# Make $1,\dots,15$ using $3, 9, 9, 9$

This was inspired by many puzzles that use three/four numbers to create other numbers. I chose these numbers in particular because of this post.

Can you find a way to make all the natural numbers from $1$ to $15$ with all four of just following numbers?

$3,9,9,9$

You are allowed to use any operation only the following operations, but can jumble up the order and even turn $9$s upside-down to make a $6$ (though that will be replacing a $9$). You can also use an operation more than once if you like.

$+\;\;\times\;\;\div\;\;-\;\;\sqrt{\cdot}\;\;0.\;\;\lfloor\rceil\;\;!\;\;\$\;\;\%\;\;(\,)\;\;\hat\,$Be as creative as you want. Why limit the mind? And$\$$does not necessarily mean dollars... You can include zeroes for decimals if you want, because really, 1=01 and 2=0002 so I see no difference. Challenge Solution: I am interested to see all the solutions, especially those containing only the mainstream operations and/or one radical and/or floor/ceiling functions. In that particular case, I myself have discovered a few solutions from 1 to 5 which means... well... I genuinely don't know if there exist these particular types of solutions for greater numbers. Accepting an Answer: The answer will be accepted to the person who finds challenge solutions from 1 to 15, and no, the first one won't be accepted... unless it is the most creative answer, because I will accept an answer if it has found all these challenge solutions and is the most creative (partly based on upvotes so the decision of accepting a certain answer is not too subjective). As for solely creativity, a 50 rep bounty will be awarded to the answer that has the most creative solutions (that might be the accepted, as well). No answer must have just one solution, especially partial answers. There must be more than one solution in the posted answer before any further progress is made. This rule just gives others time to come up with solutions themselves without being tempted to look at an answer! Enjoy!$$$$P.S. If you like mathematical challenges, go here! • @WeatherVane yes, you must use all four numbers :) Sep 8 '18 at 8:58 • I'm voting to close as too broad due to the creativity aspect. Questions on puzzling should have a single, well-defined answer, or objective criteria to determine why an answer is better than another. – ffao Sep 8 '18 at 9:23 • It should be easily fixable by specifying a fixed set of operations that are allowed and eliminating the creativity criteria; surely whoever is working on it would rather post their answer on a modified question than have a closed question they can't post to? – ffao Sep 8 '18 at 9:38 • can I use logarithms? can I? can I? please, pretty please Sep 12 '18 at 11:37 • @user477343 Done. But please don't take the tick away from Oray. He deserves it because he followed the specifications. I didn't. and I don't want to be a thief. Sep 12 '18 at 12:52 ## 8 Answers 1 \frac{3}{\sqrt{9}!+\sqrt{9}!-9} 2 \frac{3!*9}{\sqrt{9}^{\sqrt{9}}} 3 \frac{3}{\frac{9!}{9!}^{9!}} 4 \frac{3*9-\sqrt{9}}{\sqrt{9}!} 5 \frac{3*9+\sqrt{9}}{\sqrt{9}!} 6 \frac{3!}{\frac{9!}{9!}^{9!}} 7 \frac{3*9-\sqrt{9}!}{\sqrt{9}} 8 \frac{3!*9-\sqrt{9}!}{\sqrt{9}!} 9 3*9-\sqrt{9}*\sqrt{9}! 10 \frac{3!}{\sqrt{9}+\sqrt{9}}+9 11 \frac{3*9+\sqrt{9}!}{\sqrt{9}} 12 \frac{3!*9}{\sqrt{9}}-\sqrt{9}! 13 \frac{3!}{\sqrt{9}!}+\sqrt{9}+9 14 \frac{3!}{\sqrt{9}}+\sqrt{9}+9 15 \frac{3!*9}{\sqrt{9}}-\sqrt{9} Bonus: 16 \frac{3!*9-\sqrt{9}!}{\sqrt{9}} 17 \frac{3!*9-\sqrt{9}}{\sqrt{9}} 18 3!*9^{\sqrt{9}/\sqrt{9}!} 19 \frac{3!*9+\sqrt{9}}{\sqrt{9}} 20 \frac{3!*9+\sqrt{9}!}{\sqrt{9}} • Nice! This is definitely creative and shows all challenge solutions! Well done! Sep 8 '18 at 9:43 • I am accepting this answer because it actually reveals that all these numbers can be made with 3, 3, 3, 9, which I find to be very interesting (and might even make a Math.SE post discussing the matter). Therefore, congratulations! \color{green}{\checkmark} Sep 12 '18 at 11:35 • Side note, some of the answers don't really need a factorial, do they? Wouldn't 3/(9/9)^9 work perfectly well? Sep 14 '18 at 11:39 • @user1717828 yes, i just did iy for fun facts :p – Oray Sep 14 '18 at 11:40 1=\dfrac{3\times6}{9+9}=\dfrac{3+9}{6+6}=\dfrac{3!}{6}\times\frac99=3!-6+\frac99 2=\dfrac{3!\times6}{9+9} 3=(-3+6)\times\frac99 4=(-3+6)+\frac99 5=\dfrac{-3+6!!}{9}=\dfrac{3\times(6+9)}{9} 6=(-3+9)\times\frac99 7=(-3+9)+\frac99 8=(3+6)-\frac99 9=(3+6)\times\frac99 10=3!+\dfrac{6\times6}{9} 11=(3!+6)-\frac99 12=(3!+6)\times\frac99 13=(3!+6)+\frac99 14=(3!+9)-\frac99 15=(3!+9)\times\frac99 • Not one radical or floor function. That is so good! I might give this a 50 rep bounty and the other challenge solution a 100 rep bounty. Unfortunately, DVL14 :\ Sep 8 '18 at 9:22 • Upvoted. (+1) :D Sep 9 '18 at 5:18 • (Adding to message of the bounty) I forgot that this answer had factorials, but that is multiplication anyway, and not really a function, just notation Sep 11 '18 at 11:04 • I've added another '5' to remove the double factorial – JMP Sep 11 '18 at 11:50 • @IanFako seems to have achieved even simpler results, so unfortunately... the bounty will be awarded to him/her at this stage :\ Sep 11 '18 at 11:51 \lfloor99\%+39\%\rfloor = 1 \lceil99\%+39\%\rceil = 2 3 + (9\% - 9\%) * 9\% = 3 9\% / 3\% + 9\% / 9\% = 4 3! - \lfloor9\% / 9\% + 9\%\rfloor = 5 3! + (9\% - 9\%) * 9\% = 6 (9\% / 3\%)! + (9\% / 9\%) = 7 9 - \lfloor99\% + 3\%\rfloor = 8 9 + \lfloor3\% + 9\% + 9\%\rfloor = 9 9 + \lceil3\% + 9\% + 9\%\rceil = 10 9 - (9\% / 9\%) + 3 = 11 9 + 3 + (9\% - 9\%) = 12 9\% / 3\% + \lceil9 + 9\%\rceil = 13 99\% / 9\% + 3 = 14 \lceil9\% / 9\% / 9\% + 3\rceil= 15 They all have a percent sign (\%). • Hahah, I love this!! (I am also trusting this is true for now, but will double-check them all to see if they are all correct.) I wish I did not reach my daily voting limit >:O Sep 8 '18 at 9:55 • Congratulations! \boxed{\color{blue}{+50}} You had the most creative answer. I was planning on giving the bounty to @JonMarkPerry 's answer, but have changed my mind, and will be giving him a bigger bounty :P Sep 12 '18 at 11:34 1 \cfrac{3!}{6} × \cfrac{9}{9} 2 \cfrac{6}{3} × \cfrac{9}{9} = \sqrt{\cfrac{9}{3} + \cfrac{9}{9}} 3 \cfrac{9}{6} × \cfrac{6}{3} = \cfrac{9}{3!} + \cfrac{9}{6} = \sqrt[3]{9 + 9 + 9} = \sqrt{\cfrac{9 + 9 + 9}{3}} = \sqrt{\sqrt{9 × (6 + 6 - 3)}} = \sqrt{\cfrac{6 × 9}{9} + 3} 4 \cfrac{9}{3} + \cfrac{9}{9} 5 3 + \cfrac{9 + 9}{9} 6 6 - 3 + 9 - 6 = \cfrac{6 + 6 + 6}{3} = (\sqrt[3]{9 + 9 + 9})! 7 \sqrt{9 × 9} - \cfrac{6}{3} 8 \sqrt{9} + \sqrt{9} + \cfrac{6}{3} 9 \cfrac{9 + 9 + 9}{3} = \sqrt{9 × (6 + 6 - 3)} = 6 × \cfrac{9}{9} + 3 10 \cfrac{9}{0.9} × \cfrac{\sqrt{9}}{3} 11 \sqrt{9 × 9} + \cfrac{6}{3} 12 9 + 9 - 9 + 3 13 9 + 3 + \cfrac{9}{9} 14 9 + \sqrt{9} + \cfrac{6}{3} = \cfrac{9!}{6! × 3! × 6} 15 9 + 9 - 6 + 3 • Hello! Welcome to the Puzzling Stack Exchange (Puzzling.SE)! I love your answer (especially the big square roots). Since you are new to Puzzling, I suggest that you visit the Help Center, particularly here, because the construction of your answer is on point; though I see that you have received a bonus +100 reputation as being trusted on another Stack Exchange site, so you will probably become familiar with this site pretty quickly. Glad you joined, keep it up and happy puzzling! :D Sep 8 '18 at 11:02 • Thanks a lot for your motivation! Btw did you mean to post another link? The one you posted is about asking, not answering :) Sep 8 '18 at 11:11 • No, I was meant to post that link. What I meant to say was that you answered well, so you don't really need to focus on that, and perhaps maybe you might need to check out the Asking section. Every comment must have at most 600 words (with a minimum of 15) so there was only so much I could write in one comment, hahah. Sorry about that :P Sep 8 '18 at 11:16 • Oh, I just saw how others answered and followed the same scheme. Nice to hear that from you! Sep 8 '18 at 11:46 • I saw @Oray 's answer and am starting to wonder if every number can be made with just 3,3,3,9 and at least one factorial (putting this link here for other users who might see this comment and not know what a factorial is). Sep 8 '18 at 11:52 1=\frac{3\times9}{9\sqrt{9}} 2=\frac{3}{\sqrt{9}}+\frac{9}{9} 3=9-\frac{9+9}{3} 4=\frac{9}{3}+\frac{9}{9} 5=\frac{9+9}{9}+3 6=\frac{9}{9}(\sqrt{9}+3) 7=\frac{9}{9}+\sqrt{9}+3 8=9-(3(9-9))! 9=\frac{3\sqrt{9\times9}}{\sqrt{9}} 10=9+(3(9-9))! 11=9+3-\frac{9}{9} 12=\frac{9}{9}(3+9) 13=9+3+\frac{9}{9} 14=\frac{99}{9}+3 15=\sqrt{9\times9}+\sqrt{9}+3 • Hello! Welcome to the Puzzling Stack Exchange (Puzzling.SE)! Great answer, and some of the solutions you have were the ones I got myself, hahah. Since you are new to the Stack Exchange community (let alone this site), I strongly suggest you visit the Help Center, particularly here then here in the Asking section (your answering has already been nailed). If you enjoyed my puzzle and like similar ones, click on the tags. Happy Puzzling! :D Sep 9 '18 at 5:31 I made shortest solutions (it uses only given numbers once and only ! / * + - \sqrt{} operators) 1 = 3/\sqrt{9}*9/9 2 = 3/\sqrt{9}+9/9 3 = 9/3*9/9 4 = 9/3+9/9 5 = 9-3-9/9 6 = 9-3+9-9 7 = 9-3+9/9 8 = 3!+\sqrt{9}-9/ 9 = 3!+\sqrt{9}*9/ 10 = 3!+\sqrt{9}-9/ 11 = 3+9-9/9 12 = 3+9-9+9 13 = 3+9+9/9 14 = 3!+9-9/9 15 = 9+9-9/3 • It's my 1st answer here. Can anyone tell why isn't it hiding my answer? Sep 12 '18 at 11:23 • No problemo! The spoiler tag/quote (the big yellow hiding block) was not hiding the answer because of how they are formatted. I fixed it now; it should look better. Anyways, welcome to the Puzzling Stack Exchange (Puzzling SE)! Since you are new to this siteI suggest you visit the Help Center, even though you've been given a +100 reputation bonus for joining another Stack Exchange community site. Happy puzzling! :D Sep 12 '18 at 11:24 Not in the allowed limits, but I got an ok from the OP to use logarithms. So here goes: x = \log_{\frac{3}{6}}\left({\log_{9}\underbrace{\sqrt{\sqrt{\dots\sqrt{9\,}\,}\,}}_\text{x square roots}}\right) This is equivalent to x = \log_{\frac{1}{2}}\left({\log_{9}{9^{\frac{1}{2^x}}}}\right) =  \log_{\frac{1}{2}}{\frac{1}{2^x}} = x So So we can generate like this any natural number X. bonus, to get all the negative numbers replace the base of the first log from {\frac{3}{6}} to {\frac{6}{3}} • Like how$$\LARGE x = \log_{\frac{\sqrt{4}}{4}}\left({\log_{4}\underbrace{\sqrt{\sqrt{\dots\sqrt{4\,}\,}\,}}_\text{x square roots}}\right) which proves the "$4$ Fours" assertion. (I also wanted to see how large I could make this, hahah). $(+1)$ in at least $13$ hours (DVL). You will probably also get a $+100$ rep bounty for finding a general solution :D Sep 13 '18 at 10:47

A simple approach, the first solution per line contains only +, -, / and flipped 9's (no multiplication), tried to flip as few 9's as possible

$1 = 3 -\dfrac{9+9}{9}$, other solutions: $9 - 9 + \dfrac{6}{3!} = \dfrac{6}{3} - \dfrac{9}{9}$

$2 = \dfrac{9}{3} - \dfrac{9}{9}$, other solutions: $\dfrac{9}{9} + \dfrac{6}{3!}$

$3 = 9 - \dfrac{9 + 9}{3}$, other solutions: $9 - 9 + 6 - 3$

$4 = \dfrac{9}{9} + \dfrac{9}{3}$, other solutions: $\dfrac{9}{9} + 6 - 3$

$5 = \dfrac{9 + 9}{9} + 3$, other solutions: $9 - \dfrac{9}{9} - 3$

$6 = 9 + 9 - 9 - 3$, other solutions: $9 - 6 + 6 - 3$

$7 = 9 - 3 + \dfrac{9}{9}$, other solutions: $\dfrac{6 * 6}{9} + 3$

$8 = 6 + 3 - \dfrac{9}{9}$

$9 = \dfrac{9+9+9}{3}$, other solutions: $9 - 9 + 6 + 3$

$10 = 6 + 3 + \dfrac{9}{9}$

$11 = 9 + 3 - \dfrac{9}{9}$, other solutions: $9 + \dfrac{6 * 3}{9}$

$12 = 9 - 9 + 9 + 3$

$13 = 9 + 3 + \dfrac{9}{9}$, other solutions: $\dfrac{66}{3} - 9 = 9 + 6 - \dfrac{6}{3}$

$14 = 6 + 6 + \dfrac{6}{3}$, other solutions: $\dfrac{99}{9} + 3$

$15 = \dfrac{9+9}{3} + 9$, other solutions: $6 + 6 + 6 - 3$

• Uh oh. Erm, @JonMarkPerry, it seems like someone else might have this bounty :P Sep 11 '18 at 11:45
• @user477343 not sure if this one deserves a bounty, not enough creativity was used, just tried to use as few operations as possible :D Sep 11 '18 at 11:52
• Well, do you want a bounty? I am happy to give it to either you or @JonMarkPerry. I would have given a bounty to the most creative, but the question was going to become closed as too broad (and then nobody could share their answers!) so I just decided to reward the opposite. Otherwise, I would award the bounty to u_ndefined's answer. Sep 11 '18 at 11:59
• @user477343 I can always need a bounty :D but that decision is up to you Sep 11 '18 at 12:02
• @user477343 so I guess no love for me^^ Sep 12 '18 at 12:04