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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.

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  • $\begingroup$ @Oray can you please prove it then? thanks! $\endgroup$ Feb 18, 2019 at 14:45
  • $\begingroup$ @Oray I think the only number we have not found now is $79$. $\endgroup$
    – hexomino
    Feb 18, 2019 at 18:39
  • $\begingroup$ @ Original proposer - Did you mean for regular factorial to be the only factorial to be used? $\endgroup$ Jan 29, 2020 at 22:12
  • $\begingroup$ And, you should not have to state it, but a finite (specific number) of square roots need to be allowed. $\endgroup$ Jan 29, 2020 at 22:15

6 Answers 6

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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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2
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Partial Answer (too lazy to type them out :P)

enter image description here

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3
  • $\begingroup$ Oops: "You have to keep the order 1,9,6,8 for all numbers." Some of your formulae are in the order 1,9,8,6 instead. $\endgroup$
    – 3D1T0R
    Feb 19, 2019 at 18:37
  • $\begingroup$ @3D1T0R Thanks for spotting this. I've edited my answer to correct for ones where the numbers were switched and had not already been covered by trolley813. $\endgroup$
    – hexomino
    Feb 20, 2019 at 17:11
  • $\begingroup$ Your handwriting is really good. :P $\endgroup$
    – Aryaman
    Mar 1, 2019 at 6:31
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Here is an answer for 39 that was missing so far:

39 = -(1*9) + (6*8)

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  • $\begingroup$ There is also one for 0. 1 - sqrt(9) - 6 + 8 $\endgroup$
    – Auroxa
    Feb 19, 2019 at 7:45
  • 1
    $\begingroup$ @ppgdev your answer for 79 has 8 and 6 switched. $\endgroup$
    – hexomino
    Feb 20, 2019 at 17:12
  • 1
    $\begingroup$ Thanks, @hexomino. My mistake. I removed 79 from my answer. So far the only 79 solution is the one with infinite sqrts, that you found. $\endgroup$
    – ppgdev
    Feb 20, 2019 at 18:10
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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 =-(1\times(\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 $
$91 = 1^9 + (6!/8)$

Omega Krypton had some of the answers with $6$ and $8$ switched. (thanks to 3D1T0R for spotting this) Here are those fixed which are not covered already by trolley813

$46 = 1 + 9 + \sqrt{\sqrt{6^8}}$
$47 = -1^9 + (6\times 8) $
$48 = 1^9 \times 6 \times 8$
$49 = 1^9 + (6\times 8) $
$50 = ((1 + (\sqrt{9})!) \times 6) + 8 $
$51 = (1 \times \sqrt{9}) + (6 \times 8) $
$ 53 = -1 + (\sqrt{9})! + (6 \times 8) $
$54 = (1 \times (\sqrt{9})!) + (6 \times 8) $
$61 = -1 - (\sqrt{9})! + 68$

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  • $\begingroup$ @Oray which numbers are impossible? $\endgroup$
    – Auroxa
    Feb 18, 2019 at 23:33
  • $\begingroup$ nice one @Auroxa $\endgroup$ Feb 18, 2019 at 23:34
  • $\begingroup$ @Auroxa I was not right since hexomino got all of them, one is just with inf square root. $\endgroup$
    – Oray
    Feb 19, 2019 at 7:15
  • $\begingroup$ I don't like the use of infinite square root for 79, as AIUI it continues to get closer and closer to 79 the deeper we calculate our infinitely nested square root, but it never quite reaches it, then we stop calculating because we can tell that it'll never pass 79, and the difference between our calculation and 79 is so infinitesimally small that it doesn't matter to us. Although it's not expressly allowed in the question, it makes more sense to express it as a limit: $$\lim_{n\rightarrow\infty}(-1+(9+\sqrt[n]6)\times8)=79$$ $\endgroup$
    – 3D1T0R
    Feb 21, 2019 at 20:09
  • $\begingroup$ Or it could be expressed as $-1+(9+\sqrt[\infty]6)\times8=79$ though that brings back the infinitesimal rounding issue. Also note: I realise that (both examples) I wrote an infinite $n$th root, but $\root\infty\of6=\sqrt{\sqrt{\ldots\sqrt 6}}$ where $\infty$ is a power of $2$, and they're both right (if we ignore the inherent infinitesimal rounding error). Also, I don't know how to write an infinite square root to go with the $\lim_{n\rightarrow\infty}$ concept. The closest I can think of is $\sqrt[2^n]6$, but while that brings us back to the explicitly allowed set of operations, it has a 2. $\endgroup$
    – 3D1T0R
    Feb 21, 2019 at 23:47
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Here is a suggestion for 79, if subfactorial (!n) and double factorial (n!!) are allowed:

$-1+(!(\sqrt 9))\times(6!!-8) = -1+2\times40 = 79$

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Partial

Finding a better 20

$1=1^{9^{6^8}}$
$2=1+\sqrt9+6-8$
$3=1^9-6+8$
$4=1+9-\sqrt{\sqrt{\sqrt{6^8}}}$
$5=19-6-8$
$6=(1^9-6+8)!$
$7={-1}^{\bigl(9^6\bigr)}+8$
$8=1^{9^6}*8$
$9=1^{9^6}+8$
$10=-1+9-6+8$
$11=1*9-6+8$
$12=1+9-6+8$
$13=1+\frac{96}{8}$
$14=???$
$15=???$
$16={(1+\sqrt{9})}^{-6+8}$
$17=19+6-8$
$18=(1*9)(-6+8)$
$19=1+(9)(-6+8)$
$20=(1*9)!!!!!!!-6+8$

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