How do I make numbers 50-100 using only the numbers 2, 0, 1, 9?
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Rules
Use ALL the digits in the year $2019$ (you may not use any other numbers except $2, 0, 1, 9$) to write mathematical expressions that give results for the numbers 50 to 100.
You may use the arithmetic operations $+$, $-$, $times$, $sqrt{}$, and $!$ (see below).
Indices or exponents may only be made from the digits $2, 0, 1,$ and $9$; for example, $(9+1)^2 $is allowed, as it has used the $9$, $1$ and $2$. Multi-digit numbers and decimals points can be used such as $20$, $102$, and $.02$, but you CANNOT make 30 by combining $(2+1)0$.
Recurring decimals can be used using the overhead dots or bar e.g. $0.bar1=0.111 ...=frac19$
Factorials are allowed
Here's how you might use factorials:
$n!=ntimes(n-1)times(n-2)timesdotstimes2times1$
For example
$(10-9+2)!=3!=3times2times1=6$
$0!=1$
calculation-puzzle
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add a comment |
$begingroup$
Rules
Use ALL the digits in the year $2019$ (you may not use any other numbers except $2, 0, 1, 9$) to write mathematical expressions that give results for the numbers 50 to 100.
You may use the arithmetic operations $+$, $-$, $times$, $sqrt{}$, and $!$ (see below).
Indices or exponents may only be made from the digits $2, 0, 1,$ and $9$; for example, $(9+1)^2 $is allowed, as it has used the $9$, $1$ and $2$. Multi-digit numbers and decimals points can be used such as $20$, $102$, and $.02$, but you CANNOT make 30 by combining $(2+1)0$.
Recurring decimals can be used using the overhead dots or bar e.g. $0.bar1=0.111 ...=frac19$
Factorials are allowed
Here's how you might use factorials:
$n!=ntimes(n-1)times(n-2)timesdotstimes2times1$
For example
$(10-9+2)!=3!=3times2times1=6$
$0!=1$
calculation-puzzle
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So...can we use the square root operator AND the cube root operator? I know that there is a three in the cube root, but there is also an implied two in square root.
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– Brandon_J
Jan 9 at 20:23
add a comment |
$begingroup$
Rules
Use ALL the digits in the year $2019$ (you may not use any other numbers except $2, 0, 1, 9$) to write mathematical expressions that give results for the numbers 50 to 100.
You may use the arithmetic operations $+$, $-$, $times$, $sqrt{}$, and $!$ (see below).
Indices or exponents may only be made from the digits $2, 0, 1,$ and $9$; for example, $(9+1)^2 $is allowed, as it has used the $9$, $1$ and $2$. Multi-digit numbers and decimals points can be used such as $20$, $102$, and $.02$, but you CANNOT make 30 by combining $(2+1)0$.
Recurring decimals can be used using the overhead dots or bar e.g. $0.bar1=0.111 ...=frac19$
Factorials are allowed
Here's how you might use factorials:
$n!=ntimes(n-1)times(n-2)timesdotstimes2times1$
For example
$(10-9+2)!=3!=3times2times1=6$
$0!=1$
calculation-puzzle
$endgroup$
Rules
Use ALL the digits in the year $2019$ (you may not use any other numbers except $2, 0, 1, 9$) to write mathematical expressions that give results for the numbers 50 to 100.
You may use the arithmetic operations $+$, $-$, $times$, $sqrt{}$, and $!$ (see below).
Indices or exponents may only be made from the digits $2, 0, 1,$ and $9$; for example, $(9+1)^2 $is allowed, as it has used the $9$, $1$ and $2$. Multi-digit numbers and decimals points can be used such as $20$, $102$, and $.02$, but you CANNOT make 30 by combining $(2+1)0$.
Recurring decimals can be used using the overhead dots or bar e.g. $0.bar1=0.111 ...=frac19$
Factorials are allowed
Here's how you might use factorials:
$n!=ntimes(n-1)times(n-2)timesdotstimes2times1$
For example
$(10-9+2)!=3!=3times2times1=6$
$0!=1$
calculation-puzzle
calculation-puzzle
edited Jan 9 at 19:23
Sesquipedalian
357
357
asked Jan 9 at 14:10
jamalewisjamalewis
261
261
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So...can we use the square root operator AND the cube root operator? I know that there is a three in the cube root, but there is also an implied two in square root.
$endgroup$
– Brandon_J
Jan 9 at 20:23
add a comment |
$begingroup$
So...can we use the square root operator AND the cube root operator? I know that there is a three in the cube root, but there is also an implied two in square root.
$endgroup$
– Brandon_J
Jan 9 at 20:23
$begingroup$
So...can we use the square root operator AND the cube root operator? I know that there is a three in the cube root, but there is also an implied two in square root.
$endgroup$
– Brandon_J
Jan 9 at 20:23
$begingroup$
So...can we use the square root operator AND the cube root operator? I know that there is a three in the cube root, but there is also an implied two in square root.
$endgroup$
– Brandon_J
Jan 9 at 20:23
add a comment |
5 Answers
5
active
oldest
votes
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Here are answers for all numbers. I'm going to try to find better answers for 76 and 86 because not only is it dumb trying to read multifactorials, but Wolfram Alpha can't even calculate anything past double factorials.
50: $10 * (sqrt9 + 2)$
51: $((sqrt9)!)!! + 2 + 1 + 0$
52: $((sqrt9)!)!! + 2 + 1 + 0!$
53: $9 * (2 + 1)! - 0!$
54: $9 * (2 + 1)! + 0$
55: $9 * (2 + 1)! + 0!$
56: $((sqrt9)!)!! + (2 + 1 + 0!)!!$
57: $19 * (2 + 0!)$
58: $10 * (sqrt9)! - 2$
59: $20 * sqrt9 - 1$
60: $20 * sqrt9 * 1$
61: $20 * sqrt9 + 1$
62: $10 * (sqrt9)! + 2$
63: $21 * sqrt9 + 0$
64: $(9 - 1)^2 + 0$
65: $2^{(sqrt9)!} + 1 + 0$
66: $2^{(sqrt9)!} + 1 + 0!$
67: $((sqrt9)!)!! + 20 - 1$
68: $((sqrt9)!)!! + 21 - 0!$
69: $90 - 21$
70: $10 * (9 - 2)$
71: $91 - 20$
72: $12 * (sqrt9)! + 0$
73: $12 * (sqrt9)! + 0!$
74: $((sqrt9)!)! * .1 + 2 + 0$
75: $((sqrt9)!)! * .1 + 2 + 0!$
76: $21!!!!!!!!!!!!!!!!! - 9 + 0!$
77: $((sqrt9)!)! * (.bar1) - 2 - 0!$
78: $90 - 12$
79: $9^2 - 1 - 0!$
80: $9^2 - 1 + 0$
81: $9^2 + (1 * 0)$
82: $9^2 + 1 + 0$
83: $9^2 + 1 + 0!$
84: $90 - (2 + 1)!$
85: $91 - (2 + 0!)!$
86: $(20 - 1)!!!!!!!!!!!!!! - 9$
87: $90 - 2 - 1$
88: $90 - (2 * 1)$
89: $90 - 2 + 1$
90: $92 - 1 - 0!$
91: $91 + (2 * 0)$
92: $92 + (1 * 0)$
93: $92 + 1 + 0$
94: $92 + 1 + 0!$
95: $((sqrt9)!)!! * 2 - 1 + 0$
96: $90 + (2 + 1)!$
97: $91 + (2 + 0!)!$
98: $((sqrt9)!)!! * 2 + 1 + 0!$
99: $(9 + 1)^2 - 0!$
100: $(9 + 1)^2 + 0$
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Yes, apologies. Though reminding me of double factorials opens up some new possibilities I hadn't thought of...
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– Meerkat
Jan 10 at 18:03
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Yes, 51 = $((sqrt 9)!)!! + 2 + 1 + 0$
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– Weather Vane
Jan 10 at 18:08
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Also solved 67 and 68, see my answer.
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– Weather Vane
Jan 10 at 18:36
add a comment |
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Here are two more solutions:
67 = $((sqrt 9)!)!! + 20 - 1$
68 = $((sqrt 9)!)!! + 21 - 0!$
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1
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This is much more elegant than my answer for those two were going to be. I had $67 = 12!!!!!!! + (sqrt9)! + 0!$ and $68 = (12 + 0!)!!!!!!!! + sqrt9$, which are terribly unreadable.
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– Meerkat
Jan 10 at 18:46
add a comment |
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some solutions without double factorials
51:
$ .02^{-1} + .bar9 $
52:
$ (sqrt9)! * (.bar1)^{-0!} - 2 $
56:
$ .02^{-1} + (sqrt9)! $
67:
$ ((sqrt9)!)! * .1 - .2^{-0!} $
76:
$ .bar1^{-2} - (sqrt9)! + 0! $
86:
$ 91 - .2^{-0!} $
95:
$ 19 * .2^{-0!} $
98:
$ .1^{-2} + 0! - sqrt9 $
New contributor
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add a comment |
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Hope few of my solutions (not mentioned before) will help you. It's not an easy task and I'm curious if it's even possible to come up with each of them.
54:
$54 = 9*1*(2+0!)!$
64:
$64 = (9-1)^2+0$
71:
$71 = 91 - 20$
76:
$76 = 19 * (2 + 1 + 0!)$
81:
$81 = sqrt{9}^{(2^{(0!+1)})}$
82:
$82 = 92 - 10$
95:
$95 = 190 / 2$
99:
$99 = (9+1)-0!$
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76 uses1
twice, and division isn't allowed, so95 = 190 / 2
won't work either. The rest do work, however.
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– Meerkat
Jan 9 at 15:55
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The last one is missing $2$ possibly just a typo.
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– Weather Vane
Jan 9 at 18:25
add a comment |
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Partial, only a few numbers so far:
50 =
51 =
52 =
53 =
54 =
55 =
56 =
57 = $19 * (2+0!)$
58 = $29 * (1+0!)$
59 =
60 =
61 =
62 =
63 =
64 =
65 =
66 =
67 =
68 =
69 =
70 = $(9-2) * 10$
71 = $91 - 20$
72 =
73 =
74 =
75 =
76 =
77 =
78 =
79 = $91 - (0! + 2)$
80 =
81 =
82 = $92 - 10$
83 =
84 = $90 - (1 + 2)!$
85 =
86 =
87 = $90 - 1 - 2$
88 = $91 - 2 - 0!$
89 = $91 - 2 - 0$
90 = $92 - (1+0!)$
91 = $91 + 0 * 2$
92 = $92 + 1 * 0$
93 = $90 + 1 + 2$
94 = $91 + 2 + 0!$
95 =
96 = $90 + (1 + 2)!$
97 = $91 + (0! + 2)!$
98 =
99 =
100 =
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1
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86 = 90 - 2 * (1+0!)
doesn't work, as you use0
twice in the formula.
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– Meerkat
Jan 9 at 16:41
1
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79 doesn't work, you use 1 twice. 78 would though 90-12 instead of 91-12
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– Yout Ried
Jan 11 at 2:09
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here are answers for all numbers. I'm going to try to find better answers for 76 and 86 because not only is it dumb trying to read multifactorials, but Wolfram Alpha can't even calculate anything past double factorials.
50: $10 * (sqrt9 + 2)$
51: $((sqrt9)!)!! + 2 + 1 + 0$
52: $((sqrt9)!)!! + 2 + 1 + 0!$
53: $9 * (2 + 1)! - 0!$
54: $9 * (2 + 1)! + 0$
55: $9 * (2 + 1)! + 0!$
56: $((sqrt9)!)!! + (2 + 1 + 0!)!!$
57: $19 * (2 + 0!)$
58: $10 * (sqrt9)! - 2$
59: $20 * sqrt9 - 1$
60: $20 * sqrt9 * 1$
61: $20 * sqrt9 + 1$
62: $10 * (sqrt9)! + 2$
63: $21 * sqrt9 + 0$
64: $(9 - 1)^2 + 0$
65: $2^{(sqrt9)!} + 1 + 0$
66: $2^{(sqrt9)!} + 1 + 0!$
67: $((sqrt9)!)!! + 20 - 1$
68: $((sqrt9)!)!! + 21 - 0!$
69: $90 - 21$
70: $10 * (9 - 2)$
71: $91 - 20$
72: $12 * (sqrt9)! + 0$
73: $12 * (sqrt9)! + 0!$
74: $((sqrt9)!)! * .1 + 2 + 0$
75: $((sqrt9)!)! * .1 + 2 + 0!$
76: $21!!!!!!!!!!!!!!!!! - 9 + 0!$
77: $((sqrt9)!)! * (.bar1) - 2 - 0!$
78: $90 - 12$
79: $9^2 - 1 - 0!$
80: $9^2 - 1 + 0$
81: $9^2 + (1 * 0)$
82: $9^2 + 1 + 0$
83: $9^2 + 1 + 0!$
84: $90 - (2 + 1)!$
85: $91 - (2 + 0!)!$
86: $(20 - 1)!!!!!!!!!!!!!! - 9$
87: $90 - 2 - 1$
88: $90 - (2 * 1)$
89: $90 - 2 + 1$
90: $92 - 1 - 0!$
91: $91 + (2 * 0)$
92: $92 + (1 * 0)$
93: $92 + 1 + 0$
94: $92 + 1 + 0!$
95: $((sqrt9)!)!! * 2 - 1 + 0$
96: $90 + (2 + 1)!$
97: $91 + (2 + 0!)!$
98: $((sqrt9)!)!! * 2 + 1 + 0!$
99: $(9 + 1)^2 - 0!$
100: $(9 + 1)^2 + 0$
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Yes, apologies. Though reminding me of double factorials opens up some new possibilities I hadn't thought of...
$endgroup$
– Meerkat
Jan 10 at 18:03
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Yes, 51 = $((sqrt 9)!)!! + 2 + 1 + 0$
$endgroup$
– Weather Vane
Jan 10 at 18:08
$begingroup$
Also solved 67 and 68, see my answer.
$endgroup$
– Weather Vane
Jan 10 at 18:36
add a comment |
$begingroup$
Here are answers for all numbers. I'm going to try to find better answers for 76 and 86 because not only is it dumb trying to read multifactorials, but Wolfram Alpha can't even calculate anything past double factorials.
50: $10 * (sqrt9 + 2)$
51: $((sqrt9)!)!! + 2 + 1 + 0$
52: $((sqrt9)!)!! + 2 + 1 + 0!$
53: $9 * (2 + 1)! - 0!$
54: $9 * (2 + 1)! + 0$
55: $9 * (2 + 1)! + 0!$
56: $((sqrt9)!)!! + (2 + 1 + 0!)!!$
57: $19 * (2 + 0!)$
58: $10 * (sqrt9)! - 2$
59: $20 * sqrt9 - 1$
60: $20 * sqrt9 * 1$
61: $20 * sqrt9 + 1$
62: $10 * (sqrt9)! + 2$
63: $21 * sqrt9 + 0$
64: $(9 - 1)^2 + 0$
65: $2^{(sqrt9)!} + 1 + 0$
66: $2^{(sqrt9)!} + 1 + 0!$
67: $((sqrt9)!)!! + 20 - 1$
68: $((sqrt9)!)!! + 21 - 0!$
69: $90 - 21$
70: $10 * (9 - 2)$
71: $91 - 20$
72: $12 * (sqrt9)! + 0$
73: $12 * (sqrt9)! + 0!$
74: $((sqrt9)!)! * .1 + 2 + 0$
75: $((sqrt9)!)! * .1 + 2 + 0!$
76: $21!!!!!!!!!!!!!!!!! - 9 + 0!$
77: $((sqrt9)!)! * (.bar1) - 2 - 0!$
78: $90 - 12$
79: $9^2 - 1 - 0!$
80: $9^2 - 1 + 0$
81: $9^2 + (1 * 0)$
82: $9^2 + 1 + 0$
83: $9^2 + 1 + 0!$
84: $90 - (2 + 1)!$
85: $91 - (2 + 0!)!$
86: $(20 - 1)!!!!!!!!!!!!!! - 9$
87: $90 - 2 - 1$
88: $90 - (2 * 1)$
89: $90 - 2 + 1$
90: $92 - 1 - 0!$
91: $91 + (2 * 0)$
92: $92 + (1 * 0)$
93: $92 + 1 + 0$
94: $92 + 1 + 0!$
95: $((sqrt9)!)!! * 2 - 1 + 0$
96: $90 + (2 + 1)!$
97: $91 + (2 + 0!)!$
98: $((sqrt9)!)!! * 2 + 1 + 0!$
99: $(9 + 1)^2 - 0!$
100: $(9 + 1)^2 + 0$
$endgroup$
$begingroup$
Yes, apologies. Though reminding me of double factorials opens up some new possibilities I hadn't thought of...
$endgroup$
– Meerkat
Jan 10 at 18:03
$begingroup$
Yes, 51 = $((sqrt 9)!)!! + 2 + 1 + 0$
$endgroup$
– Weather Vane
Jan 10 at 18:08
$begingroup$
Also solved 67 and 68, see my answer.
$endgroup$
– Weather Vane
Jan 10 at 18:36
add a comment |
$begingroup$
Here are answers for all numbers. I'm going to try to find better answers for 76 and 86 because not only is it dumb trying to read multifactorials, but Wolfram Alpha can't even calculate anything past double factorials.
50: $10 * (sqrt9 + 2)$
51: $((sqrt9)!)!! + 2 + 1 + 0$
52: $((sqrt9)!)!! + 2 + 1 + 0!$
53: $9 * (2 + 1)! - 0!$
54: $9 * (2 + 1)! + 0$
55: $9 * (2 + 1)! + 0!$
56: $((sqrt9)!)!! + (2 + 1 + 0!)!!$
57: $19 * (2 + 0!)$
58: $10 * (sqrt9)! - 2$
59: $20 * sqrt9 - 1$
60: $20 * sqrt9 * 1$
61: $20 * sqrt9 + 1$
62: $10 * (sqrt9)! + 2$
63: $21 * sqrt9 + 0$
64: $(9 - 1)^2 + 0$
65: $2^{(sqrt9)!} + 1 + 0$
66: $2^{(sqrt9)!} + 1 + 0!$
67: $((sqrt9)!)!! + 20 - 1$
68: $((sqrt9)!)!! + 21 - 0!$
69: $90 - 21$
70: $10 * (9 - 2)$
71: $91 - 20$
72: $12 * (sqrt9)! + 0$
73: $12 * (sqrt9)! + 0!$
74: $((sqrt9)!)! * .1 + 2 + 0$
75: $((sqrt9)!)! * .1 + 2 + 0!$
76: $21!!!!!!!!!!!!!!!!! - 9 + 0!$
77: $((sqrt9)!)! * (.bar1) - 2 - 0!$
78: $90 - 12$
79: $9^2 - 1 - 0!$
80: $9^2 - 1 + 0$
81: $9^2 + (1 * 0)$
82: $9^2 + 1 + 0$
83: $9^2 + 1 + 0!$
84: $90 - (2 + 1)!$
85: $91 - (2 + 0!)!$
86: $(20 - 1)!!!!!!!!!!!!!! - 9$
87: $90 - 2 - 1$
88: $90 - (2 * 1)$
89: $90 - 2 + 1$
90: $92 - 1 - 0!$
91: $91 + (2 * 0)$
92: $92 + (1 * 0)$
93: $92 + 1 + 0$
94: $92 + 1 + 0!$
95: $((sqrt9)!)!! * 2 - 1 + 0$
96: $90 + (2 + 1)!$
97: $91 + (2 + 0!)!$
98: $((sqrt9)!)!! * 2 + 1 + 0!$
99: $(9 + 1)^2 - 0!$
100: $(9 + 1)^2 + 0$
$endgroup$
Here are answers for all numbers. I'm going to try to find better answers for 76 and 86 because not only is it dumb trying to read multifactorials, but Wolfram Alpha can't even calculate anything past double factorials.
50: $10 * (sqrt9 + 2)$
51: $((sqrt9)!)!! + 2 + 1 + 0$
52: $((sqrt9)!)!! + 2 + 1 + 0!$
53: $9 * (2 + 1)! - 0!$
54: $9 * (2 + 1)! + 0$
55: $9 * (2 + 1)! + 0!$
56: $((sqrt9)!)!! + (2 + 1 + 0!)!!$
57: $19 * (2 + 0!)$
58: $10 * (sqrt9)! - 2$
59: $20 * sqrt9 - 1$
60: $20 * sqrt9 * 1$
61: $20 * sqrt9 + 1$
62: $10 * (sqrt9)! + 2$
63: $21 * sqrt9 + 0$
64: $(9 - 1)^2 + 0$
65: $2^{(sqrt9)!} + 1 + 0$
66: $2^{(sqrt9)!} + 1 + 0!$
67: $((sqrt9)!)!! + 20 - 1$
68: $((sqrt9)!)!! + 21 - 0!$
69: $90 - 21$
70: $10 * (9 - 2)$
71: $91 - 20$
72: $12 * (sqrt9)! + 0$
73: $12 * (sqrt9)! + 0!$
74: $((sqrt9)!)! * .1 + 2 + 0$
75: $((sqrt9)!)! * .1 + 2 + 0!$
76: $21!!!!!!!!!!!!!!!!! - 9 + 0!$
77: $((sqrt9)!)! * (.bar1) - 2 - 0!$
78: $90 - 12$
79: $9^2 - 1 - 0!$
80: $9^2 - 1 + 0$
81: $9^2 + (1 * 0)$
82: $9^2 + 1 + 0$
83: $9^2 + 1 + 0!$
84: $90 - (2 + 1)!$
85: $91 - (2 + 0!)!$
86: $(20 - 1)!!!!!!!!!!!!!! - 9$
87: $90 - 2 - 1$
88: $90 - (2 * 1)$
89: $90 - 2 + 1$
90: $92 - 1 - 0!$
91: $91 + (2 * 0)$
92: $92 + (1 * 0)$
93: $92 + 1 + 0$
94: $92 + 1 + 0!$
95: $((sqrt9)!)!! * 2 - 1 + 0$
96: $90 + (2 + 1)!$
97: $91 + (2 + 0!)!$
98: $((sqrt9)!)!! * 2 + 1 + 0!$
99: $(9 + 1)^2 - 0!$
100: $(9 + 1)^2 + 0$
edited Jan 10 at 18:49
answered Jan 9 at 16:04
MeerkatMeerkat
35011
35011
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Yes, apologies. Though reminding me of double factorials opens up some new possibilities I hadn't thought of...
$endgroup$
– Meerkat
Jan 10 at 18:03
$begingroup$
Yes, 51 = $((sqrt 9)!)!! + 2 + 1 + 0$
$endgroup$
– Weather Vane
Jan 10 at 18:08
$begingroup$
Also solved 67 and 68, see my answer.
$endgroup$
– Weather Vane
Jan 10 at 18:36
add a comment |
$begingroup$
Yes, apologies. Though reminding me of double factorials opens up some new possibilities I hadn't thought of...
$endgroup$
– Meerkat
Jan 10 at 18:03
$begingroup$
Yes, 51 = $((sqrt 9)!)!! + 2 + 1 + 0$
$endgroup$
– Weather Vane
Jan 10 at 18:08
$begingroup$
Also solved 67 and 68, see my answer.
$endgroup$
– Weather Vane
Jan 10 at 18:36
$begingroup$
Yes, apologies. Though reminding me of double factorials opens up some new possibilities I hadn't thought of...
$endgroup$
– Meerkat
Jan 10 at 18:03
$begingroup$
Yes, apologies. Though reminding me of double factorials opens up some new possibilities I hadn't thought of...
$endgroup$
– Meerkat
Jan 10 at 18:03
$begingroup$
Yes, 51 = $((sqrt 9)!)!! + 2 + 1 + 0$
$endgroup$
– Weather Vane
Jan 10 at 18:08
$begingroup$
Yes, 51 = $((sqrt 9)!)!! + 2 + 1 + 0$
$endgroup$
– Weather Vane
Jan 10 at 18:08
$begingroup$
Also solved 67 and 68, see my answer.
$endgroup$
– Weather Vane
Jan 10 at 18:36
$begingroup$
Also solved 67 and 68, see my answer.
$endgroup$
– Weather Vane
Jan 10 at 18:36
add a comment |
$begingroup$
Here are two more solutions:
67 = $((sqrt 9)!)!! + 20 - 1$
68 = $((sqrt 9)!)!! + 21 - 0!$
$endgroup$
1
$begingroup$
This is much more elegant than my answer for those two were going to be. I had $67 = 12!!!!!!! + (sqrt9)! + 0!$ and $68 = (12 + 0!)!!!!!!!! + sqrt9$, which are terribly unreadable.
$endgroup$
– Meerkat
Jan 10 at 18:46
add a comment |
$begingroup$
Here are two more solutions:
67 = $((sqrt 9)!)!! + 20 - 1$
68 = $((sqrt 9)!)!! + 21 - 0!$
$endgroup$
1
$begingroup$
This is much more elegant than my answer for those two were going to be. I had $67 = 12!!!!!!! + (sqrt9)! + 0!$ and $68 = (12 + 0!)!!!!!!!! + sqrt9$, which are terribly unreadable.
$endgroup$
– Meerkat
Jan 10 at 18:46
add a comment |
$begingroup$
Here are two more solutions:
67 = $((sqrt 9)!)!! + 20 - 1$
68 = $((sqrt 9)!)!! + 21 - 0!$
$endgroup$
Here are two more solutions:
67 = $((sqrt 9)!)!! + 20 - 1$
68 = $((sqrt 9)!)!! + 21 - 0!$
answered Jan 10 at 18:35
Weather VaneWeather Vane
1,17719
1,17719
1
$begingroup$
This is much more elegant than my answer for those two were going to be. I had $67 = 12!!!!!!! + (sqrt9)! + 0!$ and $68 = (12 + 0!)!!!!!!!! + sqrt9$, which are terribly unreadable.
$endgroup$
– Meerkat
Jan 10 at 18:46
add a comment |
1
$begingroup$
This is much more elegant than my answer for those two were going to be. I had $67 = 12!!!!!!! + (sqrt9)! + 0!$ and $68 = (12 + 0!)!!!!!!!! + sqrt9$, which are terribly unreadable.
$endgroup$
– Meerkat
Jan 10 at 18:46
1
1
$begingroup$
This is much more elegant than my answer for those two were going to be. I had $67 = 12!!!!!!! + (sqrt9)! + 0!$ and $68 = (12 + 0!)!!!!!!!! + sqrt9$, which are terribly unreadable.
$endgroup$
– Meerkat
Jan 10 at 18:46
$begingroup$
This is much more elegant than my answer for those two were going to be. I had $67 = 12!!!!!!! + (sqrt9)! + 0!$ and $68 = (12 + 0!)!!!!!!!! + sqrt9$, which are terribly unreadable.
$endgroup$
– Meerkat
Jan 10 at 18:46
add a comment |
$begingroup$
some solutions without double factorials
51:
$ .02^{-1} + .bar9 $
52:
$ (sqrt9)! * (.bar1)^{-0!} - 2 $
56:
$ .02^{-1} + (sqrt9)! $
67:
$ ((sqrt9)!)! * .1 - .2^{-0!} $
76:
$ .bar1^{-2} - (sqrt9)! + 0! $
86:
$ 91 - .2^{-0!} $
95:
$ 19 * .2^{-0!} $
98:
$ .1^{-2} + 0! - sqrt9 $
New contributor
$endgroup$
add a comment |
$begingroup$
some solutions without double factorials
51:
$ .02^{-1} + .bar9 $
52:
$ (sqrt9)! * (.bar1)^{-0!} - 2 $
56:
$ .02^{-1} + (sqrt9)! $
67:
$ ((sqrt9)!)! * .1 - .2^{-0!} $
76:
$ .bar1^{-2} - (sqrt9)! + 0! $
86:
$ 91 - .2^{-0!} $
95:
$ 19 * .2^{-0!} $
98:
$ .1^{-2} + 0! - sqrt9 $
New contributor
$endgroup$
add a comment |
$begingroup$
some solutions without double factorials
51:
$ .02^{-1} + .bar9 $
52:
$ (sqrt9)! * (.bar1)^{-0!} - 2 $
56:
$ .02^{-1} + (sqrt9)! $
67:
$ ((sqrt9)!)! * .1 - .2^{-0!} $
76:
$ .bar1^{-2} - (sqrt9)! + 0! $
86:
$ 91 - .2^{-0!} $
95:
$ 19 * .2^{-0!} $
98:
$ .1^{-2} + 0! - sqrt9 $
New contributor
$endgroup$
some solutions without double factorials
51:
$ .02^{-1} + .bar9 $
52:
$ (sqrt9)! * (.bar1)^{-0!} - 2 $
56:
$ .02^{-1} + (sqrt9)! $
67:
$ ((sqrt9)!)! * .1 - .2^{-0!} $
76:
$ .bar1^{-2} - (sqrt9)! + 0! $
86:
$ 91 - .2^{-0!} $
95:
$ 19 * .2^{-0!} $
98:
$ .1^{-2} + 0! - sqrt9 $
New contributor
edited 8 hours ago
New contributor
answered Jan 12 at 4:22
Jonathan LangJonathan Lang
213
213
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
Hope few of my solutions (not mentioned before) will help you. It's not an easy task and I'm curious if it's even possible to come up with each of them.
54:
$54 = 9*1*(2+0!)!$
64:
$64 = (9-1)^2+0$
71:
$71 = 91 - 20$
76:
$76 = 19 * (2 + 1 + 0!)$
81:
$81 = sqrt{9}^{(2^{(0!+1)})}$
82:
$82 = 92 - 10$
95:
$95 = 190 / 2$
99:
$99 = (9+1)-0!$
$endgroup$
$begingroup$
76 uses1
twice, and division isn't allowed, so95 = 190 / 2
won't work either. The rest do work, however.
$endgroup$
– Meerkat
Jan 9 at 15:55
$begingroup$
The last one is missing $2$ possibly just a typo.
$endgroup$
– Weather Vane
Jan 9 at 18:25
add a comment |
$begingroup$
Hope few of my solutions (not mentioned before) will help you. It's not an easy task and I'm curious if it's even possible to come up with each of them.
54:
$54 = 9*1*(2+0!)!$
64:
$64 = (9-1)^2+0$
71:
$71 = 91 - 20$
76:
$76 = 19 * (2 + 1 + 0!)$
81:
$81 = sqrt{9}^{(2^{(0!+1)})}$
82:
$82 = 92 - 10$
95:
$95 = 190 / 2$
99:
$99 = (9+1)-0!$
$endgroup$
$begingroup$
76 uses1
twice, and division isn't allowed, so95 = 190 / 2
won't work either. The rest do work, however.
$endgroup$
– Meerkat
Jan 9 at 15:55
$begingroup$
The last one is missing $2$ possibly just a typo.
$endgroup$
– Weather Vane
Jan 9 at 18:25
add a comment |
$begingroup$
Hope few of my solutions (not mentioned before) will help you. It's not an easy task and I'm curious if it's even possible to come up with each of them.
54:
$54 = 9*1*(2+0!)!$
64:
$64 = (9-1)^2+0$
71:
$71 = 91 - 20$
76:
$76 = 19 * (2 + 1 + 0!)$
81:
$81 = sqrt{9}^{(2^{(0!+1)})}$
82:
$82 = 92 - 10$
95:
$95 = 190 / 2$
99:
$99 = (9+1)-0!$
$endgroup$
Hope few of my solutions (not mentioned before) will help you. It's not an easy task and I'm curious if it's even possible to come up with each of them.
54:
$54 = 9*1*(2+0!)!$
64:
$64 = (9-1)^2+0$
71:
$71 = 91 - 20$
76:
$76 = 19 * (2 + 1 + 0!)$
81:
$81 = sqrt{9}^{(2^{(0!+1)})}$
82:
$82 = 92 - 10$
95:
$95 = 190 / 2$
99:
$99 = (9+1)-0!$
edited Jan 9 at 20:38
a stone arachnid
1707
1707
answered Jan 9 at 15:47
Rafał KlatRafał Klat
11
11
$begingroup$
76 uses1
twice, and division isn't allowed, so95 = 190 / 2
won't work either. The rest do work, however.
$endgroup$
– Meerkat
Jan 9 at 15:55
$begingroup$
The last one is missing $2$ possibly just a typo.
$endgroup$
– Weather Vane
Jan 9 at 18:25
add a comment |
$begingroup$
76 uses1
twice, and division isn't allowed, so95 = 190 / 2
won't work either. The rest do work, however.
$endgroup$
– Meerkat
Jan 9 at 15:55
$begingroup$
The last one is missing $2$ possibly just a typo.
$endgroup$
– Weather Vane
Jan 9 at 18:25
$begingroup$
76 uses
1
twice, and division isn't allowed, so 95 = 190 / 2
won't work either. The rest do work, however.$endgroup$
– Meerkat
Jan 9 at 15:55
$begingroup$
76 uses
1
twice, and division isn't allowed, so 95 = 190 / 2
won't work either. The rest do work, however.$endgroup$
– Meerkat
Jan 9 at 15:55
$begingroup$
The last one is missing $2$ possibly just a typo.
$endgroup$
– Weather Vane
Jan 9 at 18:25
$begingroup$
The last one is missing $2$ possibly just a typo.
$endgroup$
– Weather Vane
Jan 9 at 18:25
add a comment |
$begingroup$
Partial, only a few numbers so far:
50 =
51 =
52 =
53 =
54 =
55 =
56 =
57 = $19 * (2+0!)$
58 = $29 * (1+0!)$
59 =
60 =
61 =
62 =
63 =
64 =
65 =
66 =
67 =
68 =
69 =
70 = $(9-2) * 10$
71 = $91 - 20$
72 =
73 =
74 =
75 =
76 =
77 =
78 =
79 = $91 - (0! + 2)$
80 =
81 =
82 = $92 - 10$
83 =
84 = $90 - (1 + 2)!$
85 =
86 =
87 = $90 - 1 - 2$
88 = $91 - 2 - 0!$
89 = $91 - 2 - 0$
90 = $92 - (1+0!)$
91 = $91 + 0 * 2$
92 = $92 + 1 * 0$
93 = $90 + 1 + 2$
94 = $91 + 2 + 0!$
95 =
96 = $90 + (1 + 2)!$
97 = $91 + (0! + 2)!$
98 =
99 =
100 =
$endgroup$
1
$begingroup$
86 = 90 - 2 * (1+0!)
doesn't work, as you use0
twice in the formula.
$endgroup$
– Meerkat
Jan 9 at 16:41
1
$begingroup$
79 doesn't work, you use 1 twice. 78 would though 90-12 instead of 91-12
$endgroup$
– Yout Ried
Jan 11 at 2:09
add a comment |
$begingroup$
Partial, only a few numbers so far:
50 =
51 =
52 =
53 =
54 =
55 =
56 =
57 = $19 * (2+0!)$
58 = $29 * (1+0!)$
59 =
60 =
61 =
62 =
63 =
64 =
65 =
66 =
67 =
68 =
69 =
70 = $(9-2) * 10$
71 = $91 - 20$
72 =
73 =
74 =
75 =
76 =
77 =
78 =
79 = $91 - (0! + 2)$
80 =
81 =
82 = $92 - 10$
83 =
84 = $90 - (1 + 2)!$
85 =
86 =
87 = $90 - 1 - 2$
88 = $91 - 2 - 0!$
89 = $91 - 2 - 0$
90 = $92 - (1+0!)$
91 = $91 + 0 * 2$
92 = $92 + 1 * 0$
93 = $90 + 1 + 2$
94 = $91 + 2 + 0!$
95 =
96 = $90 + (1 + 2)!$
97 = $91 + (0! + 2)!$
98 =
99 =
100 =
$endgroup$
1
$begingroup$
86 = 90 - 2 * (1+0!)
doesn't work, as you use0
twice in the formula.
$endgroup$
– Meerkat
Jan 9 at 16:41
1
$begingroup$
79 doesn't work, you use 1 twice. 78 would though 90-12 instead of 91-12
$endgroup$
– Yout Ried
Jan 11 at 2:09
add a comment |
$begingroup$
Partial, only a few numbers so far:
50 =
51 =
52 =
53 =
54 =
55 =
56 =
57 = $19 * (2+0!)$
58 = $29 * (1+0!)$
59 =
60 =
61 =
62 =
63 =
64 =
65 =
66 =
67 =
68 =
69 =
70 = $(9-2) * 10$
71 = $91 - 20$
72 =
73 =
74 =
75 =
76 =
77 =
78 =
79 = $91 - (0! + 2)$
80 =
81 =
82 = $92 - 10$
83 =
84 = $90 - (1 + 2)!$
85 =
86 =
87 = $90 - 1 - 2$
88 = $91 - 2 - 0!$
89 = $91 - 2 - 0$
90 = $92 - (1+0!)$
91 = $91 + 0 * 2$
92 = $92 + 1 * 0$
93 = $90 + 1 + 2$
94 = $91 + 2 + 0!$
95 =
96 = $90 + (1 + 2)!$
97 = $91 + (0! + 2)!$
98 =
99 =
100 =
$endgroup$
Partial, only a few numbers so far:
50 =
51 =
52 =
53 =
54 =
55 =
56 =
57 = $19 * (2+0!)$
58 = $29 * (1+0!)$
59 =
60 =
61 =
62 =
63 =
64 =
65 =
66 =
67 =
68 =
69 =
70 = $(9-2) * 10$
71 = $91 - 20$
72 =
73 =
74 =
75 =
76 =
77 =
78 =
79 = $91 - (0! + 2)$
80 =
81 =
82 = $92 - 10$
83 =
84 = $90 - (1 + 2)!$
85 =
86 =
87 = $90 - 1 - 2$
88 = $91 - 2 - 0!$
89 = $91 - 2 - 0$
90 = $92 - (1+0!)$
91 = $91 + 0 * 2$
92 = $92 + 1 * 0$
93 = $90 + 1 + 2$
94 = $91 + 2 + 0!$
95 =
96 = $90 + (1 + 2)!$
97 = $91 + (0! + 2)!$
98 =
99 =
100 =
edited Jan 11 at 2:32
answered Jan 9 at 15:29
rhsquaredrhsquared
8,14021849
8,14021849
1
$begingroup$
86 = 90 - 2 * (1+0!)
doesn't work, as you use0
twice in the formula.
$endgroup$
– Meerkat
Jan 9 at 16:41
1
$begingroup$
79 doesn't work, you use 1 twice. 78 would though 90-12 instead of 91-12
$endgroup$
– Yout Ried
Jan 11 at 2:09
add a comment |
1
$begingroup$
86 = 90 - 2 * (1+0!)
doesn't work, as you use0
twice in the formula.
$endgroup$
– Meerkat
Jan 9 at 16:41
1
$begingroup$
79 doesn't work, you use 1 twice. 78 would though 90-12 instead of 91-12
$endgroup$
– Yout Ried
Jan 11 at 2:09
1
1
$begingroup$
86 = 90 - 2 * (1+0!)
doesn't work, as you use 0
twice in the formula.$endgroup$
– Meerkat
Jan 9 at 16:41
$begingroup$
86 = 90 - 2 * (1+0!)
doesn't work, as you use 0
twice in the formula.$endgroup$
– Meerkat
Jan 9 at 16:41
1
1
$begingroup$
79 doesn't work, you use 1 twice. 78 would though 90-12 instead of 91-12
$endgroup$
– Yout Ried
Jan 11 at 2:09
$begingroup$
79 doesn't work, you use 1 twice. 78 would though 90-12 instead of 91-12
$endgroup$
– Yout Ried
Jan 11 at 2:09
add a comment |
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$begingroup$
So...can we use the square root operator AND the cube root operator? I know that there is a three in the cube root, but there is also an implied two in square root.
$endgroup$
– Brandon_J
Jan 9 at 20:23