Create Numbers 1 - 100 using 1,9,6,8

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.

• @Oray can you please prove it then? thanks! – Omega Krypton Feb 18 at 14:45
• @Oray I think the only number we have not found now is $79$. – hexomino Feb 18 at 18:39

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

• 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. – 3D1T0R Feb 19 at 18:37
• @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. – hexomino Feb 20 at 17:11
• Your handwriting is really good. :P – Aryaman Mar 1 at 6:31

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

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

• @Oray which numbers are impossible? – Auroxa Feb 18 at 23:33
• nice one @Auroxa – Omega Krypton Feb 18 at 23:34
• @Auroxa I was not right since hexomino got all of them, one is just with inf square root. – Oray Feb 19 at 7:15
• 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: $$\lim_{n\rightarrow\infty}(-1+(9+\sqrt[n]6)\times8)=79$$ – 3D1T0R Feb 21 at 20:09
• 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. – 3D1T0R Feb 21 at 23:47

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

Here is an answer for 39 that was missing so far:

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

• There is also one for 0. 1 - sqrt(9) - 6 + 8 – Auroxa Feb 19 at 7:45
• @ppgdev your answer for 79 has 8 and 6 switched. – hexomino Feb 20 at 17:12
• 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. – ppgdev Feb 20 at 18:10

Here is a suggestion for 79, if subfactorial (!n) and double factorial (n!!) are allowed:

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