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

Question

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

Answer 1

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: $ $

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

Adding some more...

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

Answer 6

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!

Answer 7

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.

Answer 8

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= $
$48= ((3*3)-1)*6$
$49= 16+33$
$50= 63-13$