This was posted on the community tab a few months ago, there it didn't explicitly say that you can't use the other digits, so in that case it was possible to just 9+9-9-9+100=100 That exploit was fixed in the more well explained version, which is this video
math is skill based. Age doesn't mean that you are or are not educated and skill in it. Simplicity is more adulting. Adding unnecessary complexity is over thinking. Remember, the KISS principle?
A agree. Any answer suddenly tossing in a period (as .9) implies you divided by 10 or some other value and hence is invalid. Hence 99/99% is allowed (as 99/.99 = 100). But (9x9+9)/.9 directly would fail. However (9-9+9)/9% would be fine.
0.9 isn't legit because the 0 isn't allowed. The question is whether .9 is a legitimate representation of the number 0.9 or not. It is commonly used and understood, so I think it pretty much has to be accepted as legit. And there's nothing wrong with using 10 implicitly. The % operator divides by 100. That's the definition of that operation. It's still just one *common symbol*. Similar for concatenation, which is defined as multiplying the first number by 10^x, where x is the number of digits in the second number, and adding the second number. And the decimal point, which divides by some multiple of 10, depending on where you put it. And factorial, which multiples together every number from 1 up to the given number. The puzzle isn't to get to 100 mathematically by only using 9s and nothing that could be interpreted as involving any other numbers. It's to get to 100 by only using 4 9s *and common mathematical symbols* (operations).
In my opinion sqrt() shouldn't be allowed as it contains a 2 that we just don't write by convention and % should definitely not be allowed because it is just a symbol for division by 100. Edit: Concatenation and stuff like .9 also feels like it's just cheating.
It's annoying when the video thumbnail is not the same as the challenge. The video thumbnail implied there must be some operator between each nine, and so I worked on it a bit before clicking. Then I felt cheated. Wasted time.
I believe i found a really good solution that doesnt 'twist' any of the rules: (9+cos(9!)) * (9+cos(9!)) = 100 because cos(9!) = 1 Or another one: arccos(9-9) + 9 + cos(9!) = 100
I like the cheekyness of the last solution as you can use it to make any number from almost any 4 copies of the same digit (only 0 and 1 fail) just by adjusting the number of square roots
In the last one, it doesn't actually matter if you use 9s or any other numbers (as long as you use the same number four times, and I actually mean number, not only digits). The trick is that pressing the sqrt function 100x ultimately gives you the 100. You could press it any number of times to get any number you like. It's sort of a magicians trick where the magician makes you think of a number and gives you instructions to compute another number from it and then proceeds to guessing that number correctly. It mostly works because the computation cleverly cancels out the number you started with and always gives the same result which the magician can then conveniently "guess". In conclusion, I consider this method cheating.
Basically we can use 3 and 6 for the 9 since • 3 = sqrt 9 • 6 = 3! = (sqrt 9)! Furthermore we can also use 1/2 because 1/2 = cos (tan-1(sqrt 3)) Finally we have (1/2)^-6 + 6*6 = 100
That last one is actually just so beautiful and can be applied to literally any number so long as they are all the same number since it derives its answer from the properties of logs rather tricks with the specific number. Technically, "(x/.x)(x/.x) will also evaluate to 100 as well regardless of x's value, but that's a little cheap since ".x" implies dividing by 10 or using a 0, plus it's not nearly as elegant.
The logarithm one at the end was my favorite, it literally just incremented to 100 in a clever way. In degrees, arccos(sin(9!))+9+9/9=100 If you allow using different meanings of common mathematical operations you can also do (9^3)*(9^3)=100 using ^ as a bitwise xor
I wonder if there is a solution using the graphic from the thumbnail, where there is a single operation between the numbers. This would remove being able to turn 2 9s into 99 and also prevent multiple operations between the numbers.
That last solution is beautiful as it’s a wonderful general case solution allowing you to replace the nice with any number as long as they are all the same as their purpose is only to cancel the logs and you can get any integer you want by just increasing or decreasing the number of sqrt in your numerator quantity
The last solution has also a nice property: It works with any base. You can use the natural log, the common log or any base b of your choice. Reflected to the calculator, it doesn't matter if the log key used the natural or the common log.
(9 + sgn(9)) * (9 + sgn(9)) Or with 3 9s, without sgn(), but with square function, which I think should be allowed if a sort is allowed: square(9 + 9/9)
I am watching you, after a lot of time... You're still the same amazing, when I used to watch you earlier... Now I think I will continue watching your content.
I would rule out the percent sign, as it literally means 1/100 making this way too easy. 9/9% = 100 already (using two more 9's without changing the result is an easy exercise). While you could see it as an "operation", I think it is more commonly viewed as *notation*, a shorthand for writing "over 100" so it's similar to why you would rule out e^x and π
This reminds me of something I do at work to entertain myself (using locomotive numbers, if anyone is curious). Given any four random digits in a given order , make an equation using any mathematical operator which does not introduce a new digit (I.e. X^2 is not allowed, but X^Y is, as is square root). I have found very few numbers for which I could not come up with something.
Logarithms can often be used to pull out how many times a square root was used, like in the last example. Seriously overpowered. There was a mathematical paper about it. And it has an implicit 2, so technically it ought to be disallowed. X^Y is still OK though.
You only need one 9 to make 100 by using the identity csc(arccot(x))=sqrt(1+x^2) csc(arccot(sqrt(9)))=sqrt(1+9)=sqrt(10) csc(arccot(sqrt(10)))=sqrt(11)...and continue until you reach sqrt(10000) Both csc(arccot(x))=sqrt(1+x^2) and sec(arctan(x))=sqrt(1+x^2) will do the task.
It'd be addition, multiplication, subtraction and division. Cosine and Sin both use ratios, which are a kind of number. So do radians. It's also bad literary theory, to try and defy the meaning that's obvious. Like, intrinsically, the square root of 9 is 3, not 9. You know? You don't treat a square root as the number being square rooted, but as its own separate number. But, even so, it's clear they meant the four standard operations, and trying to be tricky and get around that is not listening to the instructions, nor trying to comprehend what is meant.
before watching the anwser: my answer is (9 9)+(9/9) = 100. just because you CAN, put an opperator between those numbers doesn't mean you have to - and i think this is the simplest solution.
i remember numberphile showing off the last solution years ago for the four 4s problem, glad to see it's still around when you said you just wanna see 100 displayed in the calculator in the intro, my immediate thought was to just keep pressing factorial and square root until it happens to get close enough to 100 to round to it in whatever floating point data type the calculator is using
Hi, Presh. Of course you are familiar with the well-known Four 4´s challenge. A long time ago, after solving it, I wondered if I could do the same using FIVE 3´s... and I succeeded. If you are interested in this, just let me know how do I send it to you. Cheers from Brazil.
The least number of keypresses would be 9/9+99. I am of the opinion that the most efficient answer, not the most creative answer, is the most intelligent one. A working solution is better than a theoretical one. Intelligence and intellectualism are distinct qualities, and not synonymous. I think over-analyzing something that can be quickly solved is a trap of intellectualism. In fact, it's one I'm engaging in right now lol. In a practical sense, though, I think that if I was able to solve this "puzzle" in a split second with the most efficient solution possible, the benefit in a practical sense is in having another 13 minutes and 30 seconds of extra time to use for some other purpose, such as answering another 200 puzzles, or getting some other task done, putting information to practical use, etc. Then I realized that I used it to compose this instead, which in retrospect is kind of ironic. So much for making good use of my time. Damnit. See what I mean?
99 != 9, .9 != 9, cos(9) != 9, arccos(9) != 9 & log(9) != 9. I believe the equation must contain 4 quantities of 9 & not just have 4 expressions of the symbol 9. Therefore there was truely only 1 equation that was presented in the video that was valid which was a genius solution btw 👍.
A bit cheeky, but my calculator can also handle gradians (as well as convert units between degrees, radians, & gradians) so: While in degrees mode: 9*(9+9/9)=90 Convert 90⁰-> 100 gradians. Done
99+9÷9=100 You just said I have to press 9 four times. Operations between all of then wasn't necessary Edit: I just watched the video and realized this answer was presented in the video
In my opinion, "common" operations are addition, subtraction, multiplication, division and maybe exponents - can this be done with those operations? and no concatenation (or however it's spelled, the creation of new numbers like 99 in this example)
By seeing the thumbnail I actually thought we must use only the basic 4 operators and for 20min straight ive been trying and I got till 99(9×9+9+9) but after watching the video, I was like "c'mon u should've told me before"
I'm seeing a lot of comments saying sqrt is cheating because it invokes 2 (which I agree with), but don't think anyone's mentioned this applies to log implicitly invoking 10.
I'd argue a decimal place is not a mathematical operation. Mathematical operations take input values (a number) and return output values (a number). The decimal place modifies the meaning of proximate symbols, but it is not transforming a number. For example,, "5" is a number, ".5" is a different number, but "..5" is undefined. 0.5 is equivalent to .5, but "0." is certainly not an operator. "%" is arguably a mathatical operation (as is per mille: ‰). "%" as a binary operator is typically taken to be modulo division: 99 % 9 = 9. Is "%" as percent a unary operator? I would guess yes -- it converts a number into another number. any number % is another number, and 5%%% could be claimed to have well-defined meaning (5/million). As to trigonometric functions: degrees are not the default unit of angle in mathemetics: radians are. You can specify degrees with a symbol, but that symbol is not an operator.
I spent a whole week thinking about this bc of the thumbnail & I was so frustrated bc I wasn’t able to solve it so today I gave up & decided to just see the answer.. this was not satisfying at all.. all this thinking was for nothing? why did you put squares between all 9s in the thumbnail😭😭😭
Before watching the full video my answer is: (cos(√(9)!!)+9)×(cos(√(9)!!)+9) = 100 I decided i wanted to find a way to use 4 of the 9s (without combining or adding decimals because those aren't 9s, they're .9s and 99s, but i also didnt wanna make it too complicated) so i did this. After about an hour of trying it for myself i finally got this. Now if i had realized i couldve done the first part twice, i probably would've done it in half that time, but i was trying too hard lol
Thumbnail made me think it has to be four nines with three common functions between each (e.g. 9+9+9+9, 9+9x9-9, etc) and of course no combination of +-÷x ends up at 100 doing it that way.
Another option is using infinitely many square roots to turn two nines into ones than add them to the other two nines to make two tens and multiply them together
What I have gotten is (9^(1/log9))*(9^(1/log9)) where log9 is log nine base 10(like in a calculator) but I cant write it like this as i can't write 1. Can anyone find a way to express it in terms of log other than the one in the vid?
I spent about 2 seconds coming up with the simplest solution 99+9/9. When you deleted the 10^x function, I immediately saw 10^(9/9+9/9) but it was gone.
Although there's no recurring function on a standard calculator, I still think that it could be considered a mathematical function. ? In which case, 99.99(recurring) = 100 (100÷3=33.33 recurring. 33.33 recurring × 3= 99.99recurring) Would like to hear arguments against this method.
I think operations like 99% aren't fair. 99 is (90×10) + (9×1), (.9) is 0.9, or 9÷10, 99% is 99÷100, and finally, square root is in fact 2√9, so it uses 2
I was able to come up with the 2nd to last solution in this video on my own, and I thought that would be the best. But that last one was really something. Kudos.
lol for the last solution it doesn't matter what number the input are, as long as they're 4 times the same. By varying the amount of square roots you can get any result. It's a pretty universal solution.
Wait, so for that last solution, does that mean you can make any number the same way? For any number "n" you just need a "n" number of square roots and everything else should follow the same, right? That's pretty cool... However, despite many of these being mindblowing, I don't agree with allowing log, square root and even the percentage sign, since all of these have numbers hidden in them. log is the inverse of 10^ , so there's a 10 in there (Log base 9 would be fair, I guess). Square root has a 2 hidden in there and the percentage sign is really 1/100.
I have it like this: arcctg[sqrt(sqrt(9))]*arctg[sqrt(sqrt(9))]/(9+9)=100 1800/18=100 Just switch to degrees when you have sqrt(3) so that the arctg and arcctg give 60 and 30
the arccos solution is very fun, definitely my favourite here it strikes a nice balance between simplicity and flashiness with minimal repeat operations also it could’ve been fun to use log base 9(9) as a 1 substitute as in the main solution tho i guess arbitrary-base log isn’t a super common operator
I like how you imply that 99 +9/9 = 100 is cheating by placing the two digits together, yet in your own rules something like factorial is allowed. Factorial introduces many more numbers since 9! Is 1x2x3x4x5x6x7x8x9. Most of these problems state that only the basic operations are allowed x + / -
When did the % button on a calculator work as percentage? It was always modulus on every physical calculator I’ve used. So confused as a kid why percent output seaming random integers
My solution: 9/9% 9/9 I put it on the calculator and the result is 100 The % is percentage and the 9 after % means multiplication. I like this solution because it fills in exactly the gaps in the thumbnail
I didn't see that you included a solution that matches what was implied by the thumbnail. Did I miss that part of the video, or just not pay enough attention? How can you do it in 8 button presses, counting the equals sign? With a single operation key pressed between each of the 9s, and only in those places?
*@ MindYourDecisions* -- If you were allowed cos, then cos(0) would be short for cos(0 radians) = 1. However, you cannot use cos(720) to mean 1, because it means cos(720 radians) ~ -0.76. Any solutions you have that rely on a trig function of an argument, where the argument would have to be in degrees, are not legitimate.
@@cheeseparis1 -- No, a degree symbol is an extra character that is required to do it for cos(720). You have to be able to write it out, and it independently work. No setting to "degree mode."
@@forcelifeforceOf the calculator, the "DRG" button is allowed. You can set the mode you want according to the math to perform, at least it's what I understood.
Technically, using a square root is still cheating because a square root is a function, not an operator. The key is understanding the difference between known values and unknown variables. A function only has ONE UNKNOWN variable, regardless of how many instances that the unknown variable is used (e.g. ((x^2)-x-1) where x is the only unknown variable that just happens to appear twice.) An operator requires TWO UNKNOWN variables where one unknown variable occurs before the operator, and the other unknown variable occurs after the operator (e.g. (a+b) where a is one unknown variable, and b is the other unknown variable.) A square root is just a function that appears in the form of a powering operation that uses (1/2) as a KNOWN constant value in the exponent (e.g. (x^(1/2)) where x is the only unknown variable.) A normal powering operation would look like this: (a^b) where a is one unknown variable, and b is the other unknown variable. So, (1/2) is being used in several of the solutions in the video, which contradicts the established rules of the problem.
log( (cosh(9/9) - exp(-9/9)) * (cosh(9 × 9 + 9 + 9) - exp(-9 * 9 - 9 - 9)) ) = 100 This was relatively easy, if you change the e to the exp() function. Also the expanded 9's don't count, since they will all collapse together into a different number.
9 back 9 * 9 + 9 = 100 Looking at the thumbnail I immediately thought of 9 < 9 * 9 + 9 = 100 That said, I don't most of these solutions, because they all feel like either loopholes to use more digits (like %) or using functions rather than operators (like square root).
