8
$\begingroup$

I’m stuck on trying get $59$ by the numbers $1,9,4,8$ in order. I have no idea if it’s possible, I made it for 1-70 just not 59.

I made it without order: $(-1+8)*9-4$

You can use not only the basic math operators, I tried factorial, powers, square roots, logs, even binomial coefficient but that we read it from top to down.

I would like for help from anyone, maybe a solution with an operator I didn’t think of.

By the way, do you think it’s possible for someone to write a program that we give couple numbers, and ask to get other number, with or not order?

EDIT: I guess there can be some solution, but in my opinion using a function symbol, or the floor function, is cheating.
The solutions I liked and I don't think we need anymore:

$-1 + ( (\sqrt{9})!)! / (4 + 8)$ and $-1 + ( (\sqrt{9})! )!! + 4 + 8$ by @ypercubeᵀᴹ
$-1 + \sqrt{9}! * (\sqrt{4} + 8)$ by @Keelhaul

$\endgroup$
  • $\begingroup$ Welcome to puzzling.SE. How do you know for sure such a solution exists? Where did you get this puzzle from? $\endgroup$ – xhienne Apr 17 '18 at 8:57
  • $\begingroup$ Hey, I have no idea if there is an answer. That’s why I came here, because maybe someone will have a nice solution. We got a “mission” from our math teacher to find this answers for 1-70 with the numbers 1,9,4,8 by order. I got stuck only on 59. $\endgroup$ – TuYu Apr 17 '18 at 9:00
  • $\begingroup$ Then this must be related to what your math teacher is teaching you at the moment, right? They wouldn't expect you to use an esoteric operator that you are not supposed to know. $\endgroup$ – xhienne Apr 17 '18 at 9:03
  • $\begingroup$ No no, our country celebrates 70 years, and it started in 1948. Tbh, I don’t even learn math in school because I learn in the university. But my friends asked for my help because the teacher is giving 5 points bonus to the recent test. $\endgroup$ – TuYu Apr 17 '18 at 9:05
  • 1
    $\begingroup$ The teacher is giving bonuspoints to a test, to those who can google this question. It is in my opinion the same as an ongoing competition. Those who google well or ask others to solve their puzzles, get points. $\endgroup$ – PL457 Apr 17 '18 at 11:34
23
$\begingroup$

This should work :

$-1 + \sqrt{9}! * (\sqrt{4} + 8)$

$\endgroup$
  • $\begingroup$ Hey, how do you find this answer? I can't even think how do you think like that? $\endgroup$ – I am the Most Stupid Person Apr 17 '18 at 9:55
  • $\begingroup$ In the most cases I’m trying to solve this type of questions, I’m looking for familiar numbers close to the target. Then I tey to reach them. But most of the time for me is just keep playing with the numbers until I find it. This is my way though, I don’t know how others do it. $\endgroup$ – TuYu Apr 17 '18 at 10:09
  • 1
    $\begingroup$ @IamtheMostStupidPerson It looks like Keelhaul tried to get to 60 and subtract 1. This leaves you with 9, 4 and 8 to get to 60, which is way easier. Then making 10 out of sqrt(4) + 8 only leaves you with the problem of getting a 6 from a 9. Pretty good algorithm! $\endgroup$ – Ian Apr 17 '18 at 10:54
  • 1
    $\begingroup$ @IamtheMostStupidPerson, this is because You are the Most Stupid Person :p. More seriously, Ian pretty much said it : I tried getting to 60 with 9 4 8, knowing that 9 could become 3 with the root squaring or 6 with the factorial of the root squaring, and 4 could easily become 2 as well. $\endgroup$ – Keelhaul Apr 17 '18 at 11:46
7
$\begingroup$

How about

$(1\times9) + \lfloor 4^{\sqrt8}\rfloor$

$\endgroup$
  • $\begingroup$ So what is the result of Sqrt(8)? $\endgroup$ – rhsquared Apr 17 '18 at 11:47
  • $\begingroup$ Irrational AFAIK. It's $2\sqrt2$ or approximately 2.8284... $\endgroup$ – Alex Hajnal Apr 17 '18 at 11:49
  • $\begingroup$ I know, so basically your answer is an approximation? $\endgroup$ – rhsquared Apr 17 '18 at 11:50
  • 4
    $\begingroup$ Sneaky use of floor(n) a.k.a. $ \lfloor n \rfloor $ (or should that be sneaky use of powers?) $\endgroup$ – Alex Hajnal Apr 17 '18 at 11:51
  • $\begingroup$ @AlexHajnal I agree that using floor is kind of cheating, but what do you mean about sneaky use of powers? $\endgroup$ – TuYu Apr 17 '18 at 15:05
3
$\begingroup$

$ 1 - \varphi(9) + \varphi(4!)\times8$

Using the special function:

https://en.wikipedia.org/wiki/Euler%27s_totient_function

$\endgroup$
  • $\begingroup$ Original, but feels like cheating $\endgroup$ – Mixxiphoid Apr 17 '18 at 14:05
  • $\begingroup$ Same as @Mixxiphoid, I think it is cheating using a function where you just use the letter that symbols it. If you wrote the definition of a function it was nice, but then you repeat numbers, as I read about the definition. $\endgroup$ – TuYu Apr 17 '18 at 15:02
  • $\begingroup$ Mind you, the same could be said for (double) factorials (or square roots for that matter) $\endgroup$ – Alex Hajnal Apr 17 '18 at 16:29
  • $\begingroup$ Since binomial coefficient was mentioned in the question i figured it was not a problem. But I do agree that this was probably not the solution that was thought of originally and a simpler solution was in mind. $\endgroup$ – Viktor Mellgren Apr 18 '18 at 8:44
1
$\begingroup$

Another one, based on Keelhaul's idea:

$-1 + ( (\sqrt{9})!)! / (4 + 8)$

And one more that uses the double factorial:

$-1 + ( (\sqrt{9})! )!! + 4 + 8$

$\endgroup$
  • $\begingroup$ Wow! I like the first solution. I tried using double factorial, but isn't your second solution wrong? 3!!=1*3=3, then -1+3+4+8=14. Am I missing something? $\endgroup$ – TuYu Apr 17 '18 at 15:00
  • $\begingroup$ Oops, my bad, missed another factorial! Should be $6!! = 2*4*6 = 48$ and you need a standard factorial to get 6. $\endgroup$ – ypercubeᵀᴹ Apr 17 '18 at 15:35

Not the answer you're looking for? Browse other questions tagged or ask your own question.