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### Re: CarTalk Puzzler

by chas (Priest)
 on Nov 17, 2005 at 03:39 UTC ( #509272=note: print w/replies, xml ) Need Help??

I would solve it as follows. Think of each light as initially off. Then the cord is pulled once for each divisor of its number. But a number of form p_1^{r_1}p_2^{r_2}...p_j^{r_j} (where the p_i are distinct primes) has (r_1-1)...(r_j-1) factors; this is easily proved. Now it is clear that the number of factors is odd exactly when the number is a perfect square (i.e. all the r_i are even.) But pulling the switch an odd number of times when it is off turns it on. (The problem states they are all initially on, but think of them as being off and turned on for the divisor 1.)
The trouble with enumerating the result - what if it had been 2,000,000,000,000,000,000,000 instead of 20,000?
chas
Update Sorry, of course the number of factors is (r_1+1)...(r_j+1), not (r_1-1)...(r_j-1). I guess I was brain dead when I first posted. The conclusion is the same, though; this is odd exactly when all the r_i are even.

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Re^2: CarTalk Puzzler
by Perl Mouse (Chaplain) on Nov 17, 2005 at 09:51 UTC
The trouble with enumerating the result - what if it had been 2,000,000,000,000,000,000,000 instead of 20,000?
2,000,000,000,000,000,000,000 == 2 * 1021 == 5 * 4 * 1020 == 5 * (2 * 1010)2.

So, all you need to do is list the squares of all numbers from 1 to sqrt(5) * 2 * 1010. It'll take a while to list, but you can do it with a one-liner.

Perl --((8:>*
I think the OP was saying "what if the original solutions posted in this thread were used for <insert really large number here> bulbs?". The solutions posted near the top of the thread relied on walking the list numerous times, which is infeasible if the list is super large. The elegance in finding the mathematical property for the bulbs being on is that it does scale.

thor

Feel the white light, the light within
Be your own disciple, fan the sparks of will
For all of us waiting, your kingdom will come

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