For r!=r+3 you can divide both sides to r (as r is obviously not equal to 0 by substution into equation), that implies (r-1)!=1+3/r and then 3/r must be an integer, so we should check the equation for r=1 & r=3 only.
For this equation, although not elegant, since factorial expands much quickly than square function, brute force is probably the quickest method to solve it
Indeed. I tried 5 and it already worked. n² has to be an odd number, because all factorials are even and if you add 1 to any factorial you'll end up with an odd number. if n² is odd, n has to be odd as well. Therefore to start with 5 is a good idea.
@mohammedlakhfif1335 The factorial function multiplies the integer by ever integer bellow it all the way down to 1 (exp: 5! = 5 * 4 * 3 * 2 * 1 = 120). Squaring just multiplies a number by itself once (exp: 5^2 = 5 * 5 = 25). As you can see, 5! is quite a bit bigger than 5^2. And if you do it with bigger and bigger numbers, the gap will get even bigger because with the factorial, you are multiplying by an entire extra term each time but with squaring, you are just multiplying 1 more higher number by itself. 6^2 = 36 6! = 720 7^2 = 49 7! = 5049 Intuitively, it should be pretty obvious that n * (n - 1) * (n - 2) * ... * 3 * 2 * 1 is bigger than n * n for big n, specifically for all n bigger than 3. So in the problem, that's pretty much what we have. We have (n - 1)! + 1 = n^2 in video. The LHS is very simular to just n!, the only difference is that you are doing it with n - 1 but it will still be really big, and then adding one onto that makes it even bigger. So for big n, the LHS will be bigger than the RHS instead of equal to so that eliminates pretty much every number above like 5.
@@mohammedlakhfif1335 what happens when f(x) turns into f(x+1)? if f(x) = x^2, then f(x+1) = (x+1)^2, which is x^2+2x+1. You *add* 2x+1 if f(x) = x!, then f(x+1) = (x+1)!, which is (x+1) * x!. You *multiply* x+1 For example, let's say x is 9. 9^2 = 81, and 10^2=100. To get from 9^2 to 10^2, you *add* 2x+1 9! = 362,880. 10! is 3,628,800. To get from 9! to 10!, you *multiply* x+1 For big numbers, adding 2x+1 is going to be much less significant than multiplying x+1.
Love your style, including double thinks and checking thinks. You are showing us your mathematical mind in action; a great way to learn! Also I think your perfect chalk on board handwriting is sexy!
I have to admit that I'm in the camp here that's less fond of problems like this; there are interesting lessons to be learned about factoring here but there are better problems for illustrating the same techniques (in particular, ones that come from more natural-looking problems). One small piece to this one that I'll note, though: once you've found the solution n=5 (whether by inspection or otherwise), it's straightforward to show there are no larger solutions: if we write f(n)=n^2 and g(n)=(n-1)!+1 then f(n+1)-f(n) = 2n+1 and g(n+1)-g(n)=n!-(n-1)! = n*(n-1)!-(n-1)! = (n-1)*(n-1)! . It's easy to show that 6*(n-1)>2n+1 for all n>2, and since (n-1)! is always at least six, then (n-1)*(n-1)! > 2n+1 for n>4. In particular this means that the right hand side of the equation will be larger than the left hand side from there on out, since every new difference we add to the right to go from n to n+1 will be bigger than the one we add to the left.
When you get to r! = r + 3, the equation becomes trivial and you can solve it by inspection. Very few factorials will have the property that they're exactly 3 more than the initial number, especially when the factorials become large. So you check the solution up to r = 3 basically and you're done. I like the idea of the method if the equation becomes more involved, but for this one it isn't necessary
Interesting approach. I solved it completely differently. (n-1)! > n^2 for all n > =6 and simply checking all other n demonstares that n=5 is the only solution
I found an easier way to do it by remembering three things; (1) n! only accepts positive integers, (2) n! > n^3 for n > 5, and (3) 0 > a-b for a 5 by (2) and (3), therefore n < 6 by (1). bruteforce, and n = 5 is the only solution.
First of, Love from India, Sir I challenge you to solve JEE ADVANCED maths 2016, one question I've in my mind and its mind boggling. Would be really appreciable if you'd solve that selected one. I visited the channel's about section on YT but there I see no mail ID attached. How one can connect with you sir? You are just awesome, generally I don't see random maths video but iff I love explanation I just binge watch uploaded videos .
Damn your videos are so great I was focusing on your teaching methodology and the board that dumb like me didn't even recognise the email written down in the starting of the video only where we generally see our subtitles.
To make clear at the beginning that its equation, not a theorem, it would be better to use ‘x’ instead of ‘n’. But anyway, presentation is very good. Thank you!
I have another way without changing the variable n: (n-1)!+1=n^2 (n-1)!=n^2-1 (n-1)(n-2)!=(n-1)(n+1) it's clear that 1 isn't a solution to that equation, so we can devise the two sides of that equation by n-1, we get: (n-2)!=n+1 Subtract (n-1) from both sides of the equation, we get: (n-2)!-(n-2)=n+1-(n-2) (n-2)(n-3)!-(n-2)=n+1-n+2 (n-2)[(n-3)!-1]=3 Either n-2=3 and (n-3)!-1=1 Or n-2=1 and (n-3)!-1=3 So for the first case we get: n=2+3 and (n-3)!=1+1 Therefore n=5 and (n-3)!=2 which is satisfied because (5-3)!=2!=2 For the second case we get: n=2+1 and (n-3)!=3+1 Therefore n=3 and (n-3)!=4 which is impossible, so the unique solution is n=5
High school _never_ mentioned factorials. But I was taking the easy route. Sigh. I'm over 50 and "retired." But not knowing this stuff still makes me feel left out of something important. Was I really deemed to be that hopeless in high school? Why did they let me graduate on the honour roll then? 🤷♂
It seems much faster to make an educated guess (factorials get really big really fast so n isn't going to be very large), plug it in and see if it works, then adjust up or down.
We can simple take x = n-2 and rewrite equation as x! -x = 3 By hit and trial we know that x = 3, as difference between a factorial and the number itself gets bigger and bigger as the number goes up. Hence n = 5
When you were checking if r was 1 or 3, would it be permissible to substitute for r in (r-1)!-1 to see if the result is 1 or 3? Or is that not allowed by a rule of logic? Thanks!
(n-1)!=n^2-1=(n-1)(n+1) we can divide by n-1 unless n=1 (n-2)!=n+1 3!=6 n=5 the solutions are n=1 and n=5 except when you plug 1 in for n it doesn't work, so n=1 is an extraneous solution 1+1≠1^2 4!+1=5^2=25
(n - 1)! + 1 = n² (n ≥ 1) (n - 1)! = n² - 1 (n - 1)! = (n - 1)(n + 1) (n - 1)! - (n - 1)(n + 1) = 0 Since n = 1 isn't a solution, n must be greater than 1, so (n - 1)(n - 2)! - (n - 1)(n + 1) = 0 (n - 1) * ((n - 2)! - (n + 1)) = 0 which leads to (n - 2)! - (n + 1) = 0 (n - 2)! - n - 1 = 0 Add 3 to both sides: (n - 2)! - n - 1 + 3 = 3 (n - 2)! - n + 2 = 3 (n - 2)! - (n - 2) = 3 Since n = 2 isn't a solution, then n > 2: (n - 2)(n - 3)! - (n - 2) = 3 (n - 2) * ((n - 3)! - 1) = 3 Now you have a product of two naturals and 3 is a prime number, so there are only two possibilities: n - 2 = 1 and (n - 3)! - 1 = 3 n = 1 + 2 and (n - 3)! = 3 + 1 n = 3 and (n - 3)! = 4 (this one doesn't satisfy the equation because (3 - 3)! ≠ 4) or n - 2 = 3 and (n - 3)! - 1 = 1 n = 3 + 2 and (n - 3)! = 1 + 1 n = 5 and (n - 3)! = 2 Since (5 - 3)! = 2, you have n = 5 as the only solution.
Obviously there is only one "n" that fulfills this equation and "n" can't be a big number, because the faculty of "n" grows much faster than the "n^2": set n=0; then (0-1)! = (-1)! ## negative faculty values are undefined! set n=1; then (1-1)! = 0; and 0! =1 ## by definition set n=2; then (2-1)! = 1 => does not equal n^2 (=4) So we need a bigger "n", try n=10 which equals 362880 => way too big. So try a value between 2 and 10, say 5: set n=5; then (5-1)! + 1 => 4! + 1 => 24 +1 = 25 so with "n=5" both sides of the equation equal "25" and thus "5" is the only solvation.
I used this (n-1)!+1= n² (n-1)(n-2)(n-3)!= n²-1 (n-2)(n-3)!= n+1 [since n≠1] (n-3)!= (n+1)/(n-2) if number is finite and determined,denominator can't be 0 and factorial can't be -ve. So n>2 ; now do hit trial till you get a factorial on LHS. n= 3❌ n=4❌ n=5✅ put n= 5 original equation (n-1)!+1=n² 4! +1 = 25 24+1=25 25=25 Hence n= 5 is the answer
There's nothing in this convoluted method which can be reapplied to any other version of the problem. So it's really no more helpful than the approach I used, which is to look at it and immediately see the obvious answer. If you want to develop a skill that you can reuse, then learn all the factorials up to 7!, all the perfect squares up to 30² and all the cubes up to 12³. This allows you to solve a lot of these clickbait problems just by looking at them.
Disagree. This method works when the solution is 500 or some huge number where brute force does not work. Number theory is all about factorization and eliminating possibilities till you have a handful you can check by hand. This example only had a handful to start, but the method works with others as well. There is a bit of finesse required in the reduction, knowing that you need to factor n-1 for example. No one said it was easy. when you do enough of these problems, things start to get obvious, like he said.
Man I am so angry, I got the first half but I did not get the idea at 5:20, even though I paused the video and thought the whole evening. The solution was so simple at htis point. Love the videos, keep up the good work.
Starting with: (n-1)! + 1 = n^2, and assuming n is a positive whole number. Subtract 1 from both sides, factor the right side: (n-1)! = n^2 -1 = (n+1)(n-1) (Edit to specify factoring using difference-of-squares) Probably the key to solving is the realization that (n-1)! divided by (n-1) is equivalent to (n-2)! (edit after watching: this assumes n-1 does not equal 0) (n-2)! = n+1 It may be conceptually better to substitute: m=n-2, so m! = m+3 (He just uses r instead of m) Considering the multiplying m(m-1)(m-2)... only increases m by 3 means that m (and n) must be pretty small. I'll initially test only m(m-1)=m+3. (Typing this out, you could say I was testing if m-2=1.) What consecutive numbers a and b (b>a) multiplied by each other equal b+3? 2*3=6. So if 3=m, then n=5. Double check: (5-1)! + 1 = 5^2 4! + 1 = 25 4*3*2 = 26 In trying to find additional whole number solutions: smaller numbers are easily disqualified, and prime numbers start getting in the way of the divisibility. And I'm not equipped to investigate factorials in any other domains.
WolframAlpha uses the gamma function (I know it's an extension of factorials for real numbers other than negative integers, but it's over my head to compute) to determine the n can also be approximately -3.02002 or 1.37439.
@@benbookworm Alpha doesn't mention them but there should be infinitely many negative solutions : for every integer n>1 the function Gamma(x) takes all values between 1 and infinity on the interval -2n < x < -2n+1 (in fact, it does so twice!) so there will be two solutions to Gamma(x)=x^2-1 in each of those intervals.
Before watching: tl;dr: Find "n" for which (n+1) can be a factorial, and check in the equation. You will find n=5 is the solution. Alright, I want to separate the factorials from the rest, so I'm gonna subtract 1 from both sides: (n-1)! = n^2 -1. Recall that (a^2-b^2)=(a+b)(a-b), then we have: (n-1)! = (n+1)(n-1) If n-1 does not equal 0, we can divide by it on both sides: (n-1)(n-2)! = (n+1)(n-1) -> (n-2)! = n+1. At this point, I want to start plugging in numbers. (Honestly I could've probably done this at the start, saved us all some time, but eh). WHy? Because (n-2)! grows far faster than n^2. Consider say, n=10. Then we have 8! vs 11. Simply the first two terms of our factorial (8)(7) when multiplied together are far larger than our right hand side, and the other terms make it even larger. This gets more pronounced as you go up in magnitude. Keep this in mind: *each time we increase n by 1 (to "m" = n+1) , we're multiplying the left hand side by (m-2) = (n-1), and adding 1 to the right hand side* . So what numbers do we use? Well, for what numbers could (n+1) equal a factorial? Well let's see...off the top of my head, n=0 would work (doesn't work here because (-2)! isn't something one typically uses). n=1 would give us n+1 = 2 which is 2! (but it doesn't work here because we'd have (1-2)! = (-1)! on the left. n=2 gives us 3, doesn't work; n=4 gives us 5, doesn't work; n=5 gives us 6, which does work, as it is 3!. So, we test n=5: (5-2)! = (5+1) -> 3! = 6 -> 6=6 A solution is n=5. Are there others? No. Not only does (n-2)! *quickly* begin to outscale n+1, but it will take us a while to find a candidate such that n+1 equals a factorial. Specifically, after n=5, it will take us until...n=23, at which point we have 21! = 24, which we can plainly see isn't the answer. We then go to n=119 and...yes, you can see there's not really any way that we're getting back to (n-2)! = n+1. Thus, *our only solution is n=5*
I just thought about which factorials were one less than a perfect square. 4! = 24 comes up straight away, so n=5 is an answer. Since n! grows much faster than n^2, there won't be any others.
The step where you got r! = r+3; where r is an positive integer,then this was obvious that this equation is true for r=3 only, a number whose factorial is 3 more than the number it self. Btw, your lengthy method to deduce r=3 taught many concepts. Thanks
I think you should be a little more careful when saying that (n-1)! = (n-1)(n-2)! because that is only true when n-1>0, i.e. n>1. And so when r! = r(r-1)! But great video!
(n-1)!=n²-1=(n-1)(n+1) As n=1 is not a solution then n-1≠0 so after dividing both sides by n-1: (n-2)!=n+1 (n-2)!-(n-2)=3 As n=2 is not a solution we can divide both sides by n-2 (n-3)!-1=3/(n-2) Notice (n-3)!-1 is an integer, so 3/(n-2) must be an integer wich is possible only for n=1 or n=3 We can check and see than neither are solution, so the equation has no solution (in N) Edit: I am stupid I didn't see that 5 fulfill the last condition, and it is a solution
The easy way: (n-1)! + 1 = n² (n-1)! = n² - 1 (n-1)(n-2)! = (n-1)(n+1) (n=1 is no solution) (n-2)! = n+1 , let t=n-2 (t)! = t+3 t (t-1)! = t (1 + 3/t) (t-1)! = (1 + 3/t) , so t must equal to 3 (t=1 is no option) if t=3, then n=t+2=5
r! = r + 3 => as we know that r! grows much faster than any addition, it is worth to try r = 1, 2, 3, ... One easily finds r = 3. As 4! is already much larger than 7, we know that there is no other solution.
The factorial of a greater than 1 is a positive number and it is quickly rising. This means that n should be odd. Since n+1 is slow rising function it is immediately seen that 5 is the solution.
n² can end with 1,4,9,5 and 6,so (n-1)! Can end with 0,3,8,4,5,it can end with 0 if n-1>=5,and will always last with it after n-1>5,so we need to check the rest,and they all are false becides 4,in which case n=5.If we check,we can see that 5 is the solution,and then put 6, the left part will grow waay more faster than the right part(120K+1=36K is just not possible with natural numbers) which means 5 is the only solution
Supposing that the solution is in IN ( n! Is not seen as the gamma function). For n=1 we have 1+1≠1² Let n≥2. If n is odd then (n-1)! Is even and (n-1)!+ 1 is odd. And on the other hand n² is odd. If n is even (n-1!+1 is odd. And n² is even. So after all n has to be an odd integer.
(n-2)!=n+1 (n-2)(n-3)!=n+1 so (n-3)!=(n+1)/(n-2) and since (n-3)! is an integer then n-2 divides n+1 so: n-2|n+1-(n-2) ==> n-2|3 and you easy see that n=5 is the only solution .
(n-1)!+1=n^2 n ≈ -3.02002626182917709988082269046865640516883553262860189729895872685004092852... n ≈ 1.37439467883103139641850845944445697788973449756654081711426492310703130551...
My sollution was: (n-1)! + 1 = n² (n-1)! = n² - 1 (n-1)(n-2)! = (n+1)(n-1) (n-2)! = n + 1 (Eq. 1) But: (n+1)! = (n+1)n(n-1)(n-2)! Applying Eq.1: (n+1)! = (n+1)²n(n-1) (n+1)n! = (n+1)²n(n-1) n! = (n+1)n(n-1) We look for a number n whose factorial is the product of 3 consecutive numbers, the largest being the one immediately greater than n itself. As 3x2 = 6, 5 is a natural attempt, which turns out to be true: 5! = 5x4x(3x2)x1 = 5x4x6
For a kid in Brooklyn, NY or midland America, your advance problems are frightening. What I have heard from a black high school student is that mathematics, i.e., Algebra, geometry, trigonometry, is too difficult for him. You can extrapolate that to other students including whites. Your problems are so difficult until I can't understand your solution processes although I took calculus I at VCU. A student must crawl before they walk. You don't want to waste time. However, you can lose time if most students can only listen to what you are saying but not understand it. Mathematics is for application not just watching or listening to as watching rich people in a movie live a lifestyle far beyond the average person.
I woud do thefoowing.: (n1)!+1=n^2 (n-1)*(n-2)!=n^2-1=(n-1)*(n+1) Since n=1 is not a soution, n--1 is different fro 0, so we can divide both sides by (n--2): (n-2)!=n+1 n=5 is a solution of this equattion, since 3!=5+1 Factoriall grows much quicker than the inear ter on thhe rightt side (at eastt for n>2), so n=5 is thhe on soution... I justsaw, that n=5 was a soution after having (n2)!=n+1But thesubstitution wasa nice tric, if ou do nott see thesollution..
a lot of mucking around! r! = r+3, so r | 3, hence r is 1 or 3 but 1! 1+3, , and 3! = 6 = 3+3 voila! so r=3, whence n=r+1=4 and of course 4!+1 = 25 = 5^2 Interestingly 4!, 5! and 7! are all one less than a square, but no other values for which n!+1 is a square are known.
For r!=r+3 you can divide both sides to r (as r is obviously not equal to 0 by substution into equation), that implies (r-1)!=1+3/r and then 3/r must be an integer, so we should check the equation for r=1 & r=3 only.
I didn't see that. Good thinking.
When I watched the video today, I saw @PrimeNewtons doing exactly that. 🤔
No he arrived there via a longer method@@Grecks75
For this equation, although not elegant, since factorial expands much quickly than square function, brute force is probably the quickest method to solve it
Indeed. I tried 5 and it already worked. n² has to be an odd number, because all factorials are even and if you add 1 to any factorial you'll end up with an odd number. if n² is odd, n has to be odd as well. Therefore to start with 5 is a good idea.
I just know that factorial grows much faster than a square so that eliminates tons of terms bigger than like 5 right off the bat.
yeah thats what i did
Please can u explain this step with more details
@mohammedlakhfif1335 The factorial function multiplies the integer by ever integer bellow it all the way down to 1 (exp: 5! = 5 * 4 * 3 * 2 * 1 = 120). Squaring just multiplies a number by itself once (exp: 5^2 = 5 * 5 = 25). As you can see, 5! is quite a bit bigger than 5^2. And if you do it with bigger and bigger numbers, the gap will get even bigger because with the factorial, you are multiplying by an entire extra term each time but with squaring, you are just multiplying 1 more higher number by itself.
6^2 = 36
6! = 720
7^2 = 49
7! = 5049
Intuitively, it should be pretty obvious that n * (n - 1) * (n - 2) * ... * 3 * 2 * 1 is bigger than n * n for big n, specifically for all n bigger than 3.
So in the problem, that's pretty much what we have. We have (n - 1)! + 1 = n^2 in video. The LHS is very simular to just n!, the only difference is that you are doing it with n - 1 but it will still be really big, and then adding one onto that makes it even bigger. So for big n, the LHS will be bigger than the RHS instead of equal to so that eliminates pretty much every number above like 5.
@@mohammedlakhfif1335 what happens when f(x) turns into f(x+1)?
if f(x) = x^2, then f(x+1) = (x+1)^2, which is x^2+2x+1. You *add* 2x+1
if f(x) = x!, then f(x+1) = (x+1)!, which is (x+1) * x!. You *multiply* x+1
For example, let's say x is 9.
9^2 = 81, and 10^2=100. To get from 9^2 to 10^2, you *add* 2x+1
9! = 362,880. 10! is 3,628,800. To get from 9! to 10!, you *multiply* x+1
For big numbers, adding 2x+1 is going to be much less significant than multiplying x+1.
n=5 ; 69 years old but not dead !! Good mathematics make a long life.
Indeed Sir.
Love your style, including double thinks and checking thinks. You are showing us your mathematical mind in action; a great way to learn! Also I think your perfect chalk on board handwriting is sexy!
Valeu!
I have to admit that I'm in the camp here that's less fond of problems like this; there are interesting lessons to be learned about factoring here but there are better problems for illustrating the same techniques (in particular, ones that come from more natural-looking problems).
One small piece to this one that I'll note, though: once you've found the solution n=5 (whether by inspection or otherwise), it's straightforward to show there are no larger solutions: if we write f(n)=n^2 and g(n)=(n-1)!+1 then f(n+1)-f(n) = 2n+1 and g(n+1)-g(n)=n!-(n-1)! = n*(n-1)!-(n-1)! = (n-1)*(n-1)! . It's easy to show that 6*(n-1)>2n+1 for all n>2, and since (n-1)! is always at least six, then (n-1)*(n-1)! > 2n+1 for n>4. In particular this means that the right hand side of the equation will be larger than the left hand side from there on out, since every new difference we add to the right to go from n to n+1 will be bigger than the one we add to the left.
Factorial grows super fast. Once can plot two charts: left side and right side; and then see where the curves intersect.
Exactly ... but keep the "+1" on the left side in mind, thus no perfect intersection ;-)
When you get to r! = r + 3, the equation becomes trivial and you can solve it by inspection. Very few factorials will have the property that they're exactly 3 more than the initial number, especially when the factorials become large. So you check the solution up to r = 3 basically and you're done. I like the idea of the method if the equation becomes more involved, but for this one it isn't necessary
Excellent job, sir! New subscriber. I just saw that you have more playlists; I look forward to viewing your additional lessons! 🏆🏆
Hey, boss. That was good especially with your fun manner of delivery!
Glad you enjoyed it!
i solved it mentally by trial and error but couldn't come up with a way to solve it on paper, great vid
Interesting approach. I solved it completely differently. (n-1)! > n^2 for all n > =6 and simply checking all other n demonstares that n=5 is the only solution
I love your clear and concise approach. Very well explained too.
I found an easier way to do it by remembering three things; (1) n! only accepts positive integers, (2) n! > n^3 for n > 5, and (3) 0 > a-b for a 5 by (2) and (3), therefore n < 6 by (1). bruteforce, and n = 5 is the only solution.
couldn't we say earlier (at moment r!=r+3) that the only numer that has its factoriel equal to the same number plus three is five?
If the highest power is 2 then mustn't there be two solutions?
First of, Love from India, Sir I challenge you to solve JEE ADVANCED maths 2016, one question I've in my mind and its mind boggling. Would be really appreciable if you'd solve that selected one. I visited the channel's about section on YT but there I see no mail ID attached. How one can connect with you sir? You are just awesome, generally I don't see random maths video but iff I love explanation I just binge watch uploaded videos .
Damn your videos are so great I was focusing on your teaching methodology and the board that dumb like me didn't even recognise the email written down in the starting of the video only where we generally see our subtitles.
Factor of 3 is not only 1 and 3 we can have 7 x 3/7 etc.
To make clear at the beginning that its equation, not a theorem, it would be better to use ‘x’ instead of ‘n’. But anyway, presentation is very good. Thank you!
I guess it depends where you're from. I've never heard of the variable "n" being specific to theorems.
@@IceMetalPunk thank you!
@@IceMetalPunkindeed, I've never heard of something like that.
I agree, I clicked on the thumbnail to check this
There's no such convention in Australia. Sounds like a US thing
I have another way without changing the variable n:
(n-1)!+1=n^2
(n-1)!=n^2-1
(n-1)(n-2)!=(n-1)(n+1)
it's clear that 1 isn't a solution to that equation, so we can devise the two sides of that equation by n-1, we get:
(n-2)!=n+1
Subtract (n-1) from both sides of the equation, we get:
(n-2)!-(n-2)=n+1-(n-2)
(n-2)(n-3)!-(n-2)=n+1-n+2
(n-2)[(n-3)!-1]=3
Either n-2=3 and (n-3)!-1=1
Or n-2=1 and (n-3)!-1=3
So for the first case we get:
n=2+3 and (n-3)!=1+1
Therefore n=5 and (n-3)!=2 which is satisfied because (5-3)!=2!=2
For the second case we get:
n=2+1 and (n-3)!=3+1
Therefore n=3 and (n-3)!=4 which is impossible, so the unique solution is n=5
High school _never_ mentioned factorials. But I was taking the easy route. Sigh. I'm over 50 and "retired." But not knowing this stuff still makes me feel left out of something important. Was I really deemed to be that hopeless in high school? Why did they let me graduate on the honour roll then? 🤷♂
Excellent video.
n = 5 is a solution. If you replace (n-1)! with Gamma(n), then can you find another solution?
It seems much faster to make an educated guess (factorials get really big really fast so n isn't going to be very large), plug it in and see if it works, then adjust up or down.
We can simple take x = n-2 and rewrite equation as x! -x = 3
By hit and trial we know that x = 3, as difference between a factorial and the number itself gets bigger and bigger as the number goes up. Hence n = 5
I learn both math and english with you professor. Thank you so much for your time.
When you were checking if r was 1 or 3, would it be permissible to substitute for r in (r-1)!-1 to see if the result is 1 or 3? Or is that not allowed by a rule of logic? Thanks!
nice explanation, awesome attitude, great teacher.
Enjoyed u video and ur laughing face with glistering eyes.
i love what you are doing.. it brings me great satisfaction when i see someone solve math problems 💪🏻💪🏻🤓
But the factorial of ~2.666 is 4, so r = 2.666 could also work. How do we know the answer is an integer?
There is no such thing as 2.666! (Factorial of 2.666)
(n-1)!=n^2-1=(n-1)(n+1)
we can divide by n-1 unless n=1
(n-2)!=n+1
3!=6
n=5
the solutions are n=1 and n=5
except when you plug 1 in for n it doesn't work, so n=1 is an extraneous solution
1+1≠1^2
4!+1=5^2=25
(n - 1)! + 1 = n² (n ≥ 1)
(n - 1)! = n² - 1
(n - 1)! = (n - 1)(n + 1)
(n - 1)! - (n - 1)(n + 1) = 0
Since n = 1 isn't a solution, n must be greater than 1, so
(n - 1)(n - 2)! - (n - 1)(n + 1) = 0
(n - 1) * ((n - 2)! - (n + 1)) = 0
which leads to
(n - 2)! - (n + 1) = 0
(n - 2)! - n - 1 = 0
Add 3 to both sides:
(n - 2)! - n - 1 + 3 = 3
(n - 2)! - n + 2 = 3
(n - 2)! - (n - 2) = 3
Since n = 2 isn't a solution, then n > 2:
(n - 2)(n - 3)! - (n - 2) = 3
(n - 2) * ((n - 3)! - 1) = 3
Now you have a product of two naturals and 3 is a prime number, so there are only two possibilities:
n - 2 = 1 and (n - 3)! - 1 = 3
n = 1 + 2 and (n - 3)! = 3 + 1
n = 3 and (n - 3)! = 4
(this one doesn't satisfy the equation because (3 - 3)! ≠ 4)
or
n - 2 = 3 and (n - 3)! - 1 = 1
n = 3 + 2 and (n - 3)! = 1 + 1
n = 5 and (n - 3)! = 2
Since (5 - 3)! = 2, you have n = 5 as the only solution.
Obviously there is only one "n" that fulfills this equation and "n" can't be a big number, because the faculty of "n" grows much faster than the "n^2":
set n=0; then (0-1)! = (-1)! ## negative faculty values are undefined!
set n=1; then (1-1)! = 0; and 0! =1 ## by definition
set n=2; then (2-1)! = 1 => does not equal n^2 (=4)
So we need a bigger "n", try n=10 which equals 362880 => way too big. So try a value between 2 and 10, say 5:
set n=5; then (5-1)! + 1 => 4! + 1 => 24 +1 = 25
so with "n=5" both sides of the equation equal "25" and thus "5" is the only solvation.
I used this
(n-1)!+1= n²
(n-1)(n-2)(n-3)!= n²-1
(n-2)(n-3)!= n+1 [since n≠1]
(n-3)!= (n+1)/(n-2)
if number is finite and determined,denominator can't be 0 and factorial can't be -ve. So n>2 ; now do hit trial till you get a factorial on LHS.
n= 3❌
n=4❌
n=5✅
put n= 5 original equation
(n-1)!+1=n²
4! +1 = 25
24+1=25
25=25
Hence n= 5 is the answer
4:45, why did you introduce the r on the left side? Doesn’t that imbalance the equation sir?
Nevermind you just extended the r factorial okk, i got confused haha!
There's nothing in this convoluted method which can be reapplied to any other version of the problem. So it's really no more helpful than the approach I used, which is to look at it and immediately see the obvious answer.
If you want to develop a skill that you can reuse, then learn all the factorials up to 7!, all the perfect squares up to 30² and all the cubes up to 12³. This allows you to solve a lot of these clickbait problems just by looking at them.
Disagree. This method works when the solution is 500 or some huge number where brute force does not work. Number theory is all about factorization and eliminating possibilities till you have a handful you can check by hand. This example only had a handful to start, but the method works with others as well. There is a bit of finesse required in the reduction, knowing that you need to factor n-1 for example. No one said it was easy. when you do enough of these problems, things start to get obvious, like he said.
Man I saw in the blink of an eye that n should be odd. I am learning!
Man I am so angry, I got the first half but I did not get the idea at 5:20, even though I paused the video and thought the whole evening. The solution was so simple at htis point. Love the videos, keep up the good work.
Starting with: (n-1)! + 1 = n^2, and assuming n is a positive whole number.
Subtract 1 from both sides, factor the right side:
(n-1)! = n^2 -1 = (n+1)(n-1)
(Edit to specify factoring using difference-of-squares)
Probably the key to solving is the realization that (n-1)! divided by (n-1) is equivalent to (n-2)!
(edit after watching: this assumes n-1 does not equal 0)
(n-2)! = n+1
It may be conceptually better to substitute: m=n-2, so
m! = m+3
(He just uses r instead of m)
Considering the multiplying m(m-1)(m-2)... only increases m by 3 means that m (and n) must be pretty small. I'll initially test only m(m-1)=m+3. (Typing this out, you could say I was testing if m-2=1.) What consecutive numbers a and b (b>a) multiplied by each other equal b+3? 2*3=6.
So if 3=m, then n=5.
Double check:
(5-1)! + 1 = 5^2
4! + 1 = 25
4*3*2 = 26
In trying to find additional whole number solutions: smaller numbers are easily disqualified, and prime numbers start getting in the way of the divisibility. And I'm not equipped to investigate factorials in any other domains.
WolframAlpha uses the gamma function (I know it's an extension of factorials for real numbers other than negative integers, but it's over my head to compute) to determine the n can also be approximately -3.02002 or 1.37439.
@@benbookworm Alpha doesn't mention them but there should be infinitely many negative solutions : for every integer n>1 the function Gamma(x) takes all values between 1 and infinity on the interval -2n < x < -2n+1 (in fact, it does so twice!) so there will be two solutions to Gamma(x)=x^2-1 in each of those intervals.
Before watching:
tl;dr: Find "n" for which (n+1) can be a factorial, and check in the equation. You will find n=5 is the solution.
Alright, I want to separate the factorials from the rest, so I'm gonna subtract 1 from both sides:
(n-1)! = n^2 -1.
Recall that (a^2-b^2)=(a+b)(a-b), then we have:
(n-1)! = (n+1)(n-1)
If n-1 does not equal 0, we can divide by it on both sides:
(n-1)(n-2)! = (n+1)(n-1) -> (n-2)! = n+1.
At this point, I want to start plugging in numbers. (Honestly I could've probably done this at the start, saved us all some time, but eh).
WHy? Because (n-2)! grows far faster than n^2. Consider say, n=10. Then we have 8! vs 11. Simply the first two terms of our factorial (8)(7) when multiplied together are far larger than our right hand side, and the other terms make it even larger. This gets more pronounced as you go up in magnitude.
Keep this in mind: *each time we increase n by 1 (to "m" = n+1) , we're multiplying the left hand side by (m-2) = (n-1), and adding 1 to the right hand side* .
So what numbers do we use? Well, for what numbers could (n+1) equal a factorial? Well let's see...off the top of my head, n=0 would work (doesn't work here because (-2)! isn't something one typically uses). n=1 would give us n+1 = 2 which is 2! (but it doesn't work here because we'd have (1-2)! = (-1)! on the left. n=2 gives us 3, doesn't work; n=4 gives us 5, doesn't work; n=5 gives us 6, which does work, as it is 3!. So, we test n=5:
(5-2)! = (5+1) -> 3! = 6 -> 6=6
A solution is n=5. Are there others?
No. Not only does (n-2)! *quickly* begin to outscale n+1, but it will take us a while to find a candidate such that n+1 equals a factorial. Specifically, after n=5, it will take us until...n=23, at which point we have 21! = 24, which we can plainly see isn't the answer. We then go to n=119 and...yes, you can see there's not really any way that we're getting back to (n-2)! = n+1.
Thus, *our only solution is n=5*
I just thought about which factorials were one less than a perfect square. 4! = 24 comes up straight away, so n=5 is an answer. Since n! grows much faster than n^2, there won't be any others.
Great job. Very clear and eloquent.
Factorials grow so much faster than square.. n=10 is already ridiculously high on left. So try a few integers and got it.
The step where you got r! = r+3; where r is an positive integer,then this was obvious that this equation is true for r=3 only, a number whose factorial is 3 more than the number it self.
Btw, your lengthy method to deduce r=3 taught many concepts. Thanks
Solution:
(n - 1)! + 1 = n² |-1
(n - 1)! = n² - 1 |a² - b² = (a - b)(a + b)
(n - 1)! = (n - 1)(n + 1) |:(n - 1)
(n - 2)! = n + 1
Since ! is only defined for positive numbers, it has to be n > 2.
n = 3 → 1! ≠ 4
n = 4 → 2! ≠ 5
n = 5 → 3! = 6
n = 6 → 4! ≠ 7
Therefore n = 5.
I think you should be a little more careful when saying that (n-1)! = (n-1)(n-2)! because that is only true when n-1>0, i.e. n>1. And so when r! = r(r-1)!
But great video!
feels like maths asmr that isn't cringe/overdone.....great videos!!
(n-1)!=n²-1=(n-1)(n+1)
As n=1 is not a solution then n-1≠0 so after dividing both sides by n-1:
(n-2)!=n+1 (n-2)!-(n-2)=3
As n=2 is not a solution we can divide both sides by n-2
(n-3)!-1=3/(n-2)
Notice (n-3)!-1 is an integer, so 3/(n-2) must be an integer wich is possible only for n=1 or n=3
We can check and see than neither are solution, so the equation has no solution (in N)
Edit: I am stupid I didn't see that 5 fulfill the last condition, and it is a solution
Almost correct. You just overlooked that your last condition can also be fulfilled by n=5; and that one is indeed a solution.
@@Grecks75 I wonder how I couldn't see it, I am blind.
If n is a naturel number by wilson's theorem you already know that n is prime because n divides (n-1)! + 1
(5-1)!+1=5^2 n=5
The easy way:
(n-1)! + 1 = n²
(n-1)! = n² - 1
(n-1)(n-2)! = (n-1)(n+1) (n=1 is no solution)
(n-2)! = n+1 , let t=n-2
(t)! = t+3
t (t-1)! = t (1 + 3/t)
(t-1)! = (1 + 3/t) , so t must equal to 3 (t=1 is no option)
if t=3, then n=t+2=5
Great solution but there's a gray area. You needed to take as granted that 0!=1. Is that in agreement to the factorial definition?
Yeah this is a definite fact
@@asparkdeity8717 What is a fact? Are you aware of the factorial definition?
@@pelasgeuspelasgeus4634 there are a million different ways of defining the factorial, and 0! = 1 is consistent with every single one
@@asparkdeity8717 Usually definition is one but can you give an example of all those definitions you know?
@@asparkdeity8717Well? 😂
r! = r + 3 => as we know that r! grows much faster than any addition, it is worth to try r = 1, 2, 3, ... One easily finds r = 3. As 4! is already much larger than 7, we know that there is no other solution.
Easily done by inspection as n = 5...
Factorials don't necessarily have to be integers.
after assumming r=3 You can simply substitute r=3 into formula
This is a different take on the fact that x(x+1)(x+2)(x+3)+1 is always a perfect square. Using that, it is easy to come up with the correct answer.
So interesting, it sounded hard at first, but pretty simple with a few of correct steps
The factorial of a greater than 1 is a positive number and it is quickly rising. This means that n should be odd. Since n+1 is slow rising function it is immediately seen that 5 is the solution.
It's obvious n has to be a quite low number; running through the factorials and squares in my head took all of 10 seconds to find n=5.
Presumably... n here is a positive integer? Otherwise, there could be a lot of solutions (Gamma function wise).
One you find the products of integers = 3 you can say r = 3 it cannot be 1 since the LHS would then be 0.
n² can end with 1,4,9,5 and 6,so (n-1)! Can end with 0,3,8,4,5,it can end with 0 if n-1>=5,and will always last with it after n-1>5,so we need to check the rest,and they all are false becides 4,in which case n=5.If we check,we can see that 5 is the solution,and then put 6, the left part will grow waay more faster than the right part(120K+1=36K is just not possible with natural numbers) which means 5 is the only solution
Man you inspire me to solve maths problems on camera, should I?
why not bro Id like to see them
I just love your videos and learn from it.. you are my math utube teacher
Hit and trial got me think if (n-2)!= N+1 then surely n is less than 6 or so as n! Would be much greater than n² so n=5
Plot Γ(n) and n^2 - 1, pictorially I can imagine only one positive real solution but infinitely many negative ones
7:54 r = 1 -> (r - 1)! -1 = (1 - 1)! - 1 = 0! - 1 = 1 - 1 = 0 ≠ 3. It's simpler and clearer.
I saw the difference of two squares and just guessed by (n-2)!=n+1 but making a new variable is quite genius
Fantastic!!!
Supposing that the solution is in IN ( n! Is not seen as the gamma function).
For n=1 we have 1+1≠1²
Let n≥2.
If n is odd then (n-1)! Is even and (n-1)!+ 1 is odd. And on the other hand n² is odd.
If n is even (n-1!+1 is odd. And n² is even. So after all n has to be an odd integer.
n=3
(n-1)!+1 = 3
n^2 = 9
n=5
(n-1)!+1
n^2
Obviously n- prime, because ((n-1)!,n)=1, and we can show, that if n≥6 (n-1)!+1>n² with induction. After we can consider 3 case, if n=2,3,5
(n-2)!=n+1
(n-2)(n-3)!=n+1 so (n-3)!=(n+1)/(n-2) and since (n-3)! is an integer then n-2 divides n+1 so:
n-2|n+1-(n-2) ==> n-2|3 and you easy see that n=5 is the only solution .
Great! Very nice writing on board !
I saw the thumbnail and thought it was a formula, checked it with a 5, got in the video and found out I've just guessed the answer of the equation...
Now try n!+1=a^2. ive been trying this on the side for a while now
The quality is simply disgusting. Dark and meoko to boot. And the example is very interesting
I stopped at (n-2)!=n+1, knowing that there is a point at which the factorial will always be bigger than the linear term and n>2, I just used a table.
Your explanation is awesome.
Man, why integer?
What about gamma function?
Because factorial is only defined on natural numbers and different from Gamma function.
Nice. Can you solve x! = (x^4) - 6*(x^3) + 11*(x^2) - 6*x
Your videos are awesome !
(n-1)!+1=n^2
n ≈ -3.02002626182917709988082269046865640516883553262860189729895872685004092852...
n ≈ 1.37439467883103139641850845944445697788973449756654081711426492310703130551...
Those are not integers. So you failed.
My sollution was:
(n-1)! + 1 = n²
(n-1)! = n² - 1
(n-1)(n-2)! = (n+1)(n-1)
(n-2)! = n + 1 (Eq. 1)
But:
(n+1)! = (n+1)n(n-1)(n-2)!
Applying Eq.1:
(n+1)! = (n+1)²n(n-1)
(n+1)n! = (n+1)²n(n-1)
n! = (n+1)n(n-1)
We look for a number n whose factorial is the product of 3 consecutive numbers, the largest being the one immediately greater than n itself. As 3x2 = 6, 5 is a natural attempt, which turns out to be true: 5! = 5x4x(3x2)x1 = 5x4x6
you are gifted whe it comes to numbers sir good job
I love your video from Korea.
Dude this is great i love math so much
Théorème de Wilson ?
Just like that, by heart n=5 , (5-1)!=24 , 5^2=25 , 24+1=25 ,
Nice one thank you keep going❤
What's your intro music?
fact(n-2) = n-1, for all n>5 LHS>RHS and for all n
For a kid in Brooklyn, NY or midland America, your advance problems are frightening. What I have heard from a black high school student is that mathematics, i.e., Algebra, geometry, trigonometry, is too difficult for him. You can extrapolate that to other students including whites. Your problems are so difficult until I can't understand your solution processes although I took calculus I at VCU. A student must crawl before they walk. You don't want to waste time. However, you can lose time if most students can only listen to what you are saying but not understand it. Mathematics is for application not just watching or listening to as watching rich people in a movie live a lifestyle far beyond the average person.
I am not being sarcastic. I don't understand your comment.
I woud do thefoowing.:
(n1)!+1=n^2
(n-1)*(n-2)!=n^2-1=(n-1)*(n+1)
Since n=1 is not a soution, n--1 is different fro 0, so we can divide both sides by (n--2):
(n-2)!=n+1
n=5 is a solution of this equattion, since 3!=5+1
Factoriall grows much quicker than the inear ter on thhe rightt side (at eastt for n>2), so n=5 is thhe on soution...
I justsaw, that n=5 was a soution after having (n2)!=n+1But thesubstitution wasa nice tric, if ou do nott see thesollution..
a lot of mucking around!
r! = r+3, so r | 3, hence r is 1 or 3
but 1! 1+3, , and 3! = 6 = 3+3 voila!
so r=3, whence n=r+1=4 and of course 4!+1 = 25 = 5^2
Interestingly 4!, 5! and 7! are all one less than a square, but no other values for which n!+1 is a square are known.
(n -1)! = (n - 1)(n + 1)
(n - 2)! = n + 1
n = 5
(n − 1)! + 1 = n²
(n − 1)! = n² − 1
(n − 1) ⋅ (n − 2)! = (n − 1) ⋅ (n + 1)
(n − 2)! = n + 1