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2019-01-10, 14:04   #34
LaurV
Romulan Interpreter

Jun 2011
Thailand

24·571 Posts

Quote:
 Originally Posted by henryzz What can be done with just the last 3 digits?
4 digits, not 3

An 11 can be made as 99/(3*3) and a 12 is easy

Last fiddled with by LaurV on 2019-01-10 at 14:05

2019-01-10, 15:30   #35
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

2×2,897 Posts

Quote:
 Originally Posted by axn I wasn't sure about the legality of unary - operator. Anywho, I had already dropped the mic. You can't go back and pick it up again
As far as I can see the aim of the list was to make it easy to append these onto the first 4 digits.

Eventually we will need more digits in order to generate a close enough number to add them to. That is why I suggested also looking at the last 3 digits.

Now I should be able to base a load more on 78 = (8-2)*(5+8)

70 = (8-2)*(5+8)-(9-9/3/3)
71 = (8-2)*(5+8)-(9/9+3+3)
72 = (8-2)*(5+8)-(9-9+3+3)
73 = (8-2)*(5+8)+(9/9-3-3)
74 = (8-2)*(5+8)-((9+9/3)/3)
75 = (8-2)*(5+8)-(9-9/3-3)
76 = (8-2)*(5+8)-(9/9+3/3)
77 = (8-2)*(5+8)-(9/9*3/3)
78 = (8-2)*(5+8)-(9-9+3-3)
79 = (8-2)*(5+8)+(9/9*3/3)
80 = (8-2)*(5+8)+(9/9+3/3)
81 = (8-2)*(5+8)+(9-9/3-3)
82 = (8-2)*(5+8)+((9+9/3)/3)
83 = (8-2)*(5+8)-(9/9-3-3)
84 = (8-2)*(5+8)+(9-9+3+3)
85 = (8-2)*(5+8)+(9/9+3+3)
86 = (8-2)*(5+8)+(9-9/3/3)
87 = (8-2)*(5+8)+(9*9/3/3)
88 = (8-2)*(5+8)+(9+9/3/3)

Working out the 78 was the hard point above. Another load can be done based on 88 but I will leave that to someone else.

@LaurV I don't think concatenation is in the list of operators(yet)

2019-01-10, 15:39   #36
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

9,257 Posts

Quote:
 Originally Posted by unconnected 58 = 8 + 8 * 5 + 2 * 9 - 9 + 3/3
Quote:
 Originally Posted by henryzz to use the numbers from the new exponent 82,589,933.
Hmmm....

 2019-01-10, 16:21 #37 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 2·2,897 Posts 57 has the same issue. 57 = (8+2)*5+8-(9/9*3/3) 58 = (8+2)*5+8+(9-9+3-3) Last fiddled with by henryzz on 2019-01-10 at 16:22
2019-01-10, 16:49   #38
bsquared

"Ben"
Feb 2007

25×3×5×7 Posts

Quote:
 Originally Posted by henryzz Working out the 78 was the hard point above. Another load can be done based on 88 but I will leave that to someone else.
Getting to 88 is easy. (At least) three four ways to do it:
88 = 8 + 2 * 5 * 8
88 = 8 * 2 * 5 + 8
88 = 8 * (-2 + 5 + 8)
88 = (8 - 2 + 5) * 8

The next increment above that seems trickier.

Last fiddled with by bsquared on 2019-01-10 at 16:51

2019-01-10, 17:15   #39
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

2·2,897 Posts

Quote:
 Originally Posted by bsquared Getting to 88 is easy. (At least) three four ways to do it: 88 = 8 + 2 * 5 * 8 88 = 8 * 2 * 5 + 8 88 = 8 * (-2 + 5 + 8) 88 = (8 - 2 + 5) * 8 The next increment above that seems trickier.
I think the next step might beyond 88 might be to rely on 81=(9*3*3) at the end leaving 5 digits to fiddle with.

 2019-01-10, 18:02 #40 bsquared     "Ben" Feb 2007 D2016 Posts 1 = 8/2+5-8 2 = 8/(2^5/8) 3 = 8*2-5-8 4 = 8/(2*5-8) 5 = -8-2+(5!)/8 6 = 8-(2*5)+8 7 = 8/2-5+8 8 = 8*(-2-5+8) 9 = 8-2-5+8 10 = 8+(2*5)-8 Ok, I "cheated" on two of them.
2019-01-10, 18:51   #41
xilman
Bamboozled!

"πΊππ·π·π­"
May 2003
Down not across

290016 Posts

Quote:
 Originally Posted by LaurV 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...
+$$\infty$$ = (2+5)/(8-8)

0 = (8-8)/(2+5)

Yes, I know, I used parentheses.

10 = 2*5 - 8 +8

2019-01-10, 20:01   #42
uau

Jan 2017

2·43 Posts

Quote:
 Originally Posted by henryzz Initially it would be nice to keep it simple just using the operators +, -, * and / along with brackets to see how far we can get. Other operators will be added later once they are needed(maybe a scoring system could be worked out). All digits must be used. Bonus points(I suspect it will be too hard very quickly) for using the digits in the correct order.
Using just those operators (including unary minus) and digits in original order, 1543 is the first impossible integer. If you allow concatenation, then 1543 can be expressed as 8*25*8-(9+9)*3-3. In this case the first impossible integer is 3422.

 2019-01-10, 20:20 #43 bsquared     "Ben" Feb 2007 25×3×5×7 Posts Do you have a program that exhaustively searches for these or did you do some sort of proof?
 2019-01-11, 09:26 #44 houding     "Adolf" Nov 2013 South Africa 758 Posts 89 =(8-2-5)*8+(9*9)*(3/3)

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