# Use 1 9 6 2 in this order to make 75

I'm looking for a solution to make number $$75$$ with numbers $$1$$ $$9$$ $$6$$ $$2$$ in that order and the same rules as in Use 2 0 1 and 8 to make 67.
Here a copy of those rules:

1. You must use all 4 digits. Only the digits $$1$$, $$9$$, $$6$$, and $$2$$ can be used in that order.
You can make multi-digit numbers out of the numbers. Examples: $$19$$, $$96.2$$

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, $$(1 + 9)^6 - 2!$$ is acceptable (if you're trying to get $$999998$$), because 1, 9, 6, and 2 are used. However, $$19 ^ 2 / 6 + 2$$ can't be used to get $$62.166...$$ because it uses an extra 2.

3. Sorry, but the integer function may NOT be used. Nor may the round, floor, ceiling, or truncate functions.

4. $$+$$, $$-$$, $$\times$$, $$\div$$ or $$\frac{\Box}{\Box}$$, $$()$$, $$!$$, $$\sqrt{\Box}$$, $${\Box}^{\Box}$$, and $$!!$$ may be used for functions.

Please no brute-force methods. Good luck.

• Questions should be self-contained, so I've edited in the rules from the linked challenge (and added the same tags). Jun 8 '19 at 22:44
• Let's please not make "near-miss", out-of-intended-order, or otherwise clearly non-pertinent answers here. They do not attempt to answer the posed puzzle, and will be deleted: puzzles with no lateral-thinking tag do not invite such answers.
– Rubio
Jun 9 '19 at 16:27
• agreed, good point @Rubio! Jun 9 '19 at 16:56
• People need to strick this bogus idea of "infinite square roots" of a number being 1. They cannot be written down. If you're going to use the square root symbol, write it a finite number of times, for starters. Oct 24 at 21:33

$$1\times 9/.6/.2$$

this is allowed right?

• Wow, this is clever. Jun 9 '19 at 1:30

Loophole that I discovered:

allowing multifactorial also allowed us to multiply $$x$$ by any integer $$ using a certain number of $$!$$s.

(1) Pretty much based on micsthepick's answer...

$$(1*9+6)/.2$$
$$=15/.2$$
$$=75$$

(2) self-innovated:

$$(1+9)!!!!!!!/(.6-.2)$$
$$=(10*3)/.4$$
$$=75$$

(3) self-innovated:

$$(-1+9)!!!-\sqrt{\sqrt{\sqrt{...\sqrt{\sqrt{\sqrt{6}}}}}}/.2$$ (infinite \sqrt{}s)
$$=(8*5*2)-1/.2$$, (yes, according to this)
$$=75$$

(4) self-innovated

$$(19+6)!!!!!!!!!!!!!!!!!!!/2$$
$$=25!^{(19)}/2$$
$$=25*6/2$$
$$=75$$

• Some notes regarding the multiple-factorials, OP didn't specify whether theya are allowed or not except for single and double factorials. Jun 9 '19 at 14:30
• i think it is fine since it is in the linked question. thanks @athin Jun 9 '19 at 14:37
• Gonna love #3 for insanity :D Good job on it! Jun 9 '19 at 19:46
• Wow, I actually thought of the infinite square roots one, but didn't post it! +1 :)
– Duck
Jun 10 '19 at 1:58
• No, "infinite square roots" is bogus. 1) They cannot be written down. 2) It is really a form/relative of a floor function. Oct 24 at 21:30

I'm not sure if this is allowed, but

$$1 9 6 2$$

$$(1 {\sqrt 9}) + (62)$$

$$(1 3) + (62)$$

$$13 + 62$$

$$75$$

• the question does say that you can make multi digit numbers, but I would argue that it does not specifically allow the concatenation operator Jun 8 '19 at 23:57
• if 1√9 = 13 then you can simplify it. (9-2)(6-1) = 75 Jun 10 '19 at 7:48
• No, that is not allowed for standard concatenation of digits. Oct 24 at 21:39

Here's one that I found:

$$.1*((\sqrt{9})!)!+(6/2)$$
$$= .1 * 720 + 3$$
$$= 72 + 3 = 75$$

Here is one similar to micsthepick's answer:

$$1*(9+6)/.2$$

And here is a hybrid of the two:

$$1*((\sqrt{9})!)!/6!!/.2$$
= $$720 / (2*4*6) / .2 = 15 / .2 = 75$$