Match the post#

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,437
Using digits 0 to 9 in ascending order and in descending order,
and using "all the tricks of the trade", create equations showing
as result 1000 + post#.

As example, this post# is 1, so the 2 equations must equal 1001.

0 + 1 + 2*(3 + 4*5 + 6*78 + 9) = 1001
987 + 6 - 5 + 4 - 3 + 2 + 10 = 1001

Btw, 0! = 1 is allowed.

I'll do the next one (further example)...then whoever is interested
can do 1003...and so on...

RULE: ALL posts require this to be done ...... OK DAN?!
 
Last edited:

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,437
0*1 + 2 * 3*(4 + 5! + 6*7 - 8 + 9) = 1002
√9 * (8 - 7 + 6*54 + 3^2 + 1 - 0!) = 1002

So whoever is next (if any!) gets to do 1003.
 
Last edited:

topsquark

Junior Member
Joined
Aug 27, 2012
Messages
245
Oh not here, too! 🦖 (stompstompstompstompstomp...)

-Dan
 

ksdhart2

Full Member
Joined
Mar 25, 2016
Messages
962
Alright, so here's my results for 1004:

  • \(\displaystyle 123 + 4(5 \cdot 6 \cdot 7 + 8) + 9 = 1004\)
  • \(\displaystyle 9 + 8(76 + 5 + 43) + 2 + 1 = 1004\)
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,437
KS, you forgot the 0's; go stand in the corner!

0 + 1*2 - 3 + 4^5 + 6 - 7 - 8 - 9 = 1005

9 - 8 - 7 - 6 + 5 - 4*3 + 2^10 = 1005
 

ksdhart2

Full Member
Joined
Mar 25, 2016
Messages
962
Okay, I'll go sit in the corner for 0 minutes.
  • \(\displaystyle 0 + 1 + 23 \cdot 4 \cdot 5 + 67 \cdot 8 + 9 = 1006\)
  • \(\displaystyle 9 \cdot 87 + 6 + 5 \cdot 43 + 2 \cdot 1 + 0 = 1006\)
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,437
0*123 + 4^5 + (6 - 7)*(8 + 9) = 1007

9 - 8 - 7 - 6 - 5*(4 - 3) + 2^10 = 1007
 
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