3 ms·
a more interesting challenge would be the biggest number using each of the ten digits only once. ready, go.
by rookie 16y ago
a more interesting challenge would be the biggest number using each of the ten digits only once.
ready, go.
- sesqu 16y ago9876543210 If I'm also allowed an operator, 3↑⁹⁸⁷⁶⁵⁴¹⁰2
- rookie 16y agoha, yes allowed operators, as many as youd like.
- sesqu 16y agoThen the best I can come up with on no notice is 2 ↑^(4 ↑^(6 ↑^(8↑^(9↑10)) 7) 5) 3. That would be trivial to increase substantially by using some other notation for the hyperoperation, but I'm going with the presumably most familiar one. I'm sure there are bigger numbers - specifically, Wikipedia mentions Conway chains, where a→b→c = a ↑^c b, and so 2→3→4→5→6→7→8→9→10 should be quite large indeed.
- baddox 16y agoExcellent! I hadn't thought about it, but at a glance I'm pretty sure your ordering and grouping does generate the maximal result. However, the parent comment said as much punctuation as you like, which allows for unbounded expressions with unary operators like the factorial!
- deleted 16y ago[deleted]
- pmb 16y ago9876543210!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
- lakeeffect 16y ago10!^2!^3!^4!^5!^6!^7!^8!^9!
- Retric 16y agoHA! 2^3!^4!^5!^6!^7!^8!^9!^10! ;-) (2! = 2) and (2^x) > (x ^2) where x > 3. Then again 2^3^4^5^6^7^8^9^(10!!!!!!!!!) is larger. Also if it's it's simply a question of symbols 9!! is larger than 9^9 so 9 followed by n ! is probably your best bet without going into higher level math like BB(BB(...(9))...)
- kenjackson 16y agoOr what if you restricted the number of symbols to be no more than 20 symbols total.
- die_sekte 16y ago10→9→8→7→6→5→4→3→2 Somehow I think that Conway's chained arrow notation is cheating.