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 Zero to 150 in 2003 (Posted on 2002-11-29)
Take the digits 2, 0, 0 and 3. Make equations equating to all the integers from 1 to 150 using these digits according to the following rules:-

a) The above digits are the only digits to be used and no other digits should appear anywhere in the equation (except on the side where the answer will be).

b) Use of any mathematical symbols are allowed.

c) The digits 2, 0, 0 and 3 should appear in the given order in the equation. e.g - 0 + 2 + 3 + 0 = 5 is not acceptable.

d) When using the mathematical symbols try using the most simplest forms as much as possible.

 Submitted by Raveen Rating: 4.0526 (19 votes) Solution: (Hide) Brian Smith came up with a solution using the basic arithmetic operations (+, -, ÷, ×), exponentiation (a^b), factorial (x!), square root (√y), trigonometric functions (sin, cos, tan), and rounding (floor, ceil). Here is the list, which is also featured on a webiste he created where you will be able to find any updates: http://www.geocities.com/brianscsmith/zeroto150in2003.html. Most recently, DJ suggested some improved methods, which are shown on that site and reflected in this list. Another more general solution was found by Federico Kereki. Notes: 0!=1. floor x is x rounded down the the nearest integer; ceil x is x rounded up to the nearest integer. Trigonometric functions are calculated in radians. Digits may be concatenated to form other numbers (20, 003..). The listed methods use the fewest operations found to obtain each number. Grouping () is not counted as an operation. ```0= 2×0×0×3 1= 20^(0×3) 2= 2+00×3 3= 20×0+3 4= 20^0+3 5= 2+00+3 6= (2+00)×3 7= 20^0+3! 8= 2+003! 9= (2+00!)×3 10= 20÷(-(0!)+3) 11= (2+0!)!-0!+3! 12= (2+0!+0!)×3 13= (2+0!)!+0!+3! 14= 2×(00!+3!) 15= ((2+0!)!-0!)×3 16= 2^(00!+3) 17= 20-03 18= 20+0!-3 19= 20-(0×3)! 20= 20+0×3 21= 20+(0×3)! 22= 20-0!+3 23= 20+03 24= 20+0!+3 25= 20-0!+3! 26= 20+03! 27= 20+0!+3! 28= (floor √20)+(0!+3)! 29= (ceil √20)+(0!+3)! 30= (2+0!)!+(0!+3)! 31= ceil [-2×(tan (0!+0!))÷sin 3] 32= 2^(0!+0!+3) 33= (2+0!)×floor [(tan (0!))÷(sin 3)] 34= 2+0!+ceil [(tan tan (0!))÷(√3)] 35= -2+floor [(tan tan (0!))÷(-(0!)+3)] 36= (2+0!)!×(0+3)! 37= ceil [cos 2+(tan tan (0!))÷(-(0!)+3)] 38= floor [√2+(tan tan (0!))÷(-(0!)+3)] 39= 2+floor [(tan tan (0!))÷(-(0!)+3)] 40= 20×(-(0!)+3) 41= -2+0+floor [(tan tan (0!))÷(√3)] 42= ((2+0!)!+0!)×3! 43= 2×0+floor [(tan tan (0!))÷(√3)] 44= 20+(0!+3)! 45= 2+floor [(tan tan (00!))÷(√3)] 46= 2×(-(0!)+(0!+3)!) 47= 2+ceil [(tan tan (0!))÷(tan (0!))]-3 48= 2×(00!+3)! 49= floor [2×(tan tan (0!))÷03] . . . . ```

 Subject Author Date Solution Justin 2005-05-15 03:35:46 Solution or wrong interpratation of problem? pavlos 2004-05-14 18:39:24 re: Easy solution e.g. 2004-01-28 15:09:09 Easy solution Federico Kereki 2003-11-19 15:26:05 re(5): did it - aka spoiler abc 2003-09-09 22:04:55 re(4): did it - aka spoiler abc 2003-09-09 21:57:59 re: using functions?? -- here is an easy way... gregada 2003-08-07 15:29:13 using functions?? gregada 2003-08-07 15:12:02 re: DJ 2003-07-12 06:10:19 No Subject Peter 2003-06-23 23:26:52 re(2): ehm.. DJ 2003-04-30 19:00:14 re: ehm.. Brian Smith 2003-04-25 04:48:01 ehm.. DJ 2003-04-24 20:04:33 re: Using trig functions: Zero to 150 completed Cory Taylor 2003-04-24 06:28:20 Using trig functions: Zero to 150 completed Brian Smith 2003-04-24 06:15:39 re: function f(x) = x+1 Cory Taylor 2003-04-15 09:36:30 function f(x) = x+1 Brian Smith 2003-04-15 04:22:56 0! DJ 2003-03-05 05:48:42 re(2): Zero to 150 in 2003 Nigel Nisbet 2003-02-16 06:42:51 re: Zero to 150 in 2003 Nigel Nisbet 2003-02-15 14:41:22 re(5): For starters... joshua imperato 2003-01-23 19:30:52 re(4): For starters... joshua imperato 2003-01-23 19:26:17 re: Zero to 150 in 2003 Cory Taylor 2003-01-10 04:05:43 Zero to 150 in 2003 Robin Gatter 2003-01-09 23:05:56 re: solution coming??? Cory Taylor 2003-01-09 12:25:18 solution coming??? ben hagen 2003-01-09 03:32:18 re: The golden killer! levik 2002-12-21 15:01:30 The golden killer! Dulanjana 2002-12-06 15:42:35 No Subject Dulanjana 2002-12-05 15:24:26 about combinatorials (re: interesting 35) TomM 2002-12-05 13:04:02 interesting 35 pleasance 2002-12-05 06:10:31 re(3): floor? levik 2002-12-05 04:49:05 Some answers that came to mind.. Dulanjana 2002-12-05 04:34:21 re(2): floor? Dulanjana 2002-12-05 04:08:50 re: floor? pleasance 2002-12-05 03:03:05 30 Happy 2002-12-03 11:49:44 re(2): floor? friedlinguini 2002-12-03 11:03:24 re: floor? levik 2002-12-03 08:44:00 floor? Happy 2002-12-03 07:21:31 re(4): solution levik 2002-12-02 10:49:57 re(3): solution Rebecca 2002-12-02 10:30:30 aaaaaaaaaaaaaaaargh HARD Matthew Bobbins 2002-12-02 07:37:03 re(2): solution Cory Taylor 2002-12-02 05:47:21 re: solution friedlinguini 2002-12-02 04:44:52 solution Cory Taylor 2002-12-02 04:40:04 re(3): cheating TomM 2002-12-01 23:55:23 re(2): cheating Cheradenine 2002-12-01 22:48:58 re(3): Some more levik 2002-12-01 18:52:43 re: For starters... James 2002-12-01 17:02:32 re(2): Some more Dulanjana 2002-11-30 16:16:30 re: Some more levik 2002-11-30 04:40:44 Some more Dulanjana 2002-11-29 21:37:27 re(4): did it - aka spoiler levik 2002-11-29 14:11:02 re(3): did it - aka spoiler Cory Taylor 2002-11-29 08:48:14 re(2): did it - aka spoiler Nick Reed 2002-11-29 08:31:02 re: did it - aka spoiler Cory Taylor 2002-11-29 08:08:31 did it Cory Taylor 2002-11-29 08:06:51 re: cheating levik 2002-11-29 08:03:32 cheating Cheradenine 2002-11-29 07:49:16 re(3): For starters... Nick Reed 2002-11-29 07:29:03 re(2): For starters... Nick Reed 2002-11-29 07:07:43 re: For starters... Cory Taylor 2002-11-29 07:01:08 How about... Raveen 2002-11-29 05:13:49 For starters... levik 2002-11-29 04:36:45 Out of turn levik 2002-11-29 04:28:17

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