A Conversation for The Myth Of 42 [(5-3+0+5) * (6+0) = 42]
1090000 solved!!!!!
mathsmoose Started conversation Mar 11, 2004
The answer involves the prime factorial function, #, which is like ! but only primes. 10#=7^5^3^2
10# / (9-0!-0!-0!-0!) = 7^5^3^2/5 = 42
1090000 solved!!!!!
Jeremy (trying to find his way back to dinner) Posted Mar 11, 2004
Why not
-1-0!+90/(0!+0!)-0! = 42
(relatively) staight and simple ... 
Jeremy
1090000 solved!!!!!
AK - fancy that! Posted Mar 11, 2004
...because you only add zeros so that it'll be 6 digits, not 7.
...because invtan(1)+0-sqrt(9)+0+0+0=42 works
1090000 solved!!!!!
Argon0 (50 and feeling it - back for a bit) Posted Jan 27, 2005
Who is user 1090000?
We only add 0s in FRONT of the number....
Can anyone point to a definition of Prime Factorial... The closest I could find was a conjecture to do with Factorials of Primes...
Not saying that a non-prime number can have a 
factorial that is made up of the Prime Numbers less than it...
Oh, and surely 7^5^3^2^1 would be 10# by your definition... = 22539340290692258087863249....
Or did you mean to type 7*5*3*2*1 = 210
1090000 solved!!!!!
Argon0 (50 and feeling it - back for a bit) Posted Jan 27, 2005
Which does work your way....
1090000 solved!!!!!
Argon0 (50 and feeling it - back for a bit) Posted Jan 28, 2005
Huh...
If there was a usernumber 1090000 that is...
Key: Complain about this post
1090000 solved!!!!!
- 1: mathsmoose (Mar 11, 2004)
- 2: Jeremy (trying to find his way back to dinner) (Mar 11, 2004)
- 3: AK - fancy that! (Mar 11, 2004)
- 4: Argon0 (50 and feeling it - back for a bit) (Jan 27, 2005)
- 5: Argon0 (50 and feeling it - back for a bit) (Jan 27, 2005)
- 6: AK - fancy that! (Jan 28, 2005)
- 7: Argon0 (50 and feeling it - back for a bit) (Jan 28, 2005)
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