Hickory-Dickory-Dock! The mouse ran up the clock. The clock struck one and down did come. Hickory-Dickory-Dock! What am I?
Answer: A guillotine.
Solution:
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Three cars had driven into a parking lot at the same time, and the three drivers left them all for the attendant to park. Unfortunately, he isn’t too good at remembering exactly which driver drove which car. However, he is sure of these 6 facts: a. Colin drove the BMW if and only if Mr. Cooper drove the Avenger. b. Alan drove the Cortina if and only if Mr. Cooper drove the BMW. c. Colin is Mr. Brown if and only if Mr. Andrews drove the BMW. d. Brian is Mr. Andrews if and only if Colin drove the BMW. e. Mr. Cooper drove the Avenger if and only if Alan is Mr. Brown. f. Colin is Mr. Brown if and only if Alan drove the Cortina. Who arrived with which car?
Answer: Brian Brown drove the BMW, Alan Andrew drove the Avenger, and Colin Cooper drove the Cortina.
Solution:
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.
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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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What am I ?