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A hunter met two shepherds, one of whom had three loaves and the other, five loaves. All the loaves were the same size. The three men agreed to share the eight loaves equally between them. After they had eaten, the hunter gave the shepherds eight bronze coins as payment for his meal. How should the two shepherds fairly divide this money?
Answer: 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.
Solution:
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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.
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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.
How can you tell which part of the turkey is the left side?
Answer: The part that was not eaten
Solution:
George, Helen, and Steve are drinking coffee. Bert, Karen, and Dave are drinking soda. Is Elizabeth drinking coffee or soda?
Answer: Elizabeth is drinking coffee. The letter E appears twice in her name, as it does in the names of the others that are drinking coffee
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Before Google, we actually had to physically look things up in this building full of knowledge.
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There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person. Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, …). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, …). This continues until all 100 people have passed through the room. What is the final state of bulb No. 64? And how many of the light bulbs are illuminated after the 100th person has passed through the room?
Answer: First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 ........ That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it's original state. So why aren't all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it's self. Clearly the sixth person will only flick the bulb once and so the pairs don't cancel. This is true of all the square numbers. There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.
Solution:
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First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 …….. That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it’s original state. So why aren’t all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it’s self. Clearly the sixth person will only flick the bulb once and so the pairs don’t cancel. This is true of all the square numbers. There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.
Show Answer
First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 …….. That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it’s original state. So why aren’t all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it’s self. Clearly the sixth person will only flick the bulb once and so the pairs don’t cancel. This is true of all the square numbers. There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.
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