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КОМЕНТАРІ: 55
@sorenpetersen570924 дні тому
It is a lot easier tonfactorize 1680, you get 2,2,2,2,3,5,7 to get 4 numbers in sequense an you know 5 and 7 are two of the, if you try 4,5,6,7, you willhave one 2 left over so you get 5,6,7,8, 5x6x7x8=1680 so x+6=8 x must be 2 done
@regard-fk2hl14 днів тому
Oui j'ai résolu comme vous
@gregwroblewski417811 днів тому
I did it pretty much the same way. Took me 20 seconds in my head, maybe less. Since it is a factorial, we know that x+ 2 is a positive integer, and we know that 1680 must be the product of four consecutive integers. It's obvious that combinations like 2-3-4-5 and 3-4-5-6 are too low, so we quickly multiply 4-5-6-7 (still too low) 5-6-7-8 (just right) . Since we know that x+6 (the highest number in the sequence) is 8, x must be 2. I like your additional bit of logic, stipulating that 5 and 7 must be two of the factors., therefore the only possibilities are 4-5-6-7 and 5-6-7-8. I didn't think of that. I think that would have slowed me down, because I can multiply faster than I factor, but well played!
@denisrenaldo35066 днів тому
I did it the same way in 1mn. It’s arithmetical not algebrical. It’s mandatory to use positive integers only. Factorization is the right approach.
@adelsamani7917День тому
It works cos the answers are whole, otherwise his method works.
@denisrenaldo3506День тому
@@adelsamani7917 I disagree. As long as you are considering factorials, the result is obviously an integer. That’s arithmetic not algebraic.
@legomensa25 днів тому
Another method: A prime factorization shows that one set of factors for 1680 is 5,6,7,8. From there, it's straightforward that x=2
@tonitalas175722 дні тому
That's how I did it! Soooo much easier 😊
@yung-kanglee468121 день тому
@@tonitalas1757 And then you missed the second answer -11!
@johnpaullogan136520 днів тому
@@yung-kanglee4681 except that factorials are only defined for non negative integers or if you get rather fancy in your math you can extend it to non negative numbers as a whole but that gets kinda funky and beyond the scope here i think. either way you's ended up with (-5)!/(-9)! plugging that in which is not defined in either the numerator or denominator and as such is not actually a solution to the problem
@yung-kanglee468119 днів тому
@@johnpaullogan1365 yes, you are so smart!!!
@abosharaf2817 днів тому
Good
@riesenbuhai25 днів тому
Thank you intelligent indian man.
@learncommunolizer25 днів тому
Thank you very much!
@antoinegrassi379624 дні тому
Why indian ? Why man ? Why intelligent ?😉🤭
@samircalifornia75002 дні тому
brilliant work
@daniilvolfengaut883814 днів тому
Didnt watch the video, but solved it like this (the way with least possible calculations): 1) It's obvious that we are looking for 4 numbers in sequence with a product of 1680. 2) Obviously, 10*11*12*13 is already too big, because it is larger than 10000. 3) Also we can easily see, that the sequence does not contain 9, because we can divide 1680 by 3 only one time. So it can only be 5*6*7*8 or lower. 4) But on the other hand, the sequence MUST contain 8, because 1680 can be divided by 2 more than 3 times. So it can only be 5*6*7*8. OR you just guess it :)
@ytubekmr2 дні тому
Since the result ends in a zero you know that one of the values must be a 5, so you only have two choices x = 1 or 2.
@akirakasinata-fk8qyДень тому
I just did the 4th root of 1680. Which gives 6.4 roughly in the middle of the row of numbers 5, 6, 7, 8 which would make x=2
@boshboshish17 днів тому
it is actually very short if we just plug and try numbers , meaning ( X+6) (X+5)(X+4)(X+3) , if we directly plug x = till we the number on the right
@antoinegrassi379624 дні тому
Une fois écrit (X+3)(X+4)(X+5)(X+6) = 1680 Peut-on écrire 1680 en produit de 4 nombres consécutifs ? 1680 = 2^4x3x5x7 = 5x(2x3)x7x(2x2x2) = 5x6x7x8 Donc X = 2. Bingo !
@Himasthla9 днів тому
It's a (provocatively?) complex solution. ;) As it has already been written here, the obvious idea is to factorize the right part. Given that x+2 is natural, we immediately come to the only solution x=2.
@talentinvein721324 дні тому
Take group of consecutive 4 numbers in multiple as remaining cancel out
@cachotrelles471525 днів тому
Sensacional, gracias 👏👏👏👍👍
@learncommunolizer25 днів тому
Thank you very much
@joeschmo62224 дні тому
Magic! I didn't do it the "mathy" way but figured that the difference in factorials is 4 contiguous numbers (6-2). Factored 1680 to 2,4,5,6,7 which is 5 numbers, but aha!, 2*4=8, so it's 5,6,7,8, so x+6=8, and so x=2. Less aggravation.
@antoinegrassi379624 дні тому
Ta méthode est parfaitement mathématique. L'autre méthode étant d'une lourdeur peu mathématiques
@user-rw3bg2oj4g4 дні тому
8P4=1680, x=2. Permutation
@mircoceccarelli668921 день тому
👍👍👍
@esmeraldadelossantos700725 днів тому
Tapos, ano nangyari kay x?
@user-dq3uh6ee5w14 днів тому
We can use the special formula too.
@aines79585 днів тому
'thank's🙏 must x#-2
@ruslanklimenko77423 дні тому
ОК.
@petrminar954025 днів тому
Good work
@learncommunolizer25 днів тому
Thank you very much!
@user-sh6yd5uu7f7 днів тому
Здравствуйте. Почему (x+2)! > или = 0? Я думала, что НЕ равен 0 и всё.
@Stanislav_Gorbachyov25 днів тому
Красиво, а главное понятно.
@learncommunolizer25 днів тому
Thank you very much
@Partozan9 днів тому
(Х+2) (x+3) (x+4) (x+5) = 1680. Подбором за 2 мин получаем 5 6 7 8. Х равно 3. 3 мин на решение.
@alekname56442 дні тому
Да, но это ответ для другого уравнения. Плюс решение подбором не засчитывается на олимпиадах (да и на обычных уроках...)