Numbers from 1,2,3,4 to 50

As an occasional diversion from work, I like to try maths puzzles, and to this end, I recently purchased a couple of high school maths textbooks (for use with students aged from about 14 to 16).  Not that I can't do the puzzles (I think my maths is okay), but because sometimes the questions and puzzles in them some interesting ideas about extension activities - for example, the recent question about the circle in the corner of a circle and a square is probably intended to be solved with trigonometry, and instead, I solved it in terms of square roots, using just Pythagoras' theorem.  That led to me solving the situation for the circle in the corner of a hexagon (which wasn't in the textbook, but which had an interesting solution too).

Anyway, I found a great question in the extension section at the end of one of the textbooks, and it goes like this:

Creating Numbers: a task requiring imagination

Your task is to create every number from 1 to 50.
You can use only the digits 1, 2, 3 and 4 once in each and the operations + - * / . 
You can use the digits as powers, and you must use all of the digits 1, 2, 3, 4

Here are some examples:

1 = (4-3) / (2-1)
20 = 4² +3 +1
68 = 34 * 2 * 1
75 = (4+1)² * 3


So, here goes... 1 to 50, using only 1, 2, 3 and 4 and the basic maths operators.  Some of the answers seem a little repetitive or derivative (look at 38 through 43), and in some cases I found alternative answers afterwards.


1 = (4-3) * (2-1)
2 = (4-3) + (2-1)

3 = (4+2) / (3-1)  
4 = (4 * 3) / (2+1)  
5 = (2 * 4) – (1 * 3)  
6 = (2 * 4) – (3-1)
7 = (3+4) * (2-1)
8 = (3+4) + (2-1)  
9 = (3+4) + (2 * 1)  
10 = 1 + 2 + 3 + 4  
11 = (4 * 3) – (2 - 1)  
12 = (4 * 3) * (2-1)  
13 = (4 * 3) + (2-1)  
14 = (4 * 3) + (2 * 1)  
15 = (4 * 3) + 2 + 1
16= 4 ^ ((3+1)/2)  
17 = 3(4+1) + 2  
18 = 4² + (3-1)  
19 = 4² + (3 * 1)  
20 = 4² + 3 + 1  
21 = (4+3) * (2+1)  
22 = (4+1)² – 3  
23 = 3² + 14  
24 = 1 * 2 * 3 * 4  
25 = 31 – (2+4)  
26 = 13 * (4-2)  
27 = 3² * (4-1)  
28 = 32 – (4 * 1)  
29 = 31 – (4 -2)  
30 = (4+1) * 3 * 2  
31 = 34 – (1+2)  
32 = 4^(3-1) * 2  
33 = 34 – (2 -1)
34 = 34 * (2-1)  
35 = 34 + (2-1)  
36 = (4 * 3) * (2+1)  
37 = 34 + 1 + 2  
38 = 42 – (3+1)  
39 = 42 – (3 * 1)
40 = 41 – (3-2)  
41 = 43 – (1 * 2)  
42 = 43 – (2-1)  
43 = 41 + (3-1)  
44 = (14 * 3) + 2  
45 = 43 + (2 * 1)  
46 = 42 + 1 + 3  
47 = 41 + (3 * 2)  
48 = 24 * (3-1)  
49 = ((4 * 1) + 3)²  
50 = 41 + 3²

As the logical extension, I attempted to carry on past 50.  It becomes increasingly difficult, since 1, 2, 3 and 4 are all small numbers, and the combinations of those small digits become less useful in making specific larger values (especially the prime numbers). 

However, if we expand the rules to allow ! (factorial) and decimal points, then this enables us to find solutions for 57 (for example).  I'd like to thank the free math help forum community (especially Denis for his initial suggestion to extend the rules), for the additional solutions, comments, corrections and suggestions.  They've been very friendly in quickly adopting my idea and sharing their comments and solutions.  An additional rule that's been introduced is the use of decimals - by doing this, we can include dividing by .2 (for example) as a way of multiplying by 5.

51 = (12 * 4) + 3  OR (4²+1) x 3
52 = 4³ - 12
53 = (1 + 4!) * 2 + 3
54 = (13 * 4) +2
55 = 34+ 21
56 = (1 + 3 + 4!)2
57 =(1+4)! / 2 - 3  OR (4 + 2) / .1 - 3
58 = (31 * 2) - 4
59 = (21 * 3) - 4
60 = 3⁴ - 21
61 = 4³ - (1 + 2)
62 = 4³ - (1 * 2)
63 = 4³ - (2 - 1)
64 = (2 - 1) * 4³
65 = (2 - 1) + 4³
66 = (2 * 1) +4³
67 = (34 * 2) -1
68 = 34 * 2 * 1
69 = (34 * 2) + 1
70 = 4³ + (1 + 2)!
71 = ((4! / 2) * 3!) - 1
72 = 24 * 3 * 1
73 = (3 * 4!) + (2 - 1)
74 = (3 * 4!) + (2 * 1)
75 = (4+1)² * 3
76 = (41 * 2) - 3!
77 = ((4! + 1) * 3) + 2
78 = (4! + 2) * 3 * 1
79 = 3⁴ - (2 * 1)
80 = 3⁴ - (2 -1)
81 = (4! + 1 + 2) * 3
82 = 3⁴ + (2 - 1)
83 = 3⁴ + (2 * 1)
84 = 3⁴ + 2 + 1
85 = (43 * 2) - 1
86 = 43 * 2 * 1
87 = (21 * 4) + 3
88 = (43 + 1) * 2
89 = (3!)! / (2 * 4) - 1
90 = (1! + 2!) * (3! + 4!)
91 = (23 * 4) - 1
92 = 23 * 4 * 1
93 = (23 * 4) + 1
94 = (1 + 3)4! - 2
95 = 3!*(2⁴) - 1
96 = (12 * 4!) /3
97 = 4(3! - 2)! + 1
98 = (1 + 3)4! + 2
99 = 123 - 4!
100 = (3 / .12) * 4


The original textbook question asked only for the numbers from 1 to 50 with only powers and basic operators, but by expanding the rules, many more numbers have become achievable.  Dare we go above 100?

Further posts:  Numbers 101-150 and Numbers 150-200.