# Use 2, 0, 1, and 9 to make 34

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.

• An extra $2$ yields $(2+0!)!^2−1-\sqrt{9}$. Commented Sep 19, 2018 at 10:11

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

with just two operations.

• Oh this is much better! I got trapped by lateral-thinking again ;p Great answer! :D Commented Sep 18, 2018 at 20:49

If we're using lateral thinking, then

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

• As always, El-Guest strikes again like a genius (+1) :D Commented Sep 19, 2018 at 2:30
• you dont have to use lateral thinking, take square root of 9 then take factorial, it gives you 6 anyway.
– Oray
Commented Sep 19, 2018 at 5:24
• @Oray that’s a good point, though I thought flipping the 9 minimized the operations used! Commented Sep 19, 2018 at 10:04

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.

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

• Clever answer mate (+1) :D Commented Sep 19, 2018 at 2:31

Also consider:

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