There is a long tradition of posing puzzles at consultation for expert positions . Despite my respect for puzzles , I do n’t believe one ’s ability to solve one quickly in a in high spirits - pressure situation provides an accurate prognosis for their future job performance . Google , once known for its notoriously unmanageable interview mystifier , retired the praxis in 2013 , with the senior VP of people operations at the meter posit that they were “ a gross thriftlessness of time . They do n’t predict anything . They function primarily to make the interviewer feel smart . ” This week , your interviewer is Elon Musk . attempt to solve his preferred puzzler to give to locomotive engineer vie for a job at SpaceX. While the puzzle is not skyrocket science , it will require some inventiveness .
Did you miss last week ’s puzzle ? see it outhere , and find its answer at the bottom of today ’s clause . Be careful not to read too far ahead if you have n’t work last workweek ’s yet !
Puzzle #14: Elon Musk’s Interview Question
You are standing somewhere on the surface of the Earth . You take the air 1 mi south , then 1 mile west , and then 1 Swedish mile north , and you end up back at the same point where you started . Where are you ? Assume the Earth is a perfect fluid sphere .
The most natural answer to this question is the North Pole . And it ’s correct ! Walking a mile south from the North Pole , then a statute mile west , and a mile magnetic north again will land you flop back at the top of the planet . According to Musk ’s biographer , Ashlee Vance , when engineers would get this correct , Musk would respond , “ Where else could you be ? ” That problem is significantly tougher , and solving it is your task for the week .
The South Pole does n’t put to work . Traveling south from the South Pole is n’t even a meaningful construct , since moving in the south is by definition moving toward the South Pole . It helps to remember that we ’re on a vainglorious sphere . remember of lines of latitude not actually as horizontal cable but as Mexican valium trace around the globe that get increasingly wide toward the equator and narrow toward the poles .
Image: Photo: Shutterstock/Graphics: Vicky Leta
Good luck work out the puzzle this hebdomad , and perch easy that your career does n’t reckon on it .
I ’ll be back next Monday with the solution and a new puzzle . Do you cognise a cool puzzle that I should report here ? send out it to me at[email protected ]
Solution to Puzzle #13: Mouse Trap
Were you able-bodied to entrap the computer mouse from last week’spuzzle ? you may vouch that you catch the mouse in six days . There are several solutions . Numbering the cabinets in rules of order from 1 to 5 , one solution is to check them in this monastic order : 2 , 3 , 4 , 2 , 3 , 4 . A clever observation is that the mouse must alternate between odd- and even - add up cabinets each day .
hypothesise the mouse begin in an even - add up cabinet . How could you catch it in this display case ? On twenty-four hours one , you open cabinet 2 . If the shiner is in there , you win . Otherwise , the mouse must have been in cabinet 4 . This means that on day two , the computer mouse will either be in cabinet 3 or cabinet 5 ( because these are the only cabinets adjacent to 4 ) . heart-to-heart storage locker 3 on day two . If you find the shiner , you win . Otherwise , it means the black eye was in locker 5 that day . But this means that on solar day three , you know the mouse will be in locker 4 ( the mouse ’s only legal move from cabinet 5 ) . So by look into the cabinets 2 , 3 , 4 , on the first three days , you are guaranteed to get a mouse that lead off in an even - numbered locker .
If these are your first three moves and you do not uncover a black eye , this entail that the computer mouse must have begun in an odd - numbered cabinet . So where is the mouse the morning of day four ? Since it started in an odd - amount cabinet and it alternates between even and odd each day , the computer mouse must be in an even - numbered console on day four . Luckily , we already have a strategy that catches the mouse when it begins in an even - numbered cabinet ! repeat it by guessing cabinets 2 , 3 , and 4 again on day four , five , and six is then guaranteed to catch the shiner .
point out that in the solution , we could launch the disceptation in reversal and start with room access 4 before carry on to 3 and 2 . We can even ruffle the two strategies . For example , the monastic order : 2 , 3 , 4 , 4 , 3 , 2 get the mouse . It turns out that this latter backtracking approach is what is necessitate when you have an even number of cabinet . In fact , this generalizes to any routine of cabinets . This intend that even if the shiner had 100 places to hide , if you simply open 2 , 3 , 4 , up to 99 , and then set out backtracking to 98 , 97 , and back down toward 2 , the mouse has no hope of evading you .
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