My reasoning even without making any calculations was "if there is an integer solution, it's gotta be 7 as it's bigger than 6 and a prime". Turns out it's 7, so maybe not a bad heuristic lol
Although the value of 7! Is = X! (5040), it doesn't mean that 10!/6!=7! This is called a Mathematical Paradox, where the value of an matematical expression misleads to a wrong interpretation of a formula or a rule..
7! It's simple, 10!/6! = 7*8*9*10. Reduce each number to product of prime factors, then rearrange and it's simple to show that it's 7! But because 7 is a prime number, then the lowest possible answer is 7!
@@petersearls4443 yes!! You're right "the values" is similar to 7!, but is because for some reason 7.8.9.10=5040 But in not because 10!/6!=7! It is a coincidence... It's called a Mathematical Paradox.... n!/(n-x)! is not = [(n-x)+1]!
I’m a mathematics tutor specializing in discrete math. My students and I use factorials all the time and often have them as numerator and denominator like they are in this problem and I have never had a student try to cancel them the way they are in this ‘solution’.
What does "Olympiad" mean? Shouldn't it be a hard problem? This was trivial. Since x! must be less than 10! and x must be an integer, you write x! = 10*9*8*7 = 2*5 * 3*3 * 2*2*2 * 7. Since a factorial is a product of consecutive integers starting with 2, use the 2 in 2*5, the three in 3*3, the 4 in 2*2*2 , the 5 in 2*5 and the 3 and 2 left over to make 6. We made the 2*3*4*5*6 from the 10*9*8.
You don't need calculate anything. Just use a simple logic. x! = 10*9*8*7 so 7 is the prime number which must be one of factors of x! and it must be the biggest factor. That's it.
@@trueriver1950 For some context, the USA Math Olympiad gives you 6 questions, split over 2 days - because you get 4.5 hours per 3 questions. Those interested in the caliber of those questions can look up the problems themselves, each year's contest is there - and no, this would not make the bar. Not even sure it'd make it to AMC 8 (the entry-level competition for middle-school).
@@lechaikuHere you are trusting the question writer. You are also using intuition and/or something you didn't bother writing out to conclude that 8! is too big. If you trust the question writer, don't have to fully justify your work (or only need the answer), and have severely limited time, you could get the answer rapidly from what you say. I'm not a huge fan of "trust the question writer" speed demon problems FWIW but if that's the game you just have to play along.
If the numerator were 16!, the biggest prime factor is 13, but the answer probably isn't 13 and the answer probably isn't a whole number. The existence of an answer then likely hinges on whether you interpret an implied gamma function.
Since 7 is there, you can guess that it is 7!
and verify by decomposing into prime factors and recompose to fill the missing numbers
Yeah, given it has a solution, it's gonna be the highest prime factor, but that feels like cheating.
That's what I said to myself!
Spoileralert :-)
My reasoning even without making any calculations was "if there is an integer solution, it's gotta be 7 as it's bigger than 6 and a prime". Turns out it's 7, so maybe not a bad heuristic lol
Total luck. If 12! is in the numerator, the solution is not so elegant.
Excellent and amene!
5040
10*9*8*7=5040
That’s a math Olympiad problem? That is too simple of a question.
hindsight
Good ❤
Although the value of 7! Is = X! (5040), it doesn't mean that 10!/6!=7!
This is called a Mathematical Paradox, where the value of an matematical expression misleads to a wrong interpretation of a formula or a rule..
7! It's simple, 10!/6! = 7*8*9*10. Reduce each number to product of prime factors, then rearrange and it's simple to show that it's 7!
But because 7 is a prime number, then the lowest possible answer is 7!
It's not 7!
It is:
X=7X8X9X10
@@mulatacatachuflawhich is 5040 and guess what 7! = 5040
@@petersearls4443 yes!! You're right "the values" is similar to 7!, but is because for some reason 7.8.9.10=5040
But in not because 10!/6!=7!
It is a coincidence... It's called a Mathematical Paradox....
n!/(n-x)! is not = [(n-x)+1]!
@@jerry2357 try 12!/6!
@@mulatacatachufla That wasn’t the question posed.
X=7
10!=6!×7×8×9×10
10!/6! =7×8×9×10
=7×2×4×3×3×2×5
=7×(2×3)×5×4×3×2×1
=7×6×5×4×3×2×1
=7!
As per question
X!= 10!/6!
X!=7!
X=7
7!
I did this in my head in about 30 seconds
5040=7!
It's quite easy to show x = 7.
5040
10x9x8x7=5040=7!
That's an old chestnut: the obvious answer is 7. Obvious from the the prime factors...
At the beginning of the video: "What the f...". At the end: "Simple, sexy math! I love it!"
If you can't do it in your head you ain't ready 😂
Well known that 6!x7!=10!
7
2.5
10factorial
7 factorial
Amusing
10！／6！＝10×9×8×7＝7×6×5×4×3×2＝7！よってX＝7……This question is very very easy ……O.K.？
7.
7! Obviously
X=10 ... (X)(X-1)(X-2)(X-3) X=7 ...(X)(X+1)(X+2)(X+3)
x = 10/6! Easy 😂
In 5 sec, I have solved x=7
Gracias
X = 7
7! × 6! = 10!
×=7!
meaningless question, it is just good for this number.
This not interesting to me, because the elegant solution is not applicable to other factorials.
Именно так. Это всего лишь частный случай. Совпадение. Учебная задачка, рассчитанная на удачу, а не на закономерность. Не очень интересно.
7,8,9,10=7,222,33,25=7,23,5,22,3,2=7,6,54,3,2=7!
подогнал под ответ, чертяка
answer is 7
5 min video for, 30 second problem. I thought new generations should be smarter.
I’m a mathematics tutor specializing in discrete math. My students and I use factorials all the time and often have them as numerator and denominator like they are in this problem and I have never had a student try to cancel them the way they are in this ‘solution’.
Result 5040
What does "Olympiad" mean? Shouldn't it be a hard problem? This was trivial. Since x! must be less than 10! and x must be an integer, you write
x! = 10*9*8*7 = 2*5 * 3*3 * 2*2*2 * 7. Since a factorial is a product of consecutive integers starting with 2, use the 2 in 2*5, the three in 3*3, the 4 in 2*2*2 , the 5 in 2*5 and the 3 and 2 left over to make 6. We made the 2*3*4*5*6 from the 10*9*8.
I think the point is to solve it quickly; taking as long as this video does is probably not the way to do it in an Olympics
You don't need calculate anything. Just use a simple logic.
x! = 10*9*8*7
so 7 is the prime number which must be one of factors of x! and it must be the biggest factor. That's it.
@@trueriver1950 For some context, the USA Math Olympiad gives you 6 questions, split over 2 days - because you get 4.5 hours per 3 questions. Those interested in the caliber of those questions can look up the problems themselves, each year's contest is there - and no, this would not make the bar. Not even sure it'd make it to AMC 8 (the entry-level competition for middle-school).
@@lechaikuHere you are trusting the question writer. You are also using intuition and/or something you didn't bother writing out to conclude that 8! is too big. If you trust the question writer, don't have to fully justify your work (or only need the answer), and have severely limited time, you could get the answer rapidly from what you say. I'm not a huge fan of "trust the question writer" speed demon problems FWIW but if that's the game you just have to play along.
If the numerator were 16!, the biggest prime factor is 13, but the answer probably isn't 13 and the answer probably isn't a whole number. The existence of an answer then likely hinges on whether you interpret an implied gamma function.
باترجمه یا زیر نویس فارسی لطفا
Math Olympiad Question?
Please aprove this was a joke.
Done it in 15 seconds in my head!
x! = 10!/6! = 10×9×8×7×6!/6!
x! = 10×9×8×7
x! = (2×5)×(3×3)×(2×4)×7
x! = 2×5×3×(3×2)×4×7
x! = 7×6×5×4×3×2×1
x! = 7!
x = 7
10×9×8×7×6!/6!=X!
5×2×3×3×4×2×7=2×3×4×5×6×7=7!=X!
X=7
7!
7!
7!