Reasoning Exercise : 1020

A Printing Machine Problem

A printing machine, capable of only printing the 10 digits, prints the right digit 12% of the time. Whenever the machine prints erroneously, all wrong possibilities are equally likely.
If a random key is pressed twice and a same digit comes out both times, what is the probability that it was the correct one?

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The Oldest Plays the Piano

Two MIT math grads bump into each other while shopping at Fry’s. They haven't seen each other in over 20 years.

First grad to the second: "How have you been?"
Second: "Great! I got married and I have three daughters now."
First: "Really? How old are they?"
Second: "Well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there..."
First: "Right, ok... Oh wait... Hmm, I still don't know."
Second: "Oh sorry, the oldest one just started to play the piano."
First: "Wonderful! My oldest is the same age!"
How old was the first grad’s daughter?


ans :  
The Oldest Plays the Piano Solution
The possible ages ( factors of 72 ) and their sums are shown below:

Ages:            Sum of ages:
1 1 72            74
1 2 36            39
1 3 24            28
1 4 18            23
1 6 12            19
1 8 9             18
2 2 18            22
2 3 12            17
2 4 9             15
2 6 6             14
3 3 8             14
3 4 6             13

We can deduce from the man’s confusion over the building number that this wasn’t enough information to solve the problem. The chart shows the sum 14 twice for two different age possibilities, which would explain how knowing the building number alone would not have given him the answer. The clue that the “oldest one” started to play the piano rules out “2 6 6” as an answer, because there is no “oldest”. Since the first grad was certain with the piano clue, the first grad’s oldest daughter is 8. I’ll leave it up to the reader to figure out why this doesn’t necessarily mean the second grad’s oldest daughter was also 8.



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100 doors in a row  aha:!
problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
question: what state are the doors in after the last pass? which are open which are closed?
solution: 100 doors in a row
problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
for example, after the first pass every door is open. on the second pass you only visit the even doors (2,4,6,8...) so now the even doors are closed and the odd ones are opened. the third time through you will close door 3 (opened from the first pass), open door 6 (closed from the second pass), etc..
question: what state are the doors in after the last pass? which are open which are closed?
solution: you can figure out that for any given door, say door #42, you will visit it for every divisor it has. so 42 has 1 & 42, 2 & 21, 3 & 14, 6 & 7. so on pass 1 i will open the door, pass 2 i will close it, pass 3 open, pass 6 close, pass 7 open, pass 14 close, pass 21 open, pass 42 close. for every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9... leaving it open at the end. only perfect square doors will be open at the end
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