perfect square. In has increased in height by an area that is exactly equal to the area of the hole.
**Problem #101: Siblings**
Todd and Kristina
*(not yet completed)*
**Solution #101: Siblings**
*(not yet completed)*
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