Create Numbers 1 - 100 using 1,9,6,8
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Create all numbers 1 - 100 using equations made up of 1,9,6,8.
Rules:
- Use all four digits exactly once
- Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root.
- Parentheses and grouping (e.g. "21") are also allowed.
- You have to keep the order 1,9,6,8 for all numbers.
- Exponentiation can only be used in the number order with the numbers provided. Eg. 1^9 + 6 + 8 is allowed. Not 1^6 + 9 + 8.
- The modulus operator is not allowed.
- Rounding is not allowed (e.g. 201/8=25).
- Decimal point is allowed.
Credit to Fitch496 for the idea.
mathematics
New contributor
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add a comment |
$begingroup$
Create all numbers 1 - 100 using equations made up of 1,9,6,8.
Rules:
- Use all four digits exactly once
- Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root.
- Parentheses and grouping (e.g. "21") are also allowed.
- You have to keep the order 1,9,6,8 for all numbers.
- Exponentiation can only be used in the number order with the numbers provided. Eg. 1^9 + 6 + 8 is allowed. Not 1^6 + 9 + 8.
- The modulus operator is not allowed.
- Rounding is not allowed (e.g. 201/8=25).
- Decimal point is allowed.
Credit to Fitch496 for the idea.
mathematics
New contributor
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it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
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– Oray
20 hours ago
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@Oray can you please prove it then? thanks!
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– Omega Krypton
16 hours ago
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@OmegaKrypton i put it into my code many numbers cannot be achieved.
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– Oray
16 hours ago
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@Oray I think the only number we have not found now is $79$.
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– hexomino
12 hours ago
add a comment |
$begingroup$
Create all numbers 1 - 100 using equations made up of 1,9,6,8.
Rules:
- Use all four digits exactly once
- Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root.
- Parentheses and grouping (e.g. "21") are also allowed.
- You have to keep the order 1,9,6,8 for all numbers.
- Exponentiation can only be used in the number order with the numbers provided. Eg. 1^9 + 6 + 8 is allowed. Not 1^6 + 9 + 8.
- The modulus operator is not allowed.
- Rounding is not allowed (e.g. 201/8=25).
- Decimal point is allowed.
Credit to Fitch496 for the idea.
mathematics
New contributor
$endgroup$
Create all numbers 1 - 100 using equations made up of 1,9,6,8.
Rules:
- Use all four digits exactly once
- Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root.
- Parentheses and grouping (e.g. "21") are also allowed.
- You have to keep the order 1,9,6,8 for all numbers.
- Exponentiation can only be used in the number order with the numbers provided. Eg. 1^9 + 6 + 8 is allowed. Not 1^6 + 9 + 8.
- The modulus operator is not allowed.
- Rounding is not allowed (e.g. 201/8=25).
- Decimal point is allowed.
Credit to Fitch496 for the idea.
mathematics
mathematics
New contributor
New contributor
edited 20 hours ago
Aryaman
1078
1078
New contributor
asked 21 hours ago
AuroxaAuroxa
141
141
New contributor
New contributor
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it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
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– Oray
20 hours ago
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@Oray can you please prove it then? thanks!
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– Omega Krypton
16 hours ago
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@OmegaKrypton i put it into my code many numbers cannot be achieved.
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– Oray
16 hours ago
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@Oray I think the only number we have not found now is $79$.
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– hexomino
12 hours ago
add a comment |
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it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
$endgroup$
– Oray
20 hours ago
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@Oray can you please prove it then? thanks!
$endgroup$
– Omega Krypton
16 hours ago
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@OmegaKrypton i put it into my code many numbers cannot be achieved.
$endgroup$
– Oray
16 hours ago
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@Oray I think the only number we have not found now is $79$.
$endgroup$
– hexomino
12 hours ago
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it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
$endgroup$
– Oray
20 hours ago
$begingroup$
it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
$endgroup$
– Oray
20 hours ago
$begingroup$
@Oray can you please prove it then? thanks!
$endgroup$
– Omega Krypton
16 hours ago
$begingroup$
@Oray can you please prove it then? thanks!
$endgroup$
– Omega Krypton
16 hours ago
$begingroup$
@OmegaKrypton i put it into my code many numbers cannot be achieved.
$endgroup$
– Oray
16 hours ago
$begingroup$
@OmegaKrypton i put it into my code many numbers cannot be achieved.
$endgroup$
– Oray
16 hours ago
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@Oray I think the only number we have not found now is $79$.
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– hexomino
12 hours ago
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@Oray I think the only number we have not found now is $79$.
$endgroup$
– hexomino
12 hours ago
add a comment |
4 Answers
4
active
oldest
votes
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Partial answer (the most obvious solutions):
1 = -1^9 - 6 + 8 = 1^968
2 = 1 + sqrt(9) + 6 - 8
5 = 19 - 6 - 8
6 = -1 + 9 + 6 - 8 = 1 - 9 + 6 + 8
7 = 1 * (9 + 6) - 8
8 = 1 + 9 + 6 - 8
9 = 1^96 + 8
11 = -1 + 96 / 8
12 = 1 * 96 / 8 = 1 + 9 - 6 + 8
13 = 1 + 96 / 8
16 = (1 - 9) * (6 - 8)
17 = 19 + 6 - 8
18 = 1 + sqrt(9) + 6 + 8
21 = 19 - 6 + 8
22 = -1 + 9 + 6 + 8
23 = 1 * (9 + 6) + 8
24 = 1 + 9 + 6 + 8
25 = -1 + sqrt(9) * 6 + 8
26 = 1 * sqrt(9) * 6 + 8
27 = 1 + sqrt(9) * 6 + 8
29 = -19 + 6 * 8
31 = 19 + 6 + 8
32 = (1 + sqrt(9)) * 6 + 8
52 = (1 + 9) * 6 - 8
57 = 1 * 9 + 6 * 8
58 = -1 - 9 + 68
59 = -1 * 9 + 68
60 = 1 - 9 + 68
62 = 1 * 9 * 6 + 8
63 = 1 + 9 * 6 + 8
67 = 19 + 6 * 8
68 = (1 + 9) * 6 + 8
76 = -1 + 9 + 68
77 = 1 * 9 + 68
78 = 1 + 9 + 68
80 = -1 - 9 + 6! / 8
81 = -1 * 9 + 6! / 8
82 = 1 - 9 + 6! / 8
87 = -1 + 96 - 8
88 = 1 * 96 - 8
89 = 1 + 96 - 8
98 = -1 + 9 + 6! / 8
99 = 1 * 9 + 6! / 8
100 = 1 + 9 + 6! / 8
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add a comment |
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Partial Answer (too lazy to type them out :P)
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add a comment |
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Here are all the numbers which have not been obtained already by trolley813, Omega Krypton and ppgdev
$28 = 1 times ((sqrt{9})! times 6) - 8$
$30 =-(1times(sqrt{9})!) + sqrt{sqrt{6^8}} $
$33 = 19 + 6 + 8$
$34 = ((1+(sqrt{9})!) times 6) - 8 $
$35 = (1 times sqrt{(sqrt{9})^6}) + 8$
$36 = (1 + sqrt{(sqrt{9})^6}) + 8$
$37 = 1^9 + sqrt{sqrt{6^8}}$
$38 = (1+sqrt{9})! + 6 + 8$
$41 = -(1 + (sqrt{9})!) + (6 times 8)$
$42 = -(1 times (sqrt{9})!) + (6 times 8)$
$43 = -1 + ((sqrt{9})! times 6) + 8$
$44 = 1 times ((sqrt{9})! times 6) + 8$
$45 = 1 + ((sqrt{9})! times 6) + 8$
$ 55 = 1 + (sqrt{9})! + (6 times 8)$
$56 = (1^9 + 6) times 8 $
$ 64 = ((-1 + sqrt{9}) + 6) times 8$
$65 = -(1 times sqrt{9}) + 68 $
$66 = 1 - sqrt{9} + 68 $
$69 = 1^9 + 68$
$70 = -1 + sqrt{9} + 68 $
$71 = -1 + ((sqrt{9} + 6)times 8)$
$72 = ((1 times sqrt{9}) + 6) times 8$
$73 = 1 + ((sqrt{9} + 6)times 8) $
$74 = (1 times (sqrt{9})!) + 68$
$75 = 1 + (sqrt{9})!) + 68$
$ 79 = -1 + ((9 + sqrt{sqrt{ldots sqrt{6}}}) times 8)$
$ 86 = -1 - sqrt{9} + (6!/8)$
$90 = (1^9 times 6!)/8 $
$91 = 1^9 + (6!/8)$
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@Oray which numbers are impossible?
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– Auroxa
7 hours ago
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nice one @Auroxa
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– Omega Krypton
7 hours ago
add a comment |
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Here is an answer for 39 that was missing so far:
39 = -(1*9) + (6*8)
I also got 79
79 = ((-1 + (sqrt(9)!))!! + (8!!)/6
New contributor
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@Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
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– ppgdev
7 hours ago
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Partial answer (the most obvious solutions):
1 = -1^9 - 6 + 8 = 1^968
2 = 1 + sqrt(9) + 6 - 8
5 = 19 - 6 - 8
6 = -1 + 9 + 6 - 8 = 1 - 9 + 6 + 8
7 = 1 * (9 + 6) - 8
8 = 1 + 9 + 6 - 8
9 = 1^96 + 8
11 = -1 + 96 / 8
12 = 1 * 96 / 8 = 1 + 9 - 6 + 8
13 = 1 + 96 / 8
16 = (1 - 9) * (6 - 8)
17 = 19 + 6 - 8
18 = 1 + sqrt(9) + 6 + 8
21 = 19 - 6 + 8
22 = -1 + 9 + 6 + 8
23 = 1 * (9 + 6) + 8
24 = 1 + 9 + 6 + 8
25 = -1 + sqrt(9) * 6 + 8
26 = 1 * sqrt(9) * 6 + 8
27 = 1 + sqrt(9) * 6 + 8
29 = -19 + 6 * 8
31 = 19 + 6 + 8
32 = (1 + sqrt(9)) * 6 + 8
52 = (1 + 9) * 6 - 8
57 = 1 * 9 + 6 * 8
58 = -1 - 9 + 68
59 = -1 * 9 + 68
60 = 1 - 9 + 68
62 = 1 * 9 * 6 + 8
63 = 1 + 9 * 6 + 8
67 = 19 + 6 * 8
68 = (1 + 9) * 6 + 8
76 = -1 + 9 + 68
77 = 1 * 9 + 68
78 = 1 + 9 + 68
80 = -1 - 9 + 6! / 8
81 = -1 * 9 + 6! / 8
82 = 1 - 9 + 6! / 8
87 = -1 + 96 - 8
88 = 1 * 96 - 8
89 = 1 + 96 - 8
98 = -1 + 9 + 6! / 8
99 = 1 * 9 + 6! / 8
100 = 1 + 9 + 6! / 8
$endgroup$
add a comment |
$begingroup$
Partial answer (the most obvious solutions):
1 = -1^9 - 6 + 8 = 1^968
2 = 1 + sqrt(9) + 6 - 8
5 = 19 - 6 - 8
6 = -1 + 9 + 6 - 8 = 1 - 9 + 6 + 8
7 = 1 * (9 + 6) - 8
8 = 1 + 9 + 6 - 8
9 = 1^96 + 8
11 = -1 + 96 / 8
12 = 1 * 96 / 8 = 1 + 9 - 6 + 8
13 = 1 + 96 / 8
16 = (1 - 9) * (6 - 8)
17 = 19 + 6 - 8
18 = 1 + sqrt(9) + 6 + 8
21 = 19 - 6 + 8
22 = -1 + 9 + 6 + 8
23 = 1 * (9 + 6) + 8
24 = 1 + 9 + 6 + 8
25 = -1 + sqrt(9) * 6 + 8
26 = 1 * sqrt(9) * 6 + 8
27 = 1 + sqrt(9) * 6 + 8
29 = -19 + 6 * 8
31 = 19 + 6 + 8
32 = (1 + sqrt(9)) * 6 + 8
52 = (1 + 9) * 6 - 8
57 = 1 * 9 + 6 * 8
58 = -1 - 9 + 68
59 = -1 * 9 + 68
60 = 1 - 9 + 68
62 = 1 * 9 * 6 + 8
63 = 1 + 9 * 6 + 8
67 = 19 + 6 * 8
68 = (1 + 9) * 6 + 8
76 = -1 + 9 + 68
77 = 1 * 9 + 68
78 = 1 + 9 + 68
80 = -1 - 9 + 6! / 8
81 = -1 * 9 + 6! / 8
82 = 1 - 9 + 6! / 8
87 = -1 + 96 - 8
88 = 1 * 96 - 8
89 = 1 + 96 - 8
98 = -1 + 9 + 6! / 8
99 = 1 * 9 + 6! / 8
100 = 1 + 9 + 6! / 8
$endgroup$
add a comment |
$begingroup$
Partial answer (the most obvious solutions):
1 = -1^9 - 6 + 8 = 1^968
2 = 1 + sqrt(9) + 6 - 8
5 = 19 - 6 - 8
6 = -1 + 9 + 6 - 8 = 1 - 9 + 6 + 8
7 = 1 * (9 + 6) - 8
8 = 1 + 9 + 6 - 8
9 = 1^96 + 8
11 = -1 + 96 / 8
12 = 1 * 96 / 8 = 1 + 9 - 6 + 8
13 = 1 + 96 / 8
16 = (1 - 9) * (6 - 8)
17 = 19 + 6 - 8
18 = 1 + sqrt(9) + 6 + 8
21 = 19 - 6 + 8
22 = -1 + 9 + 6 + 8
23 = 1 * (9 + 6) + 8
24 = 1 + 9 + 6 + 8
25 = -1 + sqrt(9) * 6 + 8
26 = 1 * sqrt(9) * 6 + 8
27 = 1 + sqrt(9) * 6 + 8
29 = -19 + 6 * 8
31 = 19 + 6 + 8
32 = (1 + sqrt(9)) * 6 + 8
52 = (1 + 9) * 6 - 8
57 = 1 * 9 + 6 * 8
58 = -1 - 9 + 68
59 = -1 * 9 + 68
60 = 1 - 9 + 68
62 = 1 * 9 * 6 + 8
63 = 1 + 9 * 6 + 8
67 = 19 + 6 * 8
68 = (1 + 9) * 6 + 8
76 = -1 + 9 + 68
77 = 1 * 9 + 68
78 = 1 + 9 + 68
80 = -1 - 9 + 6! / 8
81 = -1 * 9 + 6! / 8
82 = 1 - 9 + 6! / 8
87 = -1 + 96 - 8
88 = 1 * 96 - 8
89 = 1 + 96 - 8
98 = -1 + 9 + 6! / 8
99 = 1 * 9 + 6! / 8
100 = 1 + 9 + 6! / 8
$endgroup$
Partial answer (the most obvious solutions):
1 = -1^9 - 6 + 8 = 1^968
2 = 1 + sqrt(9) + 6 - 8
5 = 19 - 6 - 8
6 = -1 + 9 + 6 - 8 = 1 - 9 + 6 + 8
7 = 1 * (9 + 6) - 8
8 = 1 + 9 + 6 - 8
9 = 1^96 + 8
11 = -1 + 96 / 8
12 = 1 * 96 / 8 = 1 + 9 - 6 + 8
13 = 1 + 96 / 8
16 = (1 - 9) * (6 - 8)
17 = 19 + 6 - 8
18 = 1 + sqrt(9) + 6 + 8
21 = 19 - 6 + 8
22 = -1 + 9 + 6 + 8
23 = 1 * (9 + 6) + 8
24 = 1 + 9 + 6 + 8
25 = -1 + sqrt(9) * 6 + 8
26 = 1 * sqrt(9) * 6 + 8
27 = 1 + sqrt(9) * 6 + 8
29 = -19 + 6 * 8
31 = 19 + 6 + 8
32 = (1 + sqrt(9)) * 6 + 8
52 = (1 + 9) * 6 - 8
57 = 1 * 9 + 6 * 8
58 = -1 - 9 + 68
59 = -1 * 9 + 68
60 = 1 - 9 + 68
62 = 1 * 9 * 6 + 8
63 = 1 + 9 * 6 + 8
67 = 19 + 6 * 8
68 = (1 + 9) * 6 + 8
76 = -1 + 9 + 68
77 = 1 * 9 + 68
78 = 1 + 9 + 68
80 = -1 - 9 + 6! / 8
81 = -1 * 9 + 6! / 8
82 = 1 - 9 + 6! / 8
87 = -1 + 96 - 8
88 = 1 * 96 - 8
89 = 1 + 96 - 8
98 = -1 + 9 + 6! / 8
99 = 1 * 9 + 6! / 8
100 = 1 + 9 + 6! / 8
answered 20 hours ago
trolley813trolley813
1,14638
1,14638
add a comment |
add a comment |
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Partial Answer (too lazy to type them out :P)
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add a comment |
$begingroup$
Partial Answer (too lazy to type them out :P)
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add a comment |
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Partial Answer (too lazy to type them out :P)
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Partial Answer (too lazy to type them out :P)
answered 16 hours ago
Omega KryptonOmega Krypton
4,2251339
4,2251339
add a comment |
add a comment |
$begingroup$
Here are all the numbers which have not been obtained already by trolley813, Omega Krypton and ppgdev
$28 = 1 times ((sqrt{9})! times 6) - 8$
$30 =-(1times(sqrt{9})!) + sqrt{sqrt{6^8}} $
$33 = 19 + 6 + 8$
$34 = ((1+(sqrt{9})!) times 6) - 8 $
$35 = (1 times sqrt{(sqrt{9})^6}) + 8$
$36 = (1 + sqrt{(sqrt{9})^6}) + 8$
$37 = 1^9 + sqrt{sqrt{6^8}}$
$38 = (1+sqrt{9})! + 6 + 8$
$41 = -(1 + (sqrt{9})!) + (6 times 8)$
$42 = -(1 times (sqrt{9})!) + (6 times 8)$
$43 = -1 + ((sqrt{9})! times 6) + 8$
$44 = 1 times ((sqrt{9})! times 6) + 8$
$45 = 1 + ((sqrt{9})! times 6) + 8$
$ 55 = 1 + (sqrt{9})! + (6 times 8)$
$56 = (1^9 + 6) times 8 $
$ 64 = ((-1 + sqrt{9}) + 6) times 8$
$65 = -(1 times sqrt{9}) + 68 $
$66 = 1 - sqrt{9} + 68 $
$69 = 1^9 + 68$
$70 = -1 + sqrt{9} + 68 $
$71 = -1 + ((sqrt{9} + 6)times 8)$
$72 = ((1 times sqrt{9}) + 6) times 8$
$73 = 1 + ((sqrt{9} + 6)times 8) $
$74 = (1 times (sqrt{9})!) + 68$
$75 = 1 + (sqrt{9})!) + 68$
$ 79 = -1 + ((9 + sqrt{sqrt{ldots sqrt{6}}}) times 8)$
$ 86 = -1 - sqrt{9} + (6!/8)$
$90 = (1^9 times 6!)/8 $
$91 = 1^9 + (6!/8)$
$endgroup$
$begingroup$
@Oray which numbers are impossible?
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– Auroxa
7 hours ago
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nice one @Auroxa
$endgroup$
– Omega Krypton
7 hours ago
add a comment |
$begingroup$
Here are all the numbers which have not been obtained already by trolley813, Omega Krypton and ppgdev
$28 = 1 times ((sqrt{9})! times 6) - 8$
$30 =-(1times(sqrt{9})!) + sqrt{sqrt{6^8}} $
$33 = 19 + 6 + 8$
$34 = ((1+(sqrt{9})!) times 6) - 8 $
$35 = (1 times sqrt{(sqrt{9})^6}) + 8$
$36 = (1 + sqrt{(sqrt{9})^6}) + 8$
$37 = 1^9 + sqrt{sqrt{6^8}}$
$38 = (1+sqrt{9})! + 6 + 8$
$41 = -(1 + (sqrt{9})!) + (6 times 8)$
$42 = -(1 times (sqrt{9})!) + (6 times 8)$
$43 = -1 + ((sqrt{9})! times 6) + 8$
$44 = 1 times ((sqrt{9})! times 6) + 8$
$45 = 1 + ((sqrt{9})! times 6) + 8$
$ 55 = 1 + (sqrt{9})! + (6 times 8)$
$56 = (1^9 + 6) times 8 $
$ 64 = ((-1 + sqrt{9}) + 6) times 8$
$65 = -(1 times sqrt{9}) + 68 $
$66 = 1 - sqrt{9} + 68 $
$69 = 1^9 + 68$
$70 = -1 + sqrt{9} + 68 $
$71 = -1 + ((sqrt{9} + 6)times 8)$
$72 = ((1 times sqrt{9}) + 6) times 8$
$73 = 1 + ((sqrt{9} + 6)times 8) $
$74 = (1 times (sqrt{9})!) + 68$
$75 = 1 + (sqrt{9})!) + 68$
$ 79 = -1 + ((9 + sqrt{sqrt{ldots sqrt{6}}}) times 8)$
$ 86 = -1 - sqrt{9} + (6!/8)$
$90 = (1^9 times 6!)/8 $
$91 = 1^9 + (6!/8)$
$endgroup$
$begingroup$
@Oray which numbers are impossible?
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– Auroxa
7 hours ago
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nice one @Auroxa
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– Omega Krypton
7 hours ago
add a comment |
$begingroup$
Here are all the numbers which have not been obtained already by trolley813, Omega Krypton and ppgdev
$28 = 1 times ((sqrt{9})! times 6) - 8$
$30 =-(1times(sqrt{9})!) + sqrt{sqrt{6^8}} $
$33 = 19 + 6 + 8$
$34 = ((1+(sqrt{9})!) times 6) - 8 $
$35 = (1 times sqrt{(sqrt{9})^6}) + 8$
$36 = (1 + sqrt{(sqrt{9})^6}) + 8$
$37 = 1^9 + sqrt{sqrt{6^8}}$
$38 = (1+sqrt{9})! + 6 + 8$
$41 = -(1 + (sqrt{9})!) + (6 times 8)$
$42 = -(1 times (sqrt{9})!) + (6 times 8)$
$43 = -1 + ((sqrt{9})! times 6) + 8$
$44 = 1 times ((sqrt{9})! times 6) + 8$
$45 = 1 + ((sqrt{9})! times 6) + 8$
$ 55 = 1 + (sqrt{9})! + (6 times 8)$
$56 = (1^9 + 6) times 8 $
$ 64 = ((-1 + sqrt{9}) + 6) times 8$
$65 = -(1 times sqrt{9}) + 68 $
$66 = 1 - sqrt{9} + 68 $
$69 = 1^9 + 68$
$70 = -1 + sqrt{9} + 68 $
$71 = -1 + ((sqrt{9} + 6)times 8)$
$72 = ((1 times sqrt{9}) + 6) times 8$
$73 = 1 + ((sqrt{9} + 6)times 8) $
$74 = (1 times (sqrt{9})!) + 68$
$75 = 1 + (sqrt{9})!) + 68$
$ 79 = -1 + ((9 + sqrt{sqrt{ldots sqrt{6}}}) times 8)$
$ 86 = -1 - sqrt{9} + (6!/8)$
$90 = (1^9 times 6!)/8 $
$91 = 1^9 + (6!/8)$
$endgroup$
Here are all the numbers which have not been obtained already by trolley813, Omega Krypton and ppgdev
$28 = 1 times ((sqrt{9})! times 6) - 8$
$30 =-(1times(sqrt{9})!) + sqrt{sqrt{6^8}} $
$33 = 19 + 6 + 8$
$34 = ((1+(sqrt{9})!) times 6) - 8 $
$35 = (1 times sqrt{(sqrt{9})^6}) + 8$
$36 = (1 + sqrt{(sqrt{9})^6}) + 8$
$37 = 1^9 + sqrt{sqrt{6^8}}$
$38 = (1+sqrt{9})! + 6 + 8$
$41 = -(1 + (sqrt{9})!) + (6 times 8)$
$42 = -(1 times (sqrt{9})!) + (6 times 8)$
$43 = -1 + ((sqrt{9})! times 6) + 8$
$44 = 1 times ((sqrt{9})! times 6) + 8$
$45 = 1 + ((sqrt{9})! times 6) + 8$
$ 55 = 1 + (sqrt{9})! + (6 times 8)$
$56 = (1^9 + 6) times 8 $
$ 64 = ((-1 + sqrt{9}) + 6) times 8$
$65 = -(1 times sqrt{9}) + 68 $
$66 = 1 - sqrt{9} + 68 $
$69 = 1^9 + 68$
$70 = -1 + sqrt{9} + 68 $
$71 = -1 + ((sqrt{9} + 6)times 8)$
$72 = ((1 times sqrt{9}) + 6) times 8$
$73 = 1 + ((sqrt{9} + 6)times 8) $
$74 = (1 times (sqrt{9})!) + 68$
$75 = 1 + (sqrt{9})!) + 68$
$ 79 = -1 + ((9 + sqrt{sqrt{ldots sqrt{6}}}) times 8)$
$ 86 = -1 - sqrt{9} + (6!/8)$
$90 = (1^9 times 6!)/8 $
$91 = 1^9 + (6!/8)$
edited 12 hours ago
answered 12 hours ago
hexominohexomino
41k3122191
41k3122191
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@Oray which numbers are impossible?
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– Auroxa
7 hours ago
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nice one @Auroxa
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– Omega Krypton
7 hours ago
add a comment |
$begingroup$
@Oray which numbers are impossible?
$endgroup$
– Auroxa
7 hours ago
$begingroup$
nice one @Auroxa
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– Omega Krypton
7 hours ago
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@Oray which numbers are impossible?
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– Auroxa
7 hours ago
$begingroup$
@Oray which numbers are impossible?
$endgroup$
– Auroxa
7 hours ago
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nice one @Auroxa
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– Omega Krypton
7 hours ago
$begingroup$
nice one @Auroxa
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– Omega Krypton
7 hours ago
add a comment |
$begingroup$
Here is an answer for 39 that was missing so far:
39 = -(1*9) + (6*8)
I also got 79
79 = ((-1 + (sqrt(9)!))!! + (8!!)/6
New contributor
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$begingroup$
@Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
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– ppgdev
7 hours ago
add a comment |
$begingroup$
Here is an answer for 39 that was missing so far:
39 = -(1*9) + (6*8)
I also got 79
79 = ((-1 + (sqrt(9)!))!! + (8!!)/6
New contributor
$endgroup$
$begingroup$
@Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
$endgroup$
– ppgdev
7 hours ago
add a comment |
$begingroup$
Here is an answer for 39 that was missing so far:
39 = -(1*9) + (6*8)
I also got 79
79 = ((-1 + (sqrt(9)!))!! + (8!!)/6
New contributor
$endgroup$
Here is an answer for 39 that was missing so far:
39 = -(1*9) + (6*8)
I also got 79
79 = ((-1 + (sqrt(9)!))!! + (8!!)/6
New contributor
edited 9 hours ago
New contributor
answered 12 hours ago
ppgdevppgdev
314
314
New contributor
New contributor
$begingroup$
@Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
$endgroup$
– ppgdev
7 hours ago
add a comment |
$begingroup$
@Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
$endgroup$
– ppgdev
7 hours ago
$begingroup$
@Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
$endgroup$
– ppgdev
7 hours ago
$begingroup$
@Auroxa for my 79 answer I assumed double factorial (!!) as defined on Wolfram Mathworld is allowed.
$endgroup$
– ppgdev
7 hours ago
add a comment |
Auroxa is a new contributor. Be nice, and check out our Code of Conduct.
Auroxa is a new contributor. Be nice, and check out our Code of Conduct.
Auroxa is a new contributor. Be nice, and check out our Code of Conduct.
Auroxa is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
it is obvious it is impossible to find all numbers from 1 to 100 with these rules :)
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– Oray
20 hours ago
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@Oray can you please prove it then? thanks!
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– Omega Krypton
16 hours ago
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@OmegaKrypton i put it into my code many numbers cannot be achieved.
$endgroup$
– Oray
16 hours ago
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@Oray I think the only number we have not found now is $79$.
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– hexomino
12 hours ago