# Create all numbers from 1-100 using 1,3,3,6

Create all the numbers from $$1$$ to $$100$$ using the numbers $$1$$,$$3$$,$$3$$, and $$6$$.

• You can only use each number once, except for the $$3$$, of which you have two.
• You can use addition ($$x+y$$), subtraction ($$x-y$$), division ($$\frac{x}{y}$$), multiplication ($$x\times y$$), exponentiation ($$x^y$$) and roots ($$\sqrt[\leftroot{-2}\uproot{2}x]{y}$$).
• You can combine numbers like $$1$$ and $$3$$ to $$13$$ etc.
• You must use all numbers. EDIT: no factorials, in squareroots 2 is hidden, no combining results of operations, you can use parentheses and start with negative numbers, no rounding and no decimal points. Good Luck
• Can we combine the results of operations? For example, is $(1+3) | 36 = 436$ (where | indicates concatenation) – user46002 Jan 27 '19 at 22:14
• If we need to take a square root, is the two implied? – user46002 Jan 27 '19 at 22:14
• can we use factorial? – Omega Krypton Jan 27 '19 at 23:23
• Factorials are most likely out, but what about parentheses, unary minus (like starting with -1) and decimal points? – Bass Jan 28 '19 at 0:25
• If decimal is allowed then round would probably valid too? – Mukyuu Jan 28 '19 at 3:31

These get harder with larger numbers, but here are the first 40 (and a couple of the easier ones after that) with the digits in order:

1 to 10

1: $$1 + 3 + 3 - 6$$
2: $$(1 + 3) \times 3 / 6$$
3: $$1*3 * 3 - 6$$
4: $$13 - 3 - 6$$
5: $$-1^{33} +6$$
6: $$1\times3-3+6$$
7: $$1 + 3 -3 +6$$
8: $$1+3/3 + 6$$
9: $$1^3 \times (3+6)$$
10: $$1 + \sqrt[3]3^6$$

11 to 20

11: $$\sqrt{1+3}+3+6$$
12: $$1\times 3 + 3 + 6$$
13: $$1 + 3+3+6$$
14: $$-1 + 3\times 3+6$$
15: $$-1\times3 + 3\times 6$$
16: $$1 - 3 + 3 \times 6$$
17: $$-1^3 +3\times 6$$
18: $$(1+3)*3+6$$
19: $$13 + \sqrt{36}$$
20: $$-1 + 3^3 - 6$$

21 to 30

21: $$1 * 3^3 - 6$$
22: $$13 + 3 + 6$$
23: $$-13+36$$
24: $$(1+3)\times\sqrt{36}$$
25: $$1 - 3 + \sqrt3^6$$
26: $$-1+33-6$$
27: $$1*33-6$$
28: $$1+33-6$$
29: $$-1 + 3 + \sqrt3^6$$
30: $$(-1+3+3)\times 6$$

31 to 40

31: $$13+3*6$$
32: $$-1+3^3+6$$
33: $$13*3-6$$
34: $$1+3^3+6$$
35: $$-1+(3+3)\times6$$
36: $$1\times(3+3)\times 6$$
37: $$1^3+36$$
38: $$\sqrt{1+3}+36$$
39: $$1\times3 + 36$$
40: $$1+33+6$$

41 to 50 (getting much harder now, so from now on, only the easier ones)

41: 
42: $$(1+3+3)\times 6$$
43: 
44: 
45: $$13\times3+6$$
46: 
47: 
48: $$(-1 + 3 \times 3) \times 6$$
49: $$13+36$$
50: 

51 to 60

51: 
52: 
53: $$-1 +3 \times 3 \times 6$$
54: $$1\times 3 \times 3 \times 6$$
55: $$1 + 3 \times 3 \times 6$$
56: 
57: 
58: $$(1+3)^3-6$$
59: 
60: $$(1+3\times3)\times6$$

61 to 70

61: 
62: 
63: 
64: $$(1+3/3)^6$$
65: 
66: 
67: 
68: 
69: 
70: $$(1+3)^3+6$$

71 to 80

71: 
72: $$(1+3)\times 3 \times 6$$
73: 
74: 
75: 
76: 
77: 
78: $$13 \times \sqrt{36}$$
79: 
80: $$-1 + 3\times\sqrt3^6$$

81 to 90

81: $$1\times3\times\sqrt3^6$$
82: $$1+3\times\sqrt3^6$$
83: 
84: 
85: 
86: 
87: 
88: 
89: 
90: 

91 to 100

91: 
92: 
93: 
94: 
95: 
96: $$(13+3)\times6$$
97: 
98: 
99: 
100: 

• Base64 because rot13 doesn't work with numbers: U29tZSBvdGhlcnM6CjQxID0gMzYgKyA2IC0gMQphbmQKNDMgPSAoM14zKSArIDE2CmFuZAo1NSA9IDYxIC0gc3FydCgzNikKYW5kCjY3ID0gNjEgLSBzcXJ0KDM2KQphbmQKNzYgPSA2MyArIDEzCmFuZAo5NCA9IDYxICsgMzM= – Outman Jan 28 '19 at 11:19

### Here's a solution for most of them. The remaining 21 are impossible.

• Most of them only use simple arithmetic.
• Some of them use exponentiation.
• Some use square roots.

Edit: Using only the binary operators (including digit concatenation), the number of possible combinations is pretty small (arrange three binary operators, then fill in the four numbers in some order) and I double-checked all of them with a computer. Most numbers are solvable with basic arithmetic, some require exponentiation, some require negation or square roots, and the rest are apparently impossible to solve without some extra operation such as rounding.

Accordingly, it is provably impossible to construct the following twenty-one numbers:

41, 44, 46, 47, 56, 68, 69, 74, 77, 79, 83, 85, 86, 89, 90, 91, 92, 95, 97, 98, 100.

The rest can be constructed as follows:

1-25 (complete)

1. (6+1)-(3+3)
2. (3*3)-(6+1)
3. (3*3)-(6*1)
4. 13-(6+3)
5. 36-31
6. 3*(3+1)-6
7. (6+1)+(3-3)
8. (6-3)3-1
9. 6
(3-1)-3
10. 3+13-6
11. 6+3+3-1
12. (6+3+3)*1
13. 6+3+3+1
14. (6*3)-(3+1)
15. 13 + (6/3)
16. 13 + (6-3)
17. 33-16
18. (1+3)*3 + 6
19. (6 + 1/3)*3
20. (3*6)+(3-1)
21. (13-6)*3
22. 13+3+6
23. 36-13
24. (6+1)*3 +3
25. 16+(3*3)

26-50 (except 41, 44, 46, 47)

26. 33 - (6+1)
27. 33 - (6*1)
28. 61 - 33
29. 31 - (6/3)
30. (6+3+1)*3
31. 13 + (6*3)
32. 63-31
33. 13*3 - 6
34. 31+(6-3)
35. 6*(3+3) -1
36. (6+3)(3+1)
37. 6
(3+3) + 1
38. 33+(6-1)
39. 33+(6*1)
40. 31+3+6
41. round[36 + √(31)]
42. (3+3+1)*6
43. 16 + (3^3)
44. round[(3-√3)^16]
45. (3*13)+6
46. round[3 * (13 + √6)]
47. round[31√3] - 6
48. (3*3 - 1) *6
49. 36+13
50. 63-13

51-75 (except 56, 68, 69, 74)

51. 16*3 + 3
52. 61 - (3*3)
53. (3*3*6)-1
54. 1*3*3*6
55. 61-(3+3)
56. √(3136)     (!)
57. (13+6)*3
58. (3+1)^3 - 6
59. 63 - (3+1)
60. (63*1) - 3
61. 61 + (3 - 3)
62. 61 + (3/3)
63. (6+1)*(3*3)
64. 63 + (1^3)
65. 63 + (3-1)
66. 63 + (3*1)
67. 36 + 31
68. round[63 + √31]
69. round[6 √133]
70. 61+(3*3)
71. (6^3)/3 - 1
72. 36 * (3-1)
73. (6^3)/3 + 1
74. round[3 √613]
75. 13*6 - 3

76-100 (except: 77, 79, 83, 85*, 86, 89,90,91,92, 95, 97, 98, 100.)

76. 63+13
77. round[61 * (3-√3)]
78. 13 * √(36)
79. round[6*13 + √3]
80. √(3^6) * 3 - 1
81. 13*6 + 3
82. √(3^6) * 3 + 1
83. round[(3+31)*√6]
84. 3^√(16) + 3
85. -----
86. round[√(6+√3) * 31]
87. (31*3) - 6
88. 61 + (3^3)
89. round[63 * √(3-1)]
90. round[13 * √(6/3)]
91. round[16 * √33]
92. round[3^√(3*6-1)]
93. 31 * (6-3)
94. 63+31
95. round[3*31 + √6]
96. (13+3)*6
97. round[(√3)^6 * √13]
98. round[36 * (1+√3)]
99. (3*31)+6
100. round[3*(31+√6)]

If you're allowed to concatenate the results of operations (e.g. (3+1)|5 = 45 ) then a solution for 85 is :

85. (3*3)|1 - 6

Python:

from itertools import permutations
from math import sqrt, floor, ceil

concat_literal_numbers_only = True

ops = { "+" : lambda a,b: a+b,
"-" : lambda a,b: a - b,
"/" : lambda a,b : a/float(b),
"*" : lambda a,b : a*b,
"^" : lambda a,b : a**b,
"C" : lambda a,b : float(str(a) + str(b)),
"n" : lambda a : -a,
"s" : lambda a : sqrt(a),
#"f" : lambda a : floor(a)
}

arity = {"+" : 2,
"-" : 2,
"/" : 2,
"*" : 2,
"^" : 2,
"C" : 2,
"n" : 1,
"s" : 1,
"f" : 1,
}

# print ops["/"](1,3)

# args: number of open args available
# nums: available digits to be used
# ops : tuple indicating commands used so far

def evaluate(cmds) :
"""Consume the list of commands in prefix notation, producing a pair (ans, unconsumed_symbols)"""

x = cmds.pop(0)
if not ops.get(x) :
return (x, cmds)
else :
args = []
for y in range(arity[x]) :
try :
(a, cmds) = evaluate(cmds)
args += [a]
except OverflowError :
return (None, None)
return (ops.get(x)(*args), cmds)

def score(ops):
ret = 0
ret += ops.count("+")
ret += 1.1*ops.count("-")
ret += 2 * ops.count("*")
ret += 3 * ops.count("/")
ret += 3 * ops.count("n")
ret += 4 * ops.count("^")
ret += 4 * ops.count("s")
ret += 4 * ops.count("f")
ret += 4 * ops.count("w")

# ret += 4 * ops.count("fs")
# ret += 4 * ops.count("cs")
return ret

agenda = [{"args" : 1, "nums" : [1,3,3,6], "ops" : []}]
seen = {}

only_search_for = None
ret = []

def finish(ops) :
global ret
global seen

ops_tmp = ops[:]
try :
n,_ = evaluate(ops_tmp)
except :
n = None

if n is None or not (0  score(ops) :
seen[n] = ops

print ops,"\t",n

while agenda :
x = agenda.pop(0)

if not x["nums"] and not x["args"] : # finished: used up all numbers; no open spaces.
finish(x["ops"])

if len(x["nums"]) == x["args"] : # fill in numbers only
for nums in set(permutations(x["nums"])) :
finish(x["ops"] + list(nums))

# print {"args" : 0,
#        "nums" : [],
#        "ops" : x["ops"] + list(nums)}

elif len(x["nums"]) > x["args"] :
# add new operators

for op in ops.keys() :
if arity[op] == 1 and x["ops"] and x["ops"][-1] == op :
continue # limit repeated unary operations
if arity[op] == 1 and x["ops"] and arity.get(x["ops"][-1]) == 1 :
continue # limit repeated unary operations

if (concat_literal_numbers_only and x["ops"] and (x["ops"][-1] == "C" or (len(x["ops"])>1 and x["ops"][-2] == "C")) and op != "C") :
continue

new_x = {"args" : x["args"] + arity[op] - 1,
"nums" : x["nums"],
"ops" : x["ops"] + [op]}
agenda = [new_x] + agenda

if x["args"] == 1 :
continue

for n in set(x["nums"]) :
new_nums = x["nums"][:]
new_nums.remove(n)
new_x = {"args" : x["args"] - 1,
"nums" : new_nums,
"ops" : x["ops"] + [n]}

agenda = [new_x] + agenda

# SHOW HOW TO MAKE ALL OF THE NUMBERS
miss = []
for i in range(0+1,100+1) :
if not seen.get(i) :
miss += [i]
print i, "\t", seen.get(i, "---")

# SHOW WHICH NUMBERS WERE MISSED
print "missed: ", miss

# IF YOU'RE LOOKING FOR ALL POSSIBLE WAYS TO MAKE SOMETHING, SHOW THEM HERE.
if only_search_for is not None :
ret = sorted(ret, key=score)
for x in ret:
print x

• Although it's unlikely that you'd find the missing solutions there, looks like the code ignores Nth roots entirely. – Bass Jan 29 '19 at 9:47
• @Bass . Thanks, yes ---- when posting the code I took out some operators that didn't add to the number of solutions. – user326210 Jan 30 '19 at 1:00

We can generate any integer using only $$1$$, $$3$$, $$3$$ and $$6$$ with the introduction of one special function.

The function in question is the Logarithm to an arbitrary base $$b$$ , or $$\log _{b} (x)$$.

To begin, let's discuss square root stacking.
$$\sqrt{\sqrt{a}}$$ is equivalent to $$\sqrt[4]{a}$$, and $$\sqrt{\sqrt{\sqrt{a}}}$$ is equivalent to $$\sqrt[8]{a}$$, which can be rewritten as $$a^\frac{1}{8}$$. This pattern continues indefinitely; $$a$$ with $$n$$ square roots stacked to it is equal to $$a^\frac{1}{2^n}$$.

The laws of logarithms state that $$\log _{b} (x^a) = a \cdot \log _{b} (x)$$. If we take the logarithm to base $$b$$ or our previous square root stack, we get $$\log _{b} (a^\frac{1}{2^n})$$, or $$\frac{1}{2^n} \cdot \log _{b}a$$. Setting $$a$$ and $$b$$ as 3 means that $$log _{3} (\sqrt{\sqrt[...]{3}})$$, with $$n$$ square roots, is equal to $$\frac{1}{2^n} \cdot \log _{3}3$$, or $$2^{-n}$$. $$(\frac{1}{a^x} = a^{-x})$$

$$\sqrt{\sqrt{16}} = 2$$, and $$\log _{b}(b^a) = a$$. As such,

$$0 = -\log _{\sqrt{\sqrt{16}}}(\log_{3}(3))$$
$$1 = -\log _{\sqrt{\sqrt{16}}}(\log_{3}(\sqrt{3}))$$
$$2 = -\log _{\sqrt{\sqrt{16}}}(\log_{3}(\sqrt{\sqrt{3}}))$$
$$3 = -\log _{\sqrt{\sqrt{16}}}(\log_{3}(\sqrt{\sqrt{\sqrt{3}}}))$$
$$4 = -\log _{\sqrt{\sqrt{16}}}(\log_{3}(\sqrt{\sqrt{\sqrt{\sqrt{3}}}}))$$
And so on, such that the amount of square roots is equal to your desired number.

This works for all integers; for negative numbers simply remove the $$-$$ at the start.

This was inspired by Numberphile's video on the four 4s.

Here are some:

1: $$3 + 3 - 6 + 1$$
2: $$3 * 3 - (6 + 1)$$
3: $$3 * 3 * 1 - 6$$
4: $$3 * 3 - 6 + 1$$
5: $$(3 * 6) / 3 - 1$$
6: $$(3 * 6) / 3 * 1$$
7: $$(3 * 6) / 3 + 1$$
8: $$3 * 3 - 1 ^ 6$$
9: $$(3 * 6) / (3 - 1)$$
10: $$3 * 3 + 1 ^ 6$$
11: $$36 / 3 - 1$$
12: $$36 / 3 * 1$$
13: $$36 / 3 + 1$$
14: $$3 * 6 - (3 + 1)$$
15: $$3 * 6 - (3 * 1)$$
16: $$3 * 6 - (3 - 1)$$
17: $$3 * 6 - 1 ^ 3$$
18: $$3 * 6 * 1 ^ 3$$
19: $$3 * 6 + 1 ^ 3$$
20: $$3 * 6 + 3 - 1$$
22: (omega kyrpton did some) $$3 * 6 + 3 + 1$$

I will do more later.

1-20: (Credits to @YoutRied)

1: $$3 + 3 - 6 + 1$$
2: $$3 * 3 - (6 + 1)$$
3: $$3 * 3 * 1 - 6$$
4: $$3 * 3 - 6 + 1$$
5: $$(3 * 6) / 3 - 1$$
6: $$(3 * 6) / 3 * 1$$
7: $$(3 * 6) / 3 + 1$$
8: $$3 * 3 - 1 ^ 6$$
9: $$(3 * 6) / (3 - 1)$$
10: $$3 * 3 + 1 ^ 6$$
11: $$36 / 3 - 1$$
12: $$36 / 3 * 1$$
13: $$36 / 3 + 1$$
14: $$3 * 6 - (3 + 1)$$
15: $$3 * 6 - (3 * 1)$$
16: $$3 * 6 - (3 - 1)$$
17: $$3 * 6 - 1 ^ 3$$
18: $$3 * 6 * 1 ^ 3$$
19: $$3 * 6 + 1 ^ 3$$
20: $$3 * 6 + 3 - 1$$

21-29

21: $$3 * 6 + 3 * 1$$
22: $$( 1 + 3 ) ! - ( 6 / 3 )$$
23: $$( 1 + 3 ) ! - ( 6 - 3 )$$
24: $$( 6 - 3 / 3 - 1 ) !$$
25: $$1 * 3 ^ 3 - floor(\sqrt{6})$$
26: $$( 6 - 3 ) ^ 3 - 1$$
27: $$( 6 - 3 ) ^ 3 * 1$$
28: $$( 6 - 3 ) ^ 3 + 1$$
29: $$31 - 6 / 3$$

41-50: (Credits to @Bass for 42, 45, 49)

41: $$(-1+3!)+36$$
42: $$(1+3+3)\times 6$$
43: $$31 + 6 * ceil(\sqrt{3})$$
44: $$floor( 1 * 3 * \sqrt{6 ^ 3})$$
45: $$13\times3+6$$
46: $$ceil(\sqrt{6 ^ 3} + 31)$$
47: $$floor(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{31!}}}}})+36$$
48: $$6 * ( 3 * 3 - 1 )$$
49: $$13+36$$
50: $$(6+1)^2 + 3 - 3$$

51-60:

51: $$( 3 * 6 - 1 ) * 3$$
52: $$( 3 + 3 + 1 ) * ceil(\sqrt{6})$$
53: $$-1+( 3 * 3 * 6 )$$
54: $$1*3 * 3 * 6$$
55: $$1+3*3*6$$
56: $$61-3!+floor(\sqrt{3})$$
57: $$1*63-3!$$
58: $$1+63-3!$$
59: $$floor(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{6^{(3-1)}}}}}}*3)$$
60: $$(1+3*3)*6$$

61-70:

61: $$63-3+1$$
62: $$63+1-ceil(\sqrt{3})$$
63: $$63-floor(\sqrt{3})+1$$
64: $$63+ceil(\sqrt{3})-1$$
65: $$63+3-1$$
66: $$63+3*1$$
67: $$63+3+1$$
68: $$61+3!+floor(\sqrt{3})$$
69: $$61+3!+ceil(\sqrt{3})$$
70: $$61+3*3$$

71-80

71: $$(3+1)!*3-floor(\sqrt{\sqrt{6}})$$
72: $$(3+1)*3*6$$
73: $$(3+1)!*3+floor(\sqrt{\sqrt{6}})$$
74: $$(3+1)!*3+floor(\sqrt{6})$$
75: $$(3+1)!*3+ceil(\sqrt{6})$$
76: $$ceil(\sqrt{\sqrt{\sqrt{ceil(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{ceil(\sqrt{3!!})!}}}}})}}})*(3*6+1)$$
76:(much simpler) $$13*6-ceil(\sqrt{3})$$
77: $$13*6-floor(\sqrt{3})$$
78: $$13*floor(\sqrt{3})*6$$
79: $$13*6+floor(\sqrt{3})$$
80: $$13*6+ceil(\sqrt{3})$$

81-90:

81: $$(6+3)^{(3-1)}$$
82: $$(6+3)^{ceil(\sqrt{3})}+1$$

• Who said you could use factorials? – Yout Ried Jan 28 '19 at 0:20
• What are number 23 (plus you probably can't use factorials and 24? I don't get them. – Yout Ried Jan 28 '19 at 1:08
• Oops forgot a ")" and maybe you're missing a factorial for number 24 – Yout Ried Jan 28 '19 at 1:16

Expanding on Bass's answer, I added some new numbers.

(I lost track on which numbers I added, though 1-40 is all Bass)

1: $$1 + 3 + 3 - 6$$
2: $$(1 + 3) \times 3 / 6$$
3: $$1*3 * 3 - 6$$
4: $$13 - 3 - 6$$
5: $$-1^{33} +6$$
6: $$1\times3-3+6$$
7: $$1 + 3 -3 +6$$
8: $$1+3/3 + 6$$
9: $$1^3 \times (3+6)$$
10: $$1 + \sqrt[3]3^6$$ 11: $$\sqrt{1+3}+3+6$$
12: $$1\times 3 + 3 + 6$$
13: $$1 + 3+3+6$$
14: $$-1 + 3\times 3+6$$
15: $$-1\times3 + 3\times 6$$
16: $$1 - 3 + 3 \times 6$$
17: $$-1^3 +3\times 6$$
18: $$(1+3)*3+6$$
19: $$13 + \sqrt{36}$$
20: $$-1 + 3^3 - 6$$
21: $$1 * 3^3 - 6$$
22: $$13 + 3 + 6$$
23: $$-13+36$$
24: $$(1+3)\times\sqrt{36}$$
25: $$1 - 3 + \sqrt3^6$$
26: $$-1+33-6$$
27: $$1*33-6$$
28: $$1+33-6$$
29: $$-1 + 3 + \sqrt3^6$$
30: $$(-1+3+3)\times 6$$
31: $$13+3*6$$
32: $$-1+3^3+6$$
33: $$13*3-6$$
34: $$1+3^3+6$$
35: $$-1+(3+3)\times6$$
36: $$1\times(3+3)\times 6$$
37: $$1^3+36$$
38: $$\sqrt{1+3}+36$$
39: $$1\times3 + 36$$
40: $$1+33+6$$
41: 
42: $$(1+3+3)\times 6$$
43: $$3^3 + 16$$
44: 
45: $$13\times3+6$$
46: 
47: 
48: $$(-1 + 3 \times 3) \times 6$$
49: $$13+36$$
50: 
51: $$16*3+3$$
52: $$61 - 3\times3$$
53: $$-1 +3 \times 3 \times 6$$
54: $$1\times 3 \times 3 \times 6$$
55: $$1 + 3 \times 3 \times 6$$
56: 
57: $$(6\times3+1)\times3$$
58: $$(1+3)^3-6$$
59: $$63 - 3 - 1$$
60: $$(1+3\times3)\times6$$
61: $$63 - 3 + 1$$
62: $$63 - 1^3$$
63: $$63 * 1^3$$
64: $$(1+3/3)^6$$
65: $$63 + 3 - 1$$
66: $$1 \times (63 + 3)$$
67: $$63 + 3 + 1$$
68: 
69: 
70: $$(1+3)^3+6$$ 71: $$6^3 / 3 - 1$$
72: $$(1+3)\times 3 \times 6$$
73: $$6^3 / 3 + 1$$
74: 
75: $$3^(3+1) - 6$$
76: $$63+13$$
77: 
78: $$13 \times \sqrt{36}$$
79: 
80: $$-1 + 3\times\sqrt3^6$$
81: $$1\times3\times\sqrt3^6$$
82: $$1+3\times\sqrt3^6$$
83: 
84: 
85: 
86: 
87: $$3^(3+1) + 6$$
88: $$61 + 3^3$$
89: 
90: 
91: 
92: 
93: 
94: $$33 + 61$$
95: 
96: $$(13+3)\times6$$
97: 
98: 
99: $$31\times3+6$$
100: 

Only need 41, 44, 46, 47, 50, 56, 68, 69, 74, 77, 79, 83, 84, 85, 86, 89, 90, 91, 92, 93, 94, 95, 97, 98, and 100 now!

Partial answer 1-50 (w/e 41,47):

$$1= 1+3+3-6$$
$$2= 1 + (\frac{6}{(3+3)})$$
$$3= 1^3+(\frac{6}{3})$$
$$4= (\frac{6}{3})+3-1$$
$$5= (\frac{6}{3})+3^1$$
$$6= 6^1+3-3$$
$$7= 6+1-3+3$$
$$8= 6 + 3 - 1^3$$
$$9= 1^3 * (3+6)$$
$$10= 1^3 +3+6$$
$$11= 13 - (\frac{6}{3})$$
$$12= 6+3+3^1$$
$$13= 6+3+3+1$$
$$14= 6*3 - 3 - 1$$
$$15= 6*3 - 3^1$$
$$16= 16 + 3 - 3$$
$$17= 16 + (\frac{3}{3})$$
$$18= (\frac{6*3}{1^3})$$
$$19= 6*3+1^3$$
$$20= 6*3+3-1$$
$$21= 6*3+3^1$$
$$22= 6*3+3+1$$
$$23= 36-13$$
$$24= 6*(3+1^3)$$
$$25= 16+(3*3)$$
$$26= 13*(\frac{6}{3})$$
$$27= 33-6^1$$
$$28= 33-6+1$$
$$29= 31-(\frac{6}{3})$$
$$30= 6*(3+3-1)$$
$$31= 13+3*6$$
$$32= 3^3+6-1$$
$$33= (\frac{33}{1^6})$$
$$34= 33+1^6$$
$$35= (3+3)*6-1$$
$$36= (3+3)^1*6$$
$$37= 1+(3+3)*6$$
$$38= 33+6-1$$
$$39= 33+6^1$$
$$40= 1+33+6$$
$$41=$$
$$42= (1+3+3)*6$$
$$43= 16+3^3$$
$$44= round(\sqrt{6^3}*3^1)$$
$$45= 3*3*(6-1)$$
$$46= ceil(\sqrt{6^3}*3)+1$$
$$47=$$
$$48= ((3*3)-1)*6$$
$$49= 16+33$$
$$50= 63-13$$

Alrighty, I'm piggybacking off of @OmegaKrypton else and adding some of my own.

1 to 10

1: $$1 + 3 + 3 - 6$$
2: $$(1 + 3) \times 3 / 6$$
3: $$1^3 +3/6$$
4: $$13 - 3 - 6$$
5: $$-1^{33} +6$$
6: $$1\times3-3+6$$
7: $$1 + 3 -3 +6$$
8: $$1+3/3 + 6$$
9: $$1^3 \times (3+6)$$
10: $$1^3 + 3+6$$

11 to 20

11: $$\sqrt{1+3}+3+6$$
12: $$1\times 3 + 3 + 6$$
13: $$1 + 3+3+6$$
14: $$-1 + 3\times 3+6$$
15: $$-1\times3 + 3\times 6$$
16: $$1 - 3 + 3 \times 6$$
17: $$-1^3 +3\times 6$$
18: $$(1+3)*3+6$$
19: $$13 + \sqrt{36}$$
20: $$-1 + 3^3 - 6$$

21 to 30

21: $$1 * 3^3 - 6$$
22: $$13 + 3 + 6$$
23: $$-13+36$$
24: $$(1+3)\times\sqrt{36}$$
25: $$3*6+3!-1$$ or $$(6-1)^(3!/3)$$
26: $$-1+33-6$$
27: $$1*33-6$$
28: $$1+33-6$$
29: $$36-3!-1$$
30: $$(-1+3+3)\times 6$$

31 to 40

31: $$13+3*6$$
32: $$-1+3^3+6$$
33: $$13*3-6$$
34: $$1+3^3+6$$
35: $$-1+(3+3)\times6$$
36: $$1\times(3+3)\times 6$$
37: $$1^3+36$$
38: $$\sqrt{1+3}+36$$
39: $$1\times3 + 36$$
40: $$1+33+6$$

41 to 50

41: 
42: $$(1+3+3)\times 6$$
43: $$3^3+16$$
44: 
45: $$13\times3+6$$
46: 
47: 
48: $$16*(3!-3)$$
49: $$13+36$$
50: $$63-13$$

I added a few. It's getting late here; will come back tomorrow.