A Nice Factorial Equation | x!=6!7!

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[mbmillermo]

If there is an answer, then we have to be able to form 8, 9, etc., from factors of 6! = 2*3*4* 5*6. Just proceed in order: 8 comes from 2*4. That leaves us with 3*5*6. Then 9 comes from 3*3 which leaves us with 5*2 which is 10. Thus 6!*7! = 10*9*8*7! = 10! That is interesting. How many other factorials can be written as products of factorials?!

[roberttelarket]

Really new factorial type problem possibly never seen before and such a great non-obvious solution! Keep them coming!

[sgjuxta]

Did anyone else nearly feel physical pain when he didn't break the factor of 6 into 2x3 from 7! and was just like "well, this way doesn't work?" 😂

[xirenzhang]

This is from Q11 of Kangaroo Math Malaysia Competition, Student division. It happened 4 days ago on May 18.

[brianjrobinson]

Assuming x! is a natural number, 6! Has a factor of five, so x must be at least 10. But it cannot have a prime factor, so it has to be less than 11. Check to see if 10 is an answer. It is. Solved.

[bjorntorlarsson]

At 2:40, the method DOES work. 4*2=8, yes. But also 3*5*6=90 and 9*10=90. So 7*(6!)^2 = 10! Because 8*9*10 happens to equal 6! None of 8, 9, 10 are a prime and their denominators are 6, 5, 4, 3 and 2. It doesn't work next time three non-primes occur in a row, though, i.e. 14*15*16 is not equal to any factorial. It's just one of those coincidences that happen when dealing with low integers, they can't be combined in that many different ways.

On the limitations in the world of integers, consider this puzzle: A student is taking on puzzles one at a time and either solves each puzzle or not (they cannot be retaken or worked on in parallell). Early on in the course, she has solved less than 75% of the puzzles thus far. But at the end of the course, she has solved more than 75% of all the course's puzzles. Is there a point in time when she has solved EXACTLY 75% of the puzzles taken that far?

Turns out that yes, there must be such a point in time! Let's say she fails the first puzzle, so that we start out with 0 out of 1 which is below 75%. Now we only need to find a way to cross over 75% without ever having solved exactly 75%. How hard can it be? Let's say she simply solves all the rest of the puzzles, so 1/2=50%. Then 2/3=66% and then 3/4 Oops! That would be 75% exactly. So she must fail that one and solve the rest. So from 2/4, 3/5, 4/6, 5/7, 6/Oops AGAIN 75%! So fail that one too and go on solving 5/8, 6/9, 7/10, 8/11, and... 9/12=75% exatly. This be damned! And so it is for any ratio in a sequence like this. One cannot go from less to more than 2/3 without at one time landing on exactly 2/3. Or 99/100 or what have you in the shape (n-1)/n. As I see it, because of the "crowded space" of integers, one can deduce more information from them than what is intuitively apparent, and what works with real numbers.

[ZipplyZane]

I find it's easiest to start with approach 1, then, once you run out of factors, you can then use prime factors for what's left.

It makes sense to split up 6! because it has fewer factors. You get 2*3*4*5*6. You pull out 2*4 to get 8. You can't get 9, so you prime factorize the remaining numbers, and get 3*2*3*5, which allows you to pull out 3*3 to get 9 and 2*5 to get 10.

[krwada]

Very clever. Prime Factorization saves the day

[kemelopharela]

6x5x3 gives u 90 which can be 9x10.
x! = 10!
x = 10

[jaggisaram]

I factorized each factorial to its prim es, took its LCM and got 10!. Hence x = 10

[varadarajcuram]

Thank you sir,
I learnt something which never occurred. I am 70years young and love maths.

[wtspman]

I saw the solution when you decomposed 7!. Breaking 6 into 2x3 gives you the 3 you needed to multiply by 3 to get 9, and the 2 you needed to multiply by 5 to get 10.

[THN--ueyn]

I realised that x is smaller than 11 as 6!*7! has no factor of 11, and any factorial bigger than 10! is a multiple of 11. It's clear that x is not 8 as 6! doesn't equal 8. This only left two options for x, either x=9 or 10. I realised that 6!=720=8*9*10, so 7!*6!= 7!*8*9*10=10!, meaning that x=10.

[RexxSchneider]

We can write x! = 7! * 8 * 9 * ... * x. But we are given x! = 7! * 6! = 7! * 720.
We can take out the 8 from the 720 and make 8! * 90
Then we can take the 9 out of the 90 and make 9! * 10 = 10! So x must equal 10. Maybe 30 seconds work?

[coreyyanofsky]

without watching:
7 is prime so let's look at 6!
6!
= 2·3·4·5·6
= 2⁴·3²·5
= 8·9·10

[ahmedavci]

Starting from 2:56 : the solution was too obvious to ignore, just factorize 6 for God’s sake!!!

[fedorkochemasov]

But you could rearrange the 7! to get a 10!, 7! = 7*6*5*4*3*2, we do need a 7, then 4*2 is 8, so we can get 6!*7*8 = 8!, that's what you had. Then you can look at 6 as 3*2, so 3*3 makes a 9 and 5*2 which is left is 10, thus 8!*9*10 = 10!.

Another way is to see that to get 10! from 8! you need a 90 (9*10) and when you get 8! from 6! and parts of 7! all you have left is 6*5*3 which is exactly 90.

[elevolucionestoica]

Sensacional! Sua explicação tocou minha alma. Desejo-lhe um sucesso constante e uma vida repleta de realizações.

[quigonkenny]

"This approach is not going to work."

It would if you broke apart the 6 into 2•3, and multiplied the 3 by 3 and the the 2 by 5, rather than just giving up...

[krishnamoyghosh]

factorial 7 can be divided into prime nos to get factorial 10