20190110, 06:19  #23 
"Ben"
Feb 2007
2^{5}×3×5×7 Posts 
Could't resist one more  no parens are fun!
56 = 8 * 2 + 5 * 8  9/9 + 3/3 
20190110, 06:53  #24 
Jan 2016
Mighty Black Stump
10_{16} Posts 
57 = 8 x 8 + 5 x 2  9  9 + 3/3

20190110, 07:35  #25 
May 2009
Russia, Moscow
2^{5}·79 Posts 
58 = 8 + 8 * 5 + 2 * 9  9 + 3/3

20190110, 07:56  #26  
Jun 2003
12EA_{16} Posts 
Quote:
(9/9*3/3) (9/9+3/3) (99/33) ((9+9/3)/3) (9/933) (99+3+3) (9/9+3+3) (99/3/3) (9*9/3/3) (9+9/3/3) *DROPS MIC* 

20190110, 08:58  #27 
Romulan Interpreter
Jun 2011
Thailand
23B0_{16} Posts 
Haha, now if we use only 8258 to make* 0, 10, 20, 30, +\(\infty\), then we are done...
If possible, only once, no parenthesis... *technically, as you posted 0 to 10 inclusive, we would only need to make 0, 11, 22, 33, 44... Last fiddled with by LaurV on 20190110 at 08:58 
20190110, 09:40  #28 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·2,897 Posts 
Actually they can be used to add or subtract 0 to 10 meaning a range of 21 not 11. Can this list be expanded? What can be done with just the last 3 digits?
59 = (8+2)*5+8 + (9/9*3/3) 60 = (8+2)*5+8 + (9/9+3/3) 61 = (8+2)*5+8 + (99/33) 62 = (8+2)*5+8 + ((9+9/3)/3) 63 = (8+2)*5+8  (9/933) 64 = (8+2)*5+8 + (99+3+3) 65 = (8+2)*5+8 + (9/9+3+3) 66 = (8+2)*5+8 + (99/3/3) 67 = (8+2)*5+8 + (9*9/3/3) 68 = (8+2)*5+8 + (9+9/3/3) The 5 axn posted was actually 5. Fortunately I checked my results. It is so easy to make a mistake with this puzzle!! 
20190110, 11:07  #29 
Jun 2003
2·3^{2}·269 Posts 

20190110, 11:16  #30  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5794_{10} Posts 
Quote:
I might consolidate correct results in the first post at somepoint. I will probably also make a list for the last 4 digits to aid with creating more. Last fiddled with by henryzz on 20190110 at 11:17 

20190110, 13:48  #31 
"Adolf"
Nov 2013
South Africa
61 Posts 
69 = ((8  2 + 5 + 8 + (9/9) + 3)) * 3

20190110, 13:57  #32 
Jun 2003
12EA_{16} Posts 

20190110, 14:03  #33 
Jun 2003
2·3^{2}·269 Posts 

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