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Zero to 150 in 2003 (Posted on 2002-11-29) Difficulty: 4 of 5
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]

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Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionJustin2005-05-15 03:35:46
SolutionSolution or wrong interpratation of problem?pavlos2004-05-14 18:39:24
Solutionre: Easy solutione.g.2004-01-28 15:09:09
SolutionEasy solutionFederico Kereki2003-11-19 15:26:05
re(5): did it - aka spoilerabc2003-09-09 22:04:55
re(4): did it - aka spoilerabc2003-09-09 21:57:59
re: using functions?? -- here is an easy way...gregada2003-08-07 15:29:13
using functions??gregada2003-08-07 15:12:02
re:DJ2003-07-12 06:10:19
Some ThoughtsNo SubjectPeter2003-06-23 23:26:52
re(2): ehm..DJ2003-04-30 19:00:14
re: ehm..Brian Smith2003-04-25 04:48:01
ehm..DJ2003-04-24 20:04:33
re: Using trig functions: Zero to 150 completedCory Taylor2003-04-24 06:28:20
SolutionUsing trig functions: Zero to 150 completedBrian Smith2003-04-24 06:15:39
re: function f(x) = x+1Cory Taylor2003-04-15 09:36:30
function f(x) = x+1Brian Smith2003-04-15 04:22:56
0!DJ2003-03-05 05:48:42
re(2): Zero to 150 in 2003Nigel Nisbet2003-02-16 06:42:51
re: Zero to 150 in 2003Nigel Nisbet2003-02-15 14:41:22
re(5): For starters...joshua imperato2003-01-23 19:30:52
re(4): For starters...joshua imperato2003-01-23 19:26:17
re: Zero to 150 in 2003Cory Taylor2003-01-10 04:05:43
Zero to 150 in 2003Robin Gatter2003-01-09 23:05:56
re: solution coming???Cory Taylor2003-01-09 12:25:18
solution coming???ben hagen2003-01-09 03:32:18
re: The golden killer!levik2002-12-21 15:01:30
The golden killer!Dulanjana2002-12-06 15:42:35
No SubjectDulanjana2002-12-05 15:24:26
about combinatorials (re: interesting 35)TomM2002-12-05 13:04:02
interesting 35pleasance2002-12-05 06:10:31
re(3): floor?levik2002-12-05 04:49:05
SolutionSome answers that came to mind..Dulanjana2002-12-05 04:34:21
re(2): floor?Dulanjana2002-12-05 04:08:50
Some Thoughtsre: floor?pleasance2002-12-05 03:03:05
30Happy2002-12-03 11:49:44
re(2): floor?friedlinguini2002-12-03 11:03:24
re: floor?levik2002-12-03 08:44:00
floor?Happy2002-12-03 07:21:31
re(4): solutionlevik2002-12-02 10:49:57
re(3): solutionRebecca2002-12-02 10:30:30
aaaaaaaaaaaaaaaargh HARDMatthew Bobbins2002-12-02 07:37:03
re(2): solutionCory Taylor2002-12-02 05:47:21
re: solutionfriedlinguini2002-12-02 04:44:52
solutionCory Taylor2002-12-02 04:40:04
re(3): cheatingTomM2002-12-01 23:55:23
re(2): cheatingCheradenine2002-12-01 22:48:58
re(3): Some morelevik2002-12-01 18:52:43
re: For starters...James2002-12-01 17:02:32
re(2): Some moreDulanjana2002-11-30 16:16:30
re: Some morelevik2002-11-30 04:40:44
Some moreDulanjana2002-11-29 21:37:27
re(4): did it - aka spoilerlevik2002-11-29 14:11:02
re(3): did it - aka spoilerCory Taylor2002-11-29 08:48:14
re(2): did it - aka spoilerNick Reed2002-11-29 08:31:02
re: did it - aka spoilerCory Taylor2002-11-29 08:08:31
did itCory Taylor2002-11-29 08:06:51
re: cheatinglevik2002-11-29 08:03:32
cheatingCheradenine2002-11-29 07:49:16
re(3): For starters...Nick Reed2002-11-29 07:29:03
re(2): For starters...Nick Reed2002-11-29 07:07:43
re: For starters...Cory Taylor2002-11-29 07:01:08
Some ThoughtsHow about...Raveen2002-11-29 05:13:49
For starters...levik2002-11-29 04:36:45
Out of turnlevik2002-11-29 04:28:17
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