5
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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:

$n!=n\times(n-1)\times(n-2)\times\dots\times2\times1$

For example
$(10-9+2)!=3!=3\times2\times1=6$
$0!=1$

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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
6
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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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  • $\begingroup$ 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
3
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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 $

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2
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Here are two more solutions:

67 = $((\sqrt 9)!)!! + 20 - 1$
68 = $((\sqrt 9)!)!! + 21 - 0!$

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  • 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
0
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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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  • $\begingroup$ 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
0
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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
    $\begingroup$ 86 = 90 - 2 * (1+0!) doesn't work, as you use 0 twice in the formula. $\endgroup$ – Meerkat Jan 9 at 16:41
  • 1
    $\begingroup$ 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

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