# Solution 90: Logic Problem on Types of Cars

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**Published on May 4, 2019**- Let's carefully reason our way through the jungle of subtle arguments to identify two non-fully defective cars.

Congratulations to Juju Mas, Jonathan Liu, driudwaker, Emmett Nelson, Bigg Barbarian, Richard Chen, Mokou Fujiwara, and Serengeti Ghasa for successfully solving this week's math challenge question! Juju Mas was the first person to solve the problem.

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Aakash KumarMonth ago^{+1}hey bro can you help me with a problem

how many functions f:N->N satisfy

lcm{ f(n),n } - hcf{ f(n),n }

steamroller82Month ago^{+2}Brilliant. Easy to follow. Thanks for posting!

Icestrike411 OfficialMonth agoez, just sell the carlos a random car and make him do it.

Jeremy WeissmannMonth ago^{+2}Did you by any chance see my more general solution which minimizes case analysis? Interested to see what you think.

Jeremy WeissmannMonth agoThanks as always for the thought provoking questions!

LetsSolveMathProblemsMonth ago^{+1}I think your generalization is correct and very insightful. Also, my brother explained to me that he specifically chose 4 to be the number of perfect cars because he thought easier solutions may be possible with 3 perfect cars, as you suggested.

Elan RothMonth ago^{+1}I found an even more simple solution: Test Cars 1 + 2, 3 + 4, and 5 + 6. This method is foolproof and there is no case in which you cannot find two non-fully defective cars.

Elan RothMonth ago@Jeremy Weissmann You make a great point, thank you. I guess this does not hold for all possibilities.

LetsSolveMathProblemsMonth ago@Jeremy Weissmann You are correct; the proposed strategy is not viable. I suppose many of us in this comment thread momentarily forgot that two perfect cars result in partly defective cars that cannot be used to create a test in which a car explodes. =)

Jeremy WeissmannMonth ago^{+1}@Elan Roth I still don't believe this solution works. Once perfect cars crash into perfect cars, they become partly defective and provide no new information.

Consider three scenarios:

Scenario A

Perfect: 1,2,3,4

Fully defective: 5,6,7

Scenario B

Perfect: 1,2,5,6

Fully defective: 3,4,7

Scenario C

Perfect: 3,4,5,6

Fully defective: 1,2,7

In each scenario:

Test 1&2, nothing happens

Test 3&4, nothing happens

(at this point Cars 1,2,3,4 are non-perfect)

Test 1&3, nothing happens

Which cars are supposed to be viable? We can't give car 1 or 2 because of Scenario C; and we can't give car 3 or 4 because of Scenario B; nor can we give car 5 or 6 because of Scenario A; and none of the Scenarios allow us to give car 7.

So if no test creates a result, no cars can be safely given.

Elan RothMonth agoJeremy Weissmann If there are no breaks after testing 1+2 and 3+4, then simply test 1 + 3 to figure out which pair is perfect Both cannot be fully defective because there are only 3 fully defective cars.

I made this edit a little early ^^^

Jeremy WeissmannMonth agoElan Roth suppose 1,2,3,4 are perfect. In your testing, nothing will happen.

Now suppose 1,2,5,6 are perfect. Also, nothing will happen. The same would be true if 3,4,5,6 were perfect.

So which cars are you supposed to pick? No combination of cars works in every scenario.

Hemanth KotagiriMonth ago^{+1}This sounds really cool.

Where do you get these interesting problems from?

MartinMonth ago^{+1}He makes them himself or some of them (like this one) are made by his brother IIRC.