0
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Use 2 0 1 and 8 to make 67

See above for the rules.

Note that in this one, the year is $2019$, not 2018

Whoever does this one will be rewarded with the answers for 1-100.

Hint

Look at the tags. Also, there is more than one solution, but the simplest one will be accepted.

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  • $\begingroup$ An extra $2$ yields $(2+0!)!^2−1-\sqrt{9}$. $\endgroup$ – Mr Pie Sep 19 '18 at 10:11
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How about:

$$\dfrac{102}{\sqrt{9}}$$

with just two operations.

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  • $\begingroup$ Oh this is much better! I got trapped by lateral-thinking again ;p Great answer! :D $\endgroup$ – El-Guest Sep 18 '18 at 20:49
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If we're using lateral thinking, then

rotate the 9 to make a 6, then we have $6^2 - 0! - 1 = 34$.

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  • 1
    $\begingroup$ As always, El-Guest strikes again like a genius (+1) :D $\endgroup$ – Kevin L Sep 19 '18 at 2:30
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    $\begingroup$ you dont have to use lateral thinking, take square root of 9 then take factorial, it gives you 6 anyway. $\endgroup$ – Oray Sep 19 '18 at 5:24
  • $\begingroup$ @Oray that’s a good point, though I thought flipping the 9 minimized the operations used! $\endgroup$ – El-Guest Sep 19 '18 at 10:04
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I noticed the lateral-thinking tag, so I decided to do it in the following way:

$$2\times (0!+16)=34.\tag{$9=6\small\rm \:when \:rotated$}$$

This uses a similar approach as what was shown in @El-Guest's answer and technically uses the required numbers in their order (namely $2$, $0$, $1$, $9$).

The following is another one less technical.

$$\underbrace{(2+0!)}_{3}\|\underbrace{(1+\sqrt{9})}_{4}=34\tag{$\|=\small\rm concatenation$}$$

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  • 1
    $\begingroup$ Clever answer mate (+1) :D $\endgroup$ – Kevin L Sep 19 '18 at 2:31
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Also consider:

$${9 \choose 2} - 1 - 0!$$

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