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Use $2$ $0$ $1 $ and $9$ to make the numbers $66, 74, 76, and$ $77$

RULES

  1. You must use all 4 digits. Only the digits 2, 0, 1, and 9 can be used. You can make multi-digit numbers out of the numbers. Examples: 20, 92, 2.9

  2. The square function may NOT be used. Nor may the cube, raise to a fourth power, or any other function that raises a number to a specific power. You may use the ^ operation if you use a digit, for example, [(9 + 1)^2 - 0!] is acceptable (if you're trying to get 99), because 2, 0, 1, and 9 is used. However, [20 ^ 2 / 9 + 1] can't be used to get 40 because it uses an extra 2.

  3. Sorry, but the integer function may NOT be used. Nor may the round, floor, ceiling, more than 2 factorials in a row, repeating symbol, or truncate functions.

  4. +, -, *, /, (), !, sqrt, ^, and !! may be used for functions.

  5. I picked these because these are the hardest for 2019. From the numbers 1-100, these are the only ones I didn't get.

Again, no brute-force methods. Good luck.

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  • $\begingroup$ Are these solutions possible? $\endgroup$
    – Goose
    Commented Sep 12, 2018 at 21:26
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    $\begingroup$ @QuantumTwinkie let's find out! :D $\endgroup$
    – Mr Pie
    Commented Sep 12, 2018 at 21:47
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    $\begingroup$ @Alto Doesn't using sqrt contradict your requirements? It's an explicit square root, i.e. it uses 2, or, even worse, it could be represented as x^1/2 which uses 1 and 2. $\endgroup$
    – rhsquared
    Commented Sep 13, 2018 at 2:36
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    $\begingroup$ @user477343 You are quite right. Maybe I won't be allowed to answer because of the implicit two in my name. $\endgroup$
    – rhsquared
    Commented Sep 13, 2018 at 6:01
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    $\begingroup$ Are bare decimal points allowed, like $.1$ and $.2$? $\endgroup$
    – nickgard
    Commented Sep 13, 2018 at 8:01

2 Answers 2

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66:

$(\sqrt{9})!\times(12-0!)$

74:

$9/.12-0!$

76:

$9/.12+0!$

77:

$((\sqrt{9})!)!\times.1+(0!/.2)$

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  • $\begingroup$ I think the square function cannot be used $\endgroup$
    – Ian Fako
    Commented Sep 13, 2018 at 9:18
  • $\begingroup$ No it can't even I tried using it. $\endgroup$
    – Quark-epoch
    Commented Sep 13, 2018 at 9:38
  • $\begingroup$ And you can't concatenate to make $12$, either, but $(+1)$ in at least $13$ hours for the effort (DVL) :D $\endgroup$
    – Mr Pie
    Commented Sep 13, 2018 at 10:44
  • $\begingroup$ @user477343 you can concatenate 1 and 2. See examples above 20, 92, 2.9 $\endgroup$
    – Ian Fako
    Commented Sep 13, 2018 at 10:52
  • $\begingroup$ @IanFako Wait... I thought it said "can't" .... dammit, I would have had more of a chance finding a solution otherwise. Eh well, everything happens for a reason... at least in my philosophy. $\endgroup$
    – Mr Pie
    Commented Sep 13, 2018 at 10:54
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$\begingroup$

Also here is another solution to 77 that might be easier- (√9!)!*.1+(0!/.2)

The equation doesn't require the use of a double factorial.

KG

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  • $\begingroup$ You should check the other answers before adding a new one. This is the same as the accepted answer for 77. $\endgroup$
    – Herb
    Commented Jan 7, 2019 at 21:52

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