You have numbers: [2,0,1,5] and basic math operators (+,-,x,/,^,!,√), try create numbers from 1 to 42.

First five:

1 = 0/(2 + 5) + 12 = 5 - (0 + 1 + 2)3 = (1 + 5) / (2 + 0)4 = (1 + 5) - (2 + 0)5 = 5 + 0 * (1+2)

Bad formatting

1 = 0/(2 + 5) + 1

2 = 5 - (0 + 1 + 2)

3 = (1 + 5) / (2 + 0)

4 = (1 + 5) - (2 + 0)

5 = 5 + 0 * (1+2)

6=2*0+1+5

7=2+0*1+5

8=2+0+1+5

9=5*2-1+0

10=5*2*1+0

11 = 5*2+1+0

12 = 15-2-0!

(Are we allowed parentheses?)

The last is not using numbers but cyphers I think it is wrong

12= 5*2+1+0!

Don't play twice!

I suppose 12 = 15-2-0! is correct. If not, there are solutions only for 1-27.

13=((5+1)*2)+0!

13=10+5-214=20-5-115=21-5-0!16=20-5+1

13=10+5-2

14=20-5-1

15=21-5-0!

16=20-5+1

Sorry Martyn, was writing while you posted.

17=2^(5-1)+0!

18=(5+1)*(2+0!)

19=20-1⁵²⁰⁼²(0!+1)*5

• 
  Carroll at Just Now

  19=20-1^5

  20=2^(0!+1)*5

21=20+1^5

22=21+(0!)^5

23=25-0!-1

24=5^2-1+0

• 25=50/2*1

• 26=52/(0!+1)

• 27=25+0!+1

• 28=(15-0!)*2

• 29=(2+0!+1)!+5

• 30=5!/(2+1+0!)

no idea what happened there… 29=(2+1+0!)!+5

30=5!/(2+1+0!)

Clearly the solutions of pugency do not satisfy the posed problem. 50 is a number put is not made from the numbers 5 and 0, but from there symbols, which is a big difference.

I will restart at

14 = (5+1+0!)*2

and propose to play only one number at a time.

(17,18,20,24,29,30, seem to be solved as well)

15=2^(5-1)-0!

I finally have got a solution for 19 but can't post it because of double post forbidden rule…

That's kinda silly since there won't be a solution for 28 and probably other numbers but oh well…

16 = 2^(5-1+0)

In http://en.wikipedia.org/wiki/Four_fours is number created from symbols accepted.

I suggest we dont create 1-27 by joining the symbols since these are possible by applying only the usual math operations (and I agree this is more cool since it's trickier), but allow it for 28-42. Would that be okay, Hjallti?

This is like a game of Equations. Richard specifically said, the numbers 0 1 2 5 not the digits.

31 = (2^{5) -1+0 **** 32 = (1x(2}5))+0   **** 33 = (2^5)+1+0

OK. You have numbers: [2,0,1,5] and basic math operators [ + - x / ^ ! √  ] try create numbers from 1 to 42. Except numbers 28 a 38.

For number 28 and 38 are needed joining symbols:

28 = 2 x (15 - 0!)

38 = 50 - 12

Or use special math operators (Gama, Decimal mark):

28 = Gama(5) + 0! + 1 + 2

38 = (5-0!)/.1 - 2

Is 19 = 2 + 0! + 1! + 5!! all right, using double factorial (n*(n-2)*..1) as stated in http://mathforum.org/yeargameWorksheets/2015/2015.manipulative.html ?

You could also get 28 this way using triple factorial:28 = (2 + 5)!!! - 0! +1

If the puzzle is supposed to be with digits, rather than numbers you may play it that way. I just didn't really consider it the task to do it with digits, as it stated numbers. I didn't think about sovability myself. If I see a puzzle like this I just think that it would be solvable.

The use of for me 3!!! = 720! rather than the explanation here.

But if you allow it like carrol than 38 is of course solvabel as it the same as

38=19!!!!!!!!!!!!!!!! (with 17 “!“)

and thus 38= (2 + 0! + 1! + 5!! )!!!!!!!!!!!!!!!!!!

38= ( 2 + 0! + 1! + 5!!  ) !!!!!!!!!!!!!!!!!

Nice!

21 = (2 + 0!)!/1 + 5!!

21= (2 + 0!)!/1 + 5!!

21 = (2 + 0!)!÷1 + 5!!

21 = (2 + 0!)! ÷1 + 5!!

34=2^5+1+0!

35=(5+1)^2-0!

36=(5+1)^2+0

37=(5+1)^2+0!

22 = (2 + 0!)! + 1 + 5!!

The rest is easy, but 42 eludes me without the multi-factorial trick.

42 = 2^5 + 10

39..41 = 5! ÷(2+1) ± 0!

12 = √( ( (2+1)! )! / 5) + 0

Using permutations

7P2 = 7!/(7-2)! = 42

(5 + 1 + 0!)!P2 = 42

numbers can exist in different basesbase six (the lowest due to the 5)42=5^2+0+1