Solution
Alright... the basis of this puzzle is that only the first door, and doors whose number are even powers of prime numbers, or two even powers of primes multiplied together, will be opened. All other doors will be hit an even number of times, and remain closed. The doors that can open are:
1^1 = 1 (1)
2^2 = 4 (1 2 4)
3^2 = 9 (1 3 9)
2^4 = 16 (1 2 4 8 16)
5^2 = 25 (1 5 25)
4*9 = 36 (1 2 3 4 6 9 12 18 36)
7^2 = 49 (1 7 49)
2^6 = 64 (1 2 4 8 16 32 64)
3^4 = 81 (1 3 9 27 81)
There's also another, being:
Since only seven of these eight doors opened, there were less than 64 doors and more than 48. So far, so good.
If only odd orbs are hit, all the original seven are still open, and newly opened doors would be the ones with an odd number of even divisors. Doors with an even number of divisors will remain unchanged, such as 4 (1, 2, 4), 16 (1, 2, 4, 8, 16), and 64 (1, 2, 4, 8, 16, 32, 64).
The other doors will be obtained by multiplying the odd powers of 2 (2, 8, 32) by the odd numbered doors that opened above. This is the only way to obtain a number with an odd number of even divisors. These numbers will be:
2^1 * 1^1 = 2 (1 2)
2^3 * 1^1 = 8 (1 2 4 8)
2^1 * 3^2 = 18 (1 2 3 6 9 18)
2^5 * 1^1 = 32 (1 2 4 8 16 32)
2^1 * 5^2 = 50 (1 2 5 10 25 50)
2^3 * 3^2 = 72 (1 2 3 4 6 8 9 12 18 24 36 72)
Since only 4 of these doors are opened, there must be less than 50 doors. Therefore, there are 49.
My turn:
This may look like a math puzzle, but it is in fact a pure logic puzzle. Substitute single digits for the letters A, B, C, D and E in the following equation such that you arrive at a valid solution:A B C D E
4
---------
E D C B A
Just to state the obvious, both As must hold the same value, as must both Bs, etc. As well, simply assigning all digits to be 0 is a cop-out.
Edit: I realised I'd made an error in my solution to Rabscuttle's puzzle, and it's now been corrected. The answer is 49.
[Edited by DiMono at
Local Time:11-07-2004 at 06:42 AM]
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