# How to get the numbers from 50 - 100 with the numbers 2, 0, 2, 4

All I need is 56-59, 69, 73, 75- 77, 79, 86, 90-94, 99. I've done the rest but I would love to hear other solutions.

Rules:

1. Use any of the following operations: basic operations (+ - x /), to the power of (^), Square root, factorial, double factorial, concatenation (e.g. 42 is acceptable).
2. You can write the numbers in any order but you may only use each number once.

Good Luck!

In a comment, the OP said, “Repeating decimals are not allowed as an answer (e.g. 92.22222 ≠ 92). But you may use repeating decimals as part of an equation (don't know any case).”

• Is triple factorial allowed? Commented Feb 5 at 8:41
• Triple factorial is not allowed
– bob
Commented Feb 5 at 8:42
• Okay. I found an answer for 56. Should I post a partial answer? Commented Feb 5 at 8:42
– bob
Commented Feb 5 at 8:44
• You cant make 41 unless you have something like 42 - 0!
– bob
Commented Feb 5 at 9:38

Work in progress

$$58 = (4+0!)! / 2 - 2$$
$$59 = ((4+0!)! - 2) / 2$$
$$86 = (42 + 0!) \times 2$$
$$94 = ((4 + 2)!! - 0! ) \times 2$$

• Thanks, did you say you had one for 56?
– bob
Commented Feb 5 at 8:48
• I said it's work in progress :). I hope I can find one for most of them Commented Feb 5 at 8:49
• @bob that was me, not him Commented Feb 5 at 8:51
• Have you found one yet?
– bob
Commented Feb 5 at 9:54
• Nope. that's all I have. I quit. It's too hard. Commented Feb 5 at 11:54

$$56 = ((2 + 0!)!)!! + 2 \times 4$$
$$58 = ((2 + 0!)!)!! + 4!! + 2$$ or $$(4!!)^2 - (2 + 0!)!$$
$$69 = \frac{0!}{.\bar{2} - .2} + 4!$$
$$79 = (4!! + 0!)^2 - 2$$ or $$79 = (2 + 0!)^4 - 2$$
$$90 = 42 + ((2 + 0!)!)!!$$
$$99 = (4!! + 2)^2 - 0!$$

• are you sure? I get 58 on this. Commented Feb 5 at 8:52
• I got 56. There might be a mistake though. Commented Feb 5 at 8:54
• This is correct, the answer is 56
– bob
Commented Feb 5 at 8:58
• @Topwizard29Thegamer 0 factorial is 1, 3 factorial is 6, 6 double factorial is 48. Commented Feb 5 at 22:00
• Wow, congrats for getting 90!
– bob
Commented Feb 7 at 8:35

$$\large 57=((2+0!)!)!!+\frac{\sqrt{4}}{.\bar2}$$

$$\large 69=((4!)-(0!)) \times \sqrt{ \frac{2}{.\bar2} }$$

$$\large 71=4! \times \sqrt{\frac{2}{.\bar2}} -0!$$

$$\large 73=4! \times \sqrt{\frac{2}{.\bar2}} +0!$$

$$\large 75=((4!)+(0!)) \times \sqrt{ \frac{2}{.\bar2} }$$

$$\large 76=2 \times (40-2)$$

$$\large 91=\frac{\sqrt{4}}{.\bar2-.2} + 0!$$

• It's just 77 and 93 remaining, right? Commented Feb 7 at 22:10
• @WOWOW I believe you are right. Hopefully someone will get them. Commented Feb 7 at 22:16
• I'm not good at math lol. I'm quite happy that I even got to write an answer. Commented Feb 7 at 22:17

Someone mentioned only 77 and 93 were remaining.

Here is 77:

$$77 = ((4!!)!!)*.2 + .2 + 0$$
or
$$77 = (((4!!)!!) + 2 - 0!) * .2$$

And here is 93:

$$93 = \frac{4!}{.\bar2} - (\frac{0!}{.2})!!$$

• +1 I especially like your first solution for 77. Almost all other answers posted so far use 0 factorial. Commented Feb 9 at 5:10
• Congratulations on getting 93. Commented Feb 12 at 5:05