November 18th, 2018, 02:03 PM  #11 
Global Moderator Joined: Dec 2006 Posts: 20,310 Thanks: 1980 
27 = (1 + 4)! * 9/2 31 = 1 * 4! + 9  2 45 = 1/.4 * 9 * 2 = 1^4 * 9/.2 57 = 1 * 4! + 9² 58 = 1  4! + 9² 69 = 1 + 4!*√9  2 70 = 1 * 4!*√9  2 79 = 1*√4 + 9² 80 = (1^4) + 9² = (1 + 4!  9)/.2 83 = 1 * √4 + 9² 
November 18th, 2018, 02:04 PM  #12 
Senior Member Joined: Aug 2012 Posts: 2,157 Thanks: 631  
November 18th, 2018, 04:06 PM  #13 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,986 Thanks: 995 
Skip, you missed 68; no charge: 1 + 4! + 9/.2 = 68 
November 18th, 2018, 04:19 PM  #14 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,986 Thanks: 995  
November 18th, 2018, 05:11 PM  #15 
Senior Member Joined: Aug 2012 Posts: 2,157 Thanks: 631  
November 18th, 2018, 05:38 PM  #16 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,986 Thanks: 995 
...so you meant "100!" ; but that's only a 158 digits number 
November 19th, 2018, 12:25 AM  #17 
Global Moderator Joined: Dec 2006 Posts: 20,310 Thanks: 1980  
November 19th, 2018, 01:07 PM  #18 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,986 Thanks: 995 
The 90's! 1 * √4 * 9 / .2 = 90 1 + √4 * 9 / .2 = 91 1 + T(4) +9^2 = 92 1 + √4 * T(9) + 2 = 93 1 + √4 * T(9) + T(2) = 94 1  4! + (√9 + 2)! = 95 1 * 4! + (√9 + 2)! = 96 1  4! + (√9 + 2)! = 97 1dozen + 4 + 9dozen  2 = 98 1 + T(4) + T(9) * 2 = 99 
November 19th, 2018, 03:22 PM  #19 
Newbie Joined: Nov 2018 From: Mississauga Posts: 5 Thanks: 0 
Thank you guys! But you missed one other number........ 55.. if you could find anything, that would be nice. There is also another way to make 98 Denis, which does not require dozen.... 1*49*2 and 94: 1* squrt4+92
Last edited by Spritofmaths; November 19th, 2018 at 03:25 PM. Reason: I forgot something 
November 19th, 2018, 04:39 PM  #20  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,986 Thanks: 995  Quote:
T(1+4+92) = 55 Last edited by skipjack; November 19th, 2018 at 07:17 PM.  

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