Why are your nose and chin always at variance?
Answer: Because words are passing between them.
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Walking home one day, you take a short cut along the train tracks. The tracks cross a narrow bridge over a deep gorge. At the point you are 3/8 of the way across the bridge, you hear the train whistle somewhere behind you. You charge across the bridge, and jump off the track as the train is about to run you down. As it happens, if you had gone the other way, you would have reached safety just before being run over as well. If you can run ten miles per hour, how fast is the train moving?
Answer: The train is moving at 40 miles per hour. Imagine that a friend is walking with you. When the train whistle blows, you head away from the train, he heads toward it. When he reaches safety, you will be 6/8 (or 3/4)of the way across the bridge, and the train will have just reached the bridge. For the train to cross 4/4 of the bridge in the time you cross the remaining 1/4, the train must be moving four times your speed.
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The train is moving at 40 miles per hour. Imagine that a friend is walking with you. When the train whistle blows, you head away from the train, he heads toward it. When he reaches safety, you will be 6/8 (or 3/4)of the way across the bridge, and the train will have just reached the bridge. For the train to cross 4/4 of the bridge in the time you cross the remaining 1/4, the train must be moving four times your speed.
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The train is moving at 40 miles per hour. Imagine that a friend is walking with you. When the train whistle blows, you head away from the train, he heads toward it. When he reaches safety, you will be 6/8 (or 3/4)of the way across the bridge, and the train will have just reached the bridge. For the train to cross 4/4 of the bridge in the time you cross the remaining 1/4, the train must be moving four times your speed.
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.
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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.
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