Nice clean solution for k >= 3. I would suggest that you need to do a better job of eliminating k = 1, 2. Of course, it's easy enough to do, because 2^k - 8 is negative in those cases.
BEFORE WATCHING: Check the small values for any cases: 0! + 8 = 9 1! + 8 = 9 2! + 8 = 10 3! + 8 = 14 4! + 8 = 32 THERE. 5! + 8 = 128 ALSO THERE. 6! + 8 = 728 I had an idea to consider the last digit of numbers earlier but by this point it hit me that we should probably try a different approach. 6! is obviously bigger than 8 so we can divide through. 6! / 8 is an even number because we can cancel 2 and 4 and still have an even 6 left over. So all n!/8 for n of at least 6 is even. But then n!/8 + 1 is always odd, not a power of 2 unless n! = 0 (which it can never be). So the only solutions are 4! + 8 = 2^5 and 5! + 8 = 2^7.
Honesty that's so fucking cool how that just makes sense. I was shuffling my head thinking of how to do it or even if there were an infinite/impossibly large number of combinations, but just looking at how you interpreted a solution is just great!
I used a little different approach. Let's factor n! = 2^t * p Thus, we have 2^t*p + 2³ = 2^k Or, rearranged: 2^(t-3) * p + 1 = 2^(k-3) From this, we can see that the left side is always odd unless t - 3 = 0, and the right side is always even unless k - 3 = 0 So the only solutions are those when both sides are even or both are odd. For even we have t = 3, this corresponds to a n! with three 2s in it: it's either 4! or 5! For odd, we have k = 3, but for that 2^t*p must = 0 which never happens So, now we have possible values 4 and 5 for n, we can plug them in and see n = 4 : 4 = 2^(k-3) => k = 5 n = 5 : 16 = 2^(k-3) => k = 7 So, the answer is 4, 5 and 5, 7
I used pretty much the same logic to solve this as presented in the video. This kind of problems can look a bit intimidating at first, but when you start solving them the solution turns out pretty easy. In fact, these are quite fun to solve.
I'm so happy I got the same result with almost the same method. But I divided both terms by 2^3 So n!/2^3 + 1 = 2^ (k-3) ODD + ODD = EVEN Also they must be integers. So, n>=4 because must be integer and n
A very elegant solution for sure! I probably would have sat down with a spreadsheet and tried various pairs only to then finally realize what is going on.
Nice! When I did it, I noticed that if k > 3, then RHS is a multiple of 16, which the LHS can't be for n > 5 (n! is a multiple of 16 starting with n = 6, but subtracting 8 makes the LHS not a multiple of 16). So then we just have to check n < 6. 4 and 5 are the only n values that are 8 less than a power of 2, which correspond to k = 5 and k = 7, respectively.
The way I would work this. (I immediately paused the video, then made this comment before watching the video). n! + 8 = 2^k n! = 2^k - 8 n! = 2^k - 2^3 Writing out the factorials up to 7 for n € N. 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = 5040 At this point, I can only see k = 5, and k = 7. If n! (4!) = 24, and for k = 5, 2^5 = 32; 32 - 8 = 24 If n! (5!) = 120, and for k = 7, 2^7 = 128; 128 - 8 = 120. The next factorial will be 6! = 720. 720 - 8 = 712. 712 cannot be written in the form 2^k where k € N. Again, the next factorial for n = 7 will be 7! = 5040. 5040 - 8 = 5032. 5032 cannot be written in the form 2^k, where k € N. Therefore, k = {5, and 7}
To show the equation has no integers solutions for n≥5 we can do this: Let's take n>5 and k≥4. Now notice that n! is a multiple of 5 so 2^k-8 must also be a multiple of 5, 8 leaves a remainder of 3 when divided by 5 so 2^k must also leave a remainder of 3 because 2^k-8 is a multiple of 5 but this means that 2^k is of the form 5n+3 which is an odd number, contradiction. So n≤5.
Damnn thanks sir because of you, i finally am able to solve mathematics with utmost proficiency and also this problem seemed quite easy for me to being able to think the steps through!! ❤🎉
I mean, yeah, but the idea is still the same so it’s not like someone is going to really want to solve 1 but not the other so it’s not an actual issue.
@@DroughtBee-- "... so it's not an actual issue." *False!* A person could waste time on the on the thumbnail,, only to to find the host worked on another problem.. Stop making excuses and being an apologist for errors such as this. It is a lazy and stupid post.
Looking at the problem, I knew that, I there are not many solutions (may be just One only) and the numbers are not large. I found one pair (5, 7). So I learn from this problem, applying logic = mathematical logic.
The fact that you said that you can’t figure out the answer to some of life’s problems in mathematics gives the reminiscence that we’ve been sailing in the same boat be it physics or math 😅
Your solution is great!! That's my solution here: Lets n>=6. So n factorial can be divisible by 16. With this fact we can factor out 8 in the left side: n!+8=8(n!/8+1) 8(n!/8+1)=2^k n!/8 is still can be divisible by 2 because n! can be divided by 16. So n!/8+1 is wont be an even number. We got that the power of 2 is divided by a not even number, and we got contradiction, so our first assumption that n>=6 was wrong and now n
Since m! ≥ 1 , (m! + 8) ≥ 9 Hence, 2ᵏ > 8 2ᵏ > 2³ k > 3 Since k > 3, k - 3 > 0 (k - 3) ∈ ℕ Hence, 2ᵏ⁻³ is even (2ᵏ⁻³ - 1) is odd And 2ᵏ⁻³ > 2⁰ 2ᵏ⁻³ > 1 (2ᵏ⁻³ - 1) > 0 m! + 8 = 2ᵏ m! = 2ᵏ - 8 m! = 2ᵏ - 2³ m! = 2³(2ᵏ⁻³ - 1) m! = 8(2ᵏ⁻³ - 1) --Eqn(1) Since (2ᵏ⁻³) - 1 ∈ ℕ, 8 | m! Hence, The smallest even integers that m! is divisible by are 2 and 4 Why? Because 2 x 4 = 8 m! is a product of positive consecutive integers starting from 1 Hence, If m! is divisible by 2 and 4, Smallest m! = 4! Hence, m = 4 Hence, m! must also be divisible by 3, which is between 2 and 4 From Eqn(1), This means that 2ᵏ⁻³ - 1 = 3 Since (2ᵏ⁻³ - 1) is odd, It is possible that 2ᵏ⁻³ - 1 = 3 Let’s check: 2ᵏ⁻³ - 1 = 3 2ᵏ⁻³ = 4 k - 3 = 2 k = 5 Hence, k ∈ ℕ ✓ Hence, (m, k) = (4, 5) If m! is divisible by 2 and 4, The next smallest m! is 5! Hence, m = 5 Hence, m! is also divisible by 3 and 5 3 x 5 = 15 15 is odd Hence, It is possible that 2ᵏ⁻³ - 1 = 15 Let’s check: 2ᵏ⁻³ - 1 = 15 2ᵏ⁻³ = 16 k - 3 = 4 k = 7 Hence, k ∈ ℕ ✓ Hence, (m, k) = (5, 7) If m > 5, m! > 5! Hence, m! is even From Eqn(1), (2ᵏ⁻³ - 1) is even because (2ᵏ⁻³ - 1) is divisible by 6 Since (2ᵏ⁻³ - 1) cannot be even, m > 5 is not possible Therefore, There are only 2 ordered pairs (m, k) Ans: (m, k) = (4, 5), (5, 7)
But you did not really prove that when n> 5 then the results will be 8 x even number…I was expecting a mathematical prove in this point since that it is an olympiad problem…
Solution: Let's divide both side by 8: n!/8 + 1 = 2^(k - 3) The right side is always even, while the left side is only even, as long as n!/8 is odd! 8 = 2 * 4, so n!/8 for n ≥ 4 can be rewritten to n!/4! * 3 * 1 or just as n!/4! * 3 Let's first check n < 4 1! + 8 = 9 → not a valid solution for 2^k 2! + 8 = 10 → not a valid solution for 2^k 3! + 8 = 14 → not a valid solution for 2^k So if there is a solution it has to be with n!/4! * 3. But even * odd = even, so n!/4! HAS to be odd. But as soon as n! has an even factor > 4, the solution of n!/4! is always even. Therefore the only two solutions for n are 4 and 5: n = 4 → n!/4! = 4!/4! = 1 → which is odd and therefore a possible valid solution n = 5 → n!/4! = 5!/4! = 5 → which is odd and therefore a possible valid solution Test n = 4: n! + 8 = 2^k 4! + 8 = 2^k 24 + 8 = 2^k 32 = 2^k 2^5 = 2^k k = 5 → valid solution Test n = 5: n! + 8 = 2^k 5! + 8 = 2^k 120 + 8 = 2^k 128 = 2^k 2^7 = 2^k k = 7 → valid solution So the only two valid pairs for (n, k) are (4, 5) and (5, 7)
Pointlessly long video. If n>6 then n!+8=8(n!/8+1) where n!/8+1 is odd and cannot be factor of 2^k. checking all integers less than 6 we get n=4 k=5 and n=5 k=7
@raivogrunbaum4801 How do you know it's only for n < 6 that make it even? Edit: Is it because if n >= 8 then iy divides with the 8 in n!/8 + 1 and which means it's a multiple of 2 (even) + 1 which makes it odd? And then you can test n = 7 to see it is odd as well.
![[Full] 4 ต่อ 4 Celebrity EP.922 | 10 พ.ย. 67 | one31](http://i.ytimg.com/vi/zmV5HRIxIBI/mqdefault.jpg)
I love the use of simple logic to solve a problem that can appear complicated.
Somehow you have the Bob Ross vibes, so relaxing, soft spoken, just with numbers instead of paint and brushes.
Wanted to tell you that you are simply amazing !!
You make number theory look so easy... if only my mind could keep up in practice.
Nice clean solution for k >= 3. I would suggest that you need to do a better job of eliminating k = 1, 2. Of course, it's easy enough to do, because 2^k - 8 is negative in those cases.
BEFORE WATCHING:
Check the small values for any cases:
0! + 8 = 9
1! + 8 = 9
2! + 8 = 10
3! + 8 = 14
4! + 8 = 32 THERE.
5! + 8 = 128 ALSO THERE.
6! + 8 = 728
I had an idea to consider the last digit of numbers earlier but by this point it hit me that we should probably try a different approach.
6! is obviously bigger than 8 so we can divide through.
6! / 8 is an even number because we can cancel 2 and 4 and still have an even 6 left over. So all n!/8 for n of at least 6 is even.
But then n!/8 + 1 is always odd, not a power of 2 unless n! = 0 (which it can never be).
So the only solutions are 4! + 8 = 2^5 and 5! + 8 = 2^7.
Honesty that's so fucking cool how that just makes sense. I was shuffling my head thinking of how to do it or even if there were an infinite/impossibly large number of combinations, but just looking at how you interpreted a solution is just great!
*Stop* with your major cursing, you ignorant jerk. It is inappropriate and needless.
Being an INDIAN i like the namaste at the end!🙏
And by the way a great solution!!
4!+8=2^5 5!+8=2^7 n! is an odd multiple of 8 The only possible values of n are n=4 and n=5
Another math puzzle solved with logic (and magic)! Math has always been my way to mentally relax.
A great solution! 👍
Far and away the best maths channel
Prime Newtons, Dude you solve the coolest problems ! Subscribed 🤗
I used a little different approach.
Let's factor n! = 2^t * p
Thus, we have 2^t*p + 2³ = 2^k
Or, rearranged:
2^(t-3) * p + 1 = 2^(k-3)
From this, we can see that the left side is always odd unless t - 3 = 0, and the right side is always even unless k - 3 = 0
So the only solutions are those when both sides are even or both are odd.
For even we have t = 3, this corresponds to a n! with three 2s in it: it's either 4! or 5!
For odd, we have k = 3, but for that 2^t*p must = 0 which never happens
So, now we have possible values 4 and 5 for n, we can plug them in and see
n = 4 : 4 = 2^(k-3) => k = 5
n = 5 : 16 = 2^(k-3) => k = 7
So, the answer is 4, 5 and 5, 7
I used pretty much the same logic to solve this as presented in the video. This kind of problems can look a bit intimidating at first, but when you start solving them the solution turns out pretty easy. In fact, these are quite fun to solve.
I'm so happy I got the same result with almost the same method. But I divided both terms by 2^3
So
n!/2^3 + 1 = 2^ (k-3)
ODD + ODD = EVEN
Also they must be integers.
So, n>=4 because must be integer and n
Intro of every video with music awesome 🤩
a flying white chalk is solving complex math questions
A very elegant solution for sure! I probably would have sat down with a spreadsheet and tried various pairs only to then finally realize what is going on.
I love the silent happy nods.
Nice! When I did it, I noticed that if k > 3, then RHS is a multiple of 16, which the LHS can't be for n > 5 (n! is a multiple of 16 starting with n = 6, but subtracting 8 makes the LHS not a multiple of 16). So then we just have to check n < 6. 4 and 5 are the only n values that are 8 less than a power of 2, which correspond to k = 5 and k = 7, respectively.
Nice.
I like your channel as the problem you presenting is a challenging one.
Intelligent way to teach maths.❤
can we use logarithms?? To sove this!
Yes, logarithms can be used 😉
@@alexandermorozov2248 Could you help me out, please? I don't see how logarithms would be helpful here.
This is one of my favourite math olympiad problems
Nicely done!
well done! cute problem.
Gorgeous thx deeply ❤
from Morocco thank you for this clear complete proof
Prime Newtons: n! + 8 = 2^k, n, k ϵ ℕ(positive integer); n, k =?
Trial-and-error math solution:
n! = 2^k - 8 = 8[2^(k - 3) - 1]; k > 3
k = 4: 4[2^(4 - 3) - 1](2) = 4(1)(2)(1) = 8; k > 4
k = 5: 4[2^(5 - 3) - 1](2) = 4(3)(2)(1) = 4!; n = 4
k = 6: 4[2^(6 - 3) - 1](2) = 4(7)(2)(1) = 56; k > 6
k = 7: 4[2^(7 - 3) - 1](2) = 4(15)(2)(1) = 5(4)(3)(2)(1) = 5!; n = 5
k = 8: 4[2^(8 - 3) - 1](2) = 4(31)(2)(1) = 248; No more answer
Final answer:
n = 4, k = 5 or n = 5, k = 7
Very interesting, thanks
The way I would work this. (I immediately paused the video, then made this comment before watching the video).
n! + 8 = 2^k
n! = 2^k - 8
n! = 2^k - 2^3
Writing out the factorials up to 7 for n € N.
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
At this point, I can only see k = 5, and k = 7.
If n! (4!) = 24,
and for k = 5, 2^5 = 32; 32 - 8 = 24
If n! (5!) = 120,
and for k = 7, 2^7 = 128; 128 - 8 = 120.
The next factorial will be 6! = 720. 720 - 8 = 712. 712 cannot be written in the form 2^k where k € N.
Again, the next factorial for n = 7 will be 7! = 5040. 5040 - 8 = 5032. 5032 cannot be written in the form 2^k, where k € N.
Therefore, k = {5, and 7}
To show the equation has no integers solutions for n≥5 we can do this:
Let's take n>5 and k≥4. Now notice that n! is a multiple of 5 so 2^k-8 must also be a multiple of 5, 8 leaves a remainder of 3 when divided by 5 so 2^k must also leave a remainder of 3 because 2^k-8 is a multiple of 5 but this means that 2^k is of the form 5n+3 which is an odd number, contradiction. So n≤5.
Damnn thanks sir because of you, i finally am able to solve mathematics with utmost proficiency and also this problem seemed quite easy for me to being able to think the steps through!! ❤🎉
The problem statement in the title of the video is different from the actual statement.
Yes, and that is bad.
I thought n! should have three 2s, since 8 in RHS composed of three 2s and 2^(k-3)-1 is odd. Therefore, possible n are 4 and 5.
There are just two cases when the factorial outcomes are odd numbers: 0! = 1, or 1! = 1. even x even = even, odd x odd = odd, even x odd = even
Before watching... rule out n>8 because (n!/8 + 1) is odd. Check n = 0 to 7, solutions are (4,5) and (5,7)
Well explained
Title and thumbnail do not agree. Title is wrong
I mean, yeah, but the idea is still the same so it’s not like someone is going to really want to solve 1 but not the other so it’s not an actual issue.
If he fixes the title, here is the original title
n! = 8 + 2^k
@@DroughtBee-- "... so it's not an actual issue." *False!* A person could waste time on the on the thumbnail,, only to to find the host worked on another problem.. Stop making excuses and being an apologist for errors such as this. It is a lazy and stupid post.
Thank you. I changed the thumbnail
@@PrimeNewtonsBut the title is the thing that’s wrong
You're a natural born teacher, my friend
math videos are cool, but this guy is such a good performer that I'm not sure if like his acting more than the math
Good job
When n=5 k=7 5!+8=2^7 128=128 solution.
Looking at the problem, I knew that, I there are not many solutions (may be just One only) and the numbers are not large. I found one pair (5, 7).
So I learn from this problem, applying logic = mathematical logic.
Did you understand later that there is another pair?
at time = 7:30 , I did not understand why beyond 6 all possible values are always even. can anyone explain that to me? thanks!
Because in n!, you will have at least one even factor, and an even number times any other number always gives an result that is even.
n! is an odd multiple of 8, so when we get 6 it becomes an even multiple of 8 because 6 is even
Can you teach is how to make math questions?
nice. very nice.
Congrats we 🔓 Namaskar
Any video for Metric spaces
Haha! Nope
@@PrimeNewtons okay .....I would like to see the approach
The fact that you said that you can’t figure out the answer to some of life’s problems in mathematics gives the reminiscence that we’ve been sailing in the same boat be it physics or math 😅
sir kindly make videos on Tayler series,Euler series ,Runge Kutta method
make a play list based om olympiad problem solving
Solutions (n,k)=(4,5),(5,7)
Put a comma between those pairs of parentheses: (4, 5), (5, 7)
n!+8=2^k --> n!+2³=2^k
(n,k)=(4,5),(5,7)
👍👍👍
❤️
2^1=2 , 2^2=4 , ..... , 2^5=32 , 2^6=64 , 2!7=128 , 1!=1 , 2!=2 , 3!=6 , 4!=24 , 5!=120 , --> 2^7=128 , 5!=120 , 120+8=128 ,
n!=120 , n=5 , 2^k=128 , k=7 ,
Number theory is a broad subject.......fr real😊
Your solution is great!! That's my solution here:
Lets n>=6. So n factorial can be divisible by 16.
With this fact we can factor out 8 in the left side:
n!+8=8(n!/8+1)
8(n!/8+1)=2^k
n!/8 is still can be divisible by 2 because n! can be divided by 16.
So n!/8+1 is wont be an even number.
We got that the power of 2 is divided by a not even number, and we got contradiction, so our first assumption that n>=6 was wrong and now n
Video title is different from the question
At first sight (n=5,k=7)
but there may be more solutions like (n=4,k=5)
And it seems that i guessed them all
When n equals 4,k equals 5. When n equals 5,k equals 7.
Put a comma after "4."
"When n equals 4, k equals 5."
Since m! ≥ 1 ,
(m! + 8) ≥ 9
Hence,
2ᵏ > 8
2ᵏ > 2³
k > 3
Since k > 3,
k - 3 > 0
(k - 3) ∈ ℕ
Hence,
2ᵏ⁻³ is even
(2ᵏ⁻³ - 1) is odd
And
2ᵏ⁻³ > 2⁰
2ᵏ⁻³ > 1
(2ᵏ⁻³ - 1) > 0
m! + 8 = 2ᵏ
m! = 2ᵏ - 8
m! = 2ᵏ - 2³
m! = 2³(2ᵏ⁻³ - 1)
m! = 8(2ᵏ⁻³ - 1) --Eqn(1)
Since
(2ᵏ⁻³) - 1 ∈ ℕ,
8 | m!
Hence,
The smallest even integers that m! is divisible by are 2 and 4
Why?
Because
2 x 4 = 8
m! is a product of positive consecutive integers starting from 1
Hence,
If m! is divisible by 2 and 4,
Smallest m! = 4!
Hence,
m = 4
Hence,
m! must also be divisible by 3, which is between 2 and 4
From Eqn(1),
This means that
2ᵏ⁻³ - 1 = 3
Since
(2ᵏ⁻³ - 1) is odd,
It is possible that
2ᵏ⁻³ - 1 = 3
Let’s check:
2ᵏ⁻³ - 1 = 3
2ᵏ⁻³ = 4
k - 3 = 2
k = 5
Hence,
k ∈ ℕ ✓
Hence,
(m, k) = (4, 5)
If m! is divisible by 2 and 4,
The next smallest m! is 5!
Hence,
m = 5
Hence,
m! is also divisible by 3 and 5
3 x 5 = 15
15 is odd
Hence,
It is possible that
2ᵏ⁻³ - 1 = 15
Let’s check:
2ᵏ⁻³ - 1 = 15
2ᵏ⁻³ = 16
k - 3 = 4
k = 7
Hence,
k ∈ ℕ ✓
Hence,
(m, k) = (5, 7)
If m > 5,
m! > 5!
Hence,
m! is even
From Eqn(1),
(2ᵏ⁻³ - 1) is even because
(2ᵏ⁻³ - 1) is divisible by 6
Since (2ᵏ⁻³ - 1) cannot be even,
m > 5 is not possible
Therefore,
There are only 2 ordered pairs
(m, k)
Ans:
(m, k)
= (4, 5),
(5, 7)
No response
😮
BY OBSERVATION
n! + 8 = 2^k
4!=24 , 24+8=32 , 32=2^5
For
n! - 8 = 2^k
4!=24 ,24-8=16=2^4 then k=4
You missed k = 7. For 2^k, 2^7 = 128. 128 - 8 = 120. 5! = 120.
But you did not really prove that when n> 5 then the results will be 8 x even number…I was expecting a mathematical prove in this point since that it is an olympiad problem…
2^k must be strictly greater than 2^3 since all factorials are strictly greater than 0.
The Title is wrong.
It should be n! + 8 = 2^k not n! = 8 + 2^k
That is correct.
sir not to be mean but the video title is wrong its n! = 8+ 2^k instead of n! + 8 = 2^k
He needs to correct these errors as soon as he is told.
Sneaky solution
4:49 EXCEPT 0
You say "n! is an odd multiple of 8" and it is not possible. I think the real situation is that n!/8 must be odd
Huh? Saying that n! is an odd multiple of 8 and saying that n!/8 is odd mean exactly the same thing.
Solution:
Let's divide both side by 8:
n!/8 + 1 = 2^(k - 3)
The right side is always even, while the left side is only even, as long as n!/8 is odd!
8 = 2 * 4, so n!/8 for n ≥ 4 can be rewritten to n!/4! * 3 * 1 or just as n!/4! * 3
Let's first check n < 4
1! + 8 = 9 → not a valid solution for 2^k
2! + 8 = 10 → not a valid solution for 2^k
3! + 8 = 14 → not a valid solution for 2^k
So if there is a solution it has to be with n!/4! * 3.
But even * odd = even, so n!/4! HAS to be odd.
But as soon as n! has an even factor > 4, the solution of n!/4! is always even.
Therefore the only two solutions for n are 4 and 5:
n = 4 → n!/4! = 4!/4! = 1 → which is odd and therefore a possible valid solution
n = 5 → n!/4! = 5!/4! = 5 → which is odd and therefore a possible valid solution
Test n = 4:
n! + 8 = 2^k
4! + 8 = 2^k
24 + 8 = 2^k
32 = 2^k
2^5 = 2^k
k = 5 → valid solution
Test n = 5:
n! + 8 = 2^k
5! + 8 = 2^k
120 + 8 = 2^k
128 = 2^k
2^7 = 2^k
k = 7 → valid solution
So the only two valid pairs for (n, k) are (4, 5) and (5, 7)
After the video:
Very different and very complex approach to the same solution...
Pointlessly long video. If n>6 then n!+8=8(n!/8+1) where n!/8+1 is odd and cannot be factor of 2^k. checking all integers less than 6 we get n=4 k=5 and n=5 k=7
n!/8 + 1 is not a factor of 2^k itself because it's odd but what does that have to do with n>6? You can say that about any n.
@@maxhagenauer24 no you cant if n
@raivogrunbaum4801 How do you know it's only for n < 6 that make it even?
Edit: Is it because if n >= 8 then iy divides with the 8 in n!/8 + 1 and which means it's a multiple of 2 (even) + 1 which makes it odd? And then you can test n = 7 to see it is odd as well.
This is high school math. He is trying to create most accessible video.
@@souverain1er I never even learned what the factorial was in high school.