You should probably specify that you’re allowed to move the numbers around. Normally for problems like this, you’re only allowed to place the operators, not actually move the numbers. Without moving the numbers or adding brackets, you can’t do the division you did for 0, 2, and 19.
Once, based on the well-known "Four 4s" challenge, I also succeeded in writing expressions for all the integers from 0 to 100, using "Five 3s". If someone is interested in having it, just let me know. Cheers from Brazil.
Hello! please check the rules at 1:20 - 1:35, fractions are allowed, and any calculations in the numerator and denominator are valid, so the method is OK. Brackets are not necessary, and threes don't need to be inline for this challenge. :)
This is how I got 19: 3! × 3 + !!3 = 19 Since !3 = 2, !!3 = !2 = 1 As far as I can tell, using '!' twice to get the subfactorial of a subfactorial is allowed by your rules.
@@guyhoghton399 Well...The concept of a double subfactorial isn't a standard term in combinatorics, but it's an interesting idea to explore. If we think of "double subfactorial" as applying the subfactorial operation twice, we might define it as the number of derangements of derangements. However, this leads to some conceptual challenges, as derangements themselves are specific permutations, and deranging them again doesn't have a clear combinatorial meaning. So to be honest, I never head someone do double subfactorial...but your idea is really interesting.
@@ThePhantomoftheMath Thanks for this feedback. I too was unaware of the existence of "double subfactorial" as a defined concept, which made me think that _!!n_ would therefore have to be unambiguously interpreted as meaning _!(!n)_ which in turn would allow me to get round the "no brackets" rule by simply suppressing them - a bit of a liberty I know! However your notes inspired me to do a little google research and it seems that a definition of the double factorial does in fact exist at least as a sequence, whether or not it has any meaning in combinatorics. Searching for "online encyclopedia of integer sequences" then on that website for "double subfactorial" gives the details, and it turns out that _!!3 ≠ 1_ so my suggestion above won't work I fear! However your solution for the value 19 is very clever.
I think that would need brackets, well, by how I would define double factorial: if !n = n! × ∑k=0 to n of ((-1)^k)/k! then it seems logical to me that !!n = n!! × ∑k=0 to n of ((-1)^k)/k!! Which I think would mean that !!3 = 1/2 But since he didn't define his terms, and I at least haven't heard of a standard way of defining it, I think your solution is still very interesting, and by the rules he did define, a possible solution! I always like to define terms in ways which will not have other ways of being written, so we can have a lot of different symbols at our disposal for different uses.
@@JL710ML Funnily enough I did reply to the professor's comment, conceding that my solution would need brackets (and is therefore not a solution according to the rules) because I found that there is indeed a definition of "double subfactorial", but for some reason my reply disappeared. I might try and post it again. The definition turns out to be close to your suggestion. Searching for "online encyclopedia of integer sequences" then on that website for "double subfactorial" gives the details. I agree with you about the benefits of standardising mathematical notation. Have a nice day.
Hello! I'm glad you liked the video! If you check the rules at 0:40, fractions are allowed, and any calculations in the numerator and denominator are valid, so the method is OK. I suppose by 'invisible brackets' you mean the inline writing of threes, but they don't need to be inline for this challenge. :) Of course, you can always use alternative solutions, like 3! - 3 - 3 = 0, since there are many ways to get these numbers. Cheers!
Sorry, but powers are not allowed for this challenge, and numbers other than three number threes. Please check the "Rules" section in the video for more detailed info!
Can I use # (Primorial) ? Or ? (Terminal) ? Or Cos and Sin ? Like … (3!)#/3 + 3 = 6#/3 + 3 = 13 Or ((3!)? + (3!)?)/3 = (6? + 6?)/3 = (21 + 21)/3 = 14 Or (3!)? - cos(3-3) = 6? - cos(0) = 21-1 = 20
ummmmm ... Your fractions have a problem here. You are representing the fractions using grouping in the numerator. I thought one of the rules for solving was to have all the number threes in line as a single string. If you do this, and have division operator, then none of your division solutions in this video will work!
Hi! Thanks so much for watching and taking the time to comment! I actually mentioned that grouping in the numerator is allowed-please check the 'Allowed' section from 1:20 to 1:35. I also didn’t specify that the numbers need to be in a single line. Hope this clears things up!
@@ThePhantomoftheMath I know, however, this grouping is the same as using parenthesis. The parenthesis is implied by writing the equation in the way you did. I got all the answers to your fine puzzle correct. However, I could not figure out how to do the division properly without grouping. To me, as a software engineer, grouping is always done in-line using parenthesis.
(3-3)/3 = 0 (3!)/(3+3) = 1 3-(3/3) = 2 3-3+3 = 3 3+(3/3) = 4 3! - (3/3) = 5 3! +3-3 = 6 3! + (3/3) = 7 !3 +3+3 = 8 3+3+3 = 9 3! + !3 + !3 = 10 !3 + 3*3 = 11 3!+3+3 = 12 3/.3 + 3 = 13 3!+3!+!3 = 14 3!+3!+3 = 15 3! + 3/.3 = 16 (3!/.3) - 3 = 17 3*(!3*3) = 18 19...best I can think of is (3! * 3)+!(!3). I have to write it that way because !!3 and !(!3) are two different things (at least I presume they are, like how 3!! and (3!)! are two different things). (3+3)/.3 = 20
@@Psykolord1989 Yeah, n!! and (n!)! are different things in math, but honestly, I’ve never heard of a 'double subfactorial.' In combinatorics, that would be the derangement of a derangement, if that makes any sense...But your idea is really interesting!
@@ninafrank9228 Hi! That is indeed 19. However, according to the rules, you can't use the power operation or any new numbers. So, we can't use 0 or any power operations. 🙃
Damn i got as far in math as linear algebra and reliability engineering and i don't think i've ever heard of subfactorial :D
@@user-DaleAcosta I know, I know. It's just a part of combinatorics. A lot of people actually don't know about the subfactorial.
You should probably specify that you’re allowed to move the numbers around. Normally for problems like this, you’re only allowed to place the operators, not actually move the numbers. Without moving the numbers or adding brackets, you can’t do the division you did for 0, 2, and 19.
@@omamba5105 Hi! Thanks for watching! I actually did specify in the "allowed" section of the rules. Check 1:20 - 1:35
@ThePhantomoftheMath Yes but dividing sum of 2 numbers with another number is same as using brackets
Once, based on the well-known "Four 4s" challenge, I also succeeded in writing expressions for all the integers from 0 to 100, using "Five 3s". If someone is interested in having it, just let me know. Cheers from Brazil.
Fun stuff. I look forward to scoping your other videos. Thanks. Subscribed. Cheers
Thanks! I'm glad you enjoyed the video!
Nice solution for 19. Thanks.
Didn't know anything about subfactorial up to mow.
Oh 19 was so clever
you broke the rules to find 19 and 0 because you used brackets without showing them (3! - .3)/.3 and (3-3)/3
Hello! please check the rules at 1:20 - 1:35, fractions are allowed, and any calculations in the numerator and denominator are valid, so the method is OK. Brackets are not necessary, and threes don't need to be inline for this challenge. :)
Great method
Cool 😎
Thanks! :)
So other number bases are up in the air??
Since they are different counting systems.
For 8 I thought 3+3+ !3 = 8 was most obvious.
This is how I got 19:
3! × 3 + !!3 = 19
Since !3 = 2, !!3 = !2 = 1
As far as I can tell, using '!' twice to get the subfactorial of a subfactorial is allowed by your rules.
@@guyhoghton399 Well...The concept of a double subfactorial isn't a standard term in combinatorics, but it's an interesting idea to explore. If we think of "double subfactorial" as applying the subfactorial operation twice, we might define it as the number of derangements of derangements. However, this leads to some conceptual challenges, as derangements themselves are specific permutations, and deranging them again doesn't have a clear combinatorial meaning.
So to be honest, I never head someone do double subfactorial...but your idea is really interesting.
@@ThePhantomoftheMath Thanks for this feedback. I too was unaware of the existence of "double subfactorial" as a defined concept, which made me think that _!!n_ would therefore have to be unambiguously interpreted as meaning _!(!n)_ which in turn would allow me to get round the "no brackets" rule by simply suppressing them - a bit of a liberty I know!
However your notes inspired me to do a little google research and it seems that a definition of the double factorial does in fact exist at least as a sequence, whether or not it has any meaning in combinatorics. Searching for "online encyclopedia of integer sequences" then on that website for "double subfactorial" gives the details, and it turns out that _!!3 ≠ 1_ so my suggestion above won't work I fear! However your solution for the value 19 is very clever.
I think that would need brackets, well, by how I would define double factorial:
if !n = n! × ∑k=0 to n of ((-1)^k)/k!
then it seems logical to me that !!n = n!! × ∑k=0 to n of ((-1)^k)/k!!
Which I think would mean that !!3 = 1/2
But since he didn't define his terms, and I at least haven't heard of a standard way of defining it, I think your solution is still very interesting, and by the rules he did define, a possible solution!
I always like to define terms in ways which will not have other ways of being written, so we can have a lot of different symbols at our disposal for different uses.
@@JL710ML Funnily enough I did reply to the professor's comment, conceding that my solution would need brackets (and is therefore not a solution according to the rules) because I found that there is indeed a definition of "double subfactorial", but for some reason my reply disappeared. I might try and post it again.
The definition turns out to be close to your suggestion. Searching for "online encyclopedia of integer sequences" then on that website for "double subfactorial" gives the details. I agree with you about the benefits of standardising mathematical notation. Have a nice day.
Doesn't the solution for 0 in fact require brackets since the brackets used are "invisible" in this representation? I do like the video, though.
Hello! I'm glad you liked the video! If you check the rules at 0:40, fractions are allowed, and any calculations in the numerator and denominator are valid, so the method is OK. I suppose by 'invisible brackets' you mean the inline writing of threes, but they don't need to be inline for this challenge. :)
Of course, you can always use alternative solutions, like 3! - 3 - 3 = 0, since there are many ways to get these numbers. Cheers!
@@ThePhantomoftheMath Thanks for the clarification. :)
yes he broke his rules and the same for 19 he used invisible brackets
@@alimetlakI believe I already explained this, so no, none of the rules are broken in this challenge
1:38 What about double factorial?
Sure! You can do double factorials!
I used to do this with 4s instead of 3s and up to 32 whilst travelling jammed packed on the London Underground.
@@CatholicSatan 🤣🤣🤣
WWWOOOWWW 👍👍👏👏👏💐
Challenge: do 3 3's for 21 to 30 with same rules
3^3 - 3^2 + 3^0 = 27 - 9 + 1 = 19. I try it in 30 seconds!
Sorry, but powers are not allowed for this challenge, and numbers other than three number threes. Please check the "Rules" section in the video for more detailed info!
Isn't factorial a function?
(3^3)/3 = 1x9
All falls apart when one insists on writing decimals less than 1 in the proper fashion: 0.3 rather than the godawful .3
For 20 I've a shorter solution:
(3 + 3)/.3 = 20
Great one!
The real question is how far can we go like this without skipping numbers?
Seems like 24 is the limit using only 3 threes and given rules.
Even if 25=3^3-!3 is accepted, 28 seems to be a wall.
there is a surprisingly high amount of combinations to get some numbers. for example, for 8:
3 + 3 + !3
3 + !3 + 3
3 * !3 + !3
3 / .3 - !3
3! / 3 + 3!
3! + 3! / 3
!3 + 3 + 3
!3 + 3 * !3
!3 * 3 + !3
!3 + !3 * 3
!3 * !3 * !3
Yes! Thank you for your contribution and for taking the time to show everyone an impressive number of combinations for just one number!
3 3 3 = 27
3 × 3 × 3 = 27
3! - 3 - !3 = 1
i like this one cuz it's a palindrome
Yeah, that's cool!
Can I use # (Primorial) ?
Or ? (Terminal) ?
Or Cos and Sin ?
Like … (3!)#/3 + 3 = 6#/3 + 3 = 13
Or ((3!)? + (3!)?)/3 = (6? + 6?)/3 = (21 + 21)/3 = 14
Or (3!)? - cos(3-3) = 6? - cos(0) = 21-1 = 20
Not for this challenge...sorry. Not even brackets. But your solutions are really nice!
*SOLUTION*
3!-3-3 = 0
3-3!/3 = 1
3-3/3 = 2
3-3+3 = 3
3+3/3 = 4
3!-3/3 = 5
3!-3+3 = 6
3!+3/3 = 7
(3+3/3)!! = 8
3+3+3 = 9
(3!-3/3)!!! = 10
33/3 = 11
3×3+3 = 12
(3!)!!!!+3/3 = 13
(3!+3/3)!!!!! = 14
3×3+3! = 15
(3!)!!!-(3!/3) = 16
(3!)!!!-(3/3) = 17
(3!)!!!-3+3 = 18
(3!)!!!+(3/3) = 19
(3!)!!!+(3!/3) = 20
I’m pretty sure it’s not allowed to do "!!!!"
@@quentind1924 it's quadruple factorial.
n!...! = n(n-x)(n-2x)
|__| (n-3x)...
x
@@manjuegazos4672 Yes, which isn’t part of the allowed operations
@@quentind1924 he said only using the symbols, so I guess the parentheses are the ones I can't use.
ummmmm ... Your fractions have a problem here. You are representing the fractions using grouping in the numerator. I thought one of the rules for solving was to have all the number threes in line as a single string. If you do this, and have division operator, then none of your division solutions in this video will work!
Hi! Thanks so much for watching and taking the time to comment! I actually mentioned that grouping in the numerator is allowed-please check the 'Allowed' section from 1:20 to 1:35. I also didn’t specify that the numbers need to be in a single line. Hope this clears things up!
@@ThePhantomoftheMath I know, however, this grouping is the same as using parenthesis. The parenthesis is implied by writing the equation in the way you did.
I got all the answers to your fine puzzle correct. However, I could not figure out how to do the division properly without grouping. To me, as a software engineer, grouping is always done in-line using parenthesis.
I completely understand. I promise the next puzzle like this will strictly be in line.
(3-3)/3 = 0
(3!)/(3+3) = 1
3-(3/3) = 2
3-3+3 = 3
3+(3/3) = 4
3! - (3/3) = 5
3! +3-3 = 6
3! + (3/3) = 7
!3 +3+3 = 8
3+3+3 = 9
3! + !3 + !3 = 10
!3 + 3*3 = 11
3!+3+3 = 12
3/.3 + 3 = 13
3!+3!+!3 = 14
3!+3!+3 = 15
3! + 3/.3 = 16
(3!/.3) - 3 = 17
3*(!3*3) = 18
19...best I can think of is (3! * 3)+!(!3). I have to write it that way because !!3 and !(!3) are two different things (at least I presume they are, like how 3!! and (3!)! are two different things).
(3+3)/.3 = 20
@@Psykolord1989 Very, very, very nice! Good job!
@@Psykolord1989 Yeah, n!! and (n!)! are different things in math, but honestly, I’ve never heard of a 'double subfactorial.' In combinatorics, that would be the derangement of a derangement, if that makes any sense...But your idea is really interesting!
19 is this: (3!-0.3)/0.3 = 6/0.3 - 0.3/0.3 = 20 - 1 = 19.
@@joemitchell5869 That works too!
1.defining a function f(x,y,z) = 19, for x,y,z in R.
2. f(3,3,3) = 19
works for other numbers than 19 too 🔥🔥🔥
bro spent all of his skill points into programming and 0 into math skills
3! x3+3^0=19
@@ninafrank9228 Hi! That is indeed 19. However, according to the rules, you can't use the power operation or any new numbers. So, we can't use 0 or any power operations. 🙃
3⁰+3²+3²=19
Exponents aren't allowed in the challenge
3^2+3^2+3^0=19
@@user-wk6je7pi4h Yeah that will give you 19, but power is not allowed, and any other number than 3 is not allowed