Two math problems from Africa.

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7 сен 2024

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Комментарии : 95

@dimy931 2 года назад

First problem alternative solution: Changing x_i from 1 to -1 can only change the sum by adding or subtracting 4. Therefore the remainder of the sum modulo 4 is an invariant of the sum. Checking the sum with all 1s gives is 2009 =1(mod4), 999 =3(mod 4). Contradiction with the assumption there is any such xs

@lefty5705 2 года назад

Same with my approach :D I imagine a circle with 2009 numbers (which is 1 or -1) are surrounding and switches 1 to -1 or -1 to 1. Then the remainder of the sum modulo 4 is not changed by any swap

@TikeMyson69 2 года назад

Nice solution

@SanketAlekar 2 года назад

Yep I did this approach as well.

@tomatrix7525 2 года назад

Lovely approach

@helo3827 2 года назад

he started singing christmas songs from 0:31

@threstytorres4306 2 года назад

IF N = 4 We Have: (4³+3)/(4²+7) (64+3)/(16+7) 67/23 Since 67 is a PRIME NUMBER, The quotient of the two is a decimal number IF N = 5, We Have: (5³+3)/(5²+7) (125+3)/(25+7)) 128/32 =4

@Pablo360able 2 года назад

What’s great about that is that, in both cases, you don’t even need to do any math once you’ve got the numerator and the denominator to find the answer. 67 is a prime and the denominator is clearly not going to be 1 or 67, so 4 is right out, and 128 and 32 are both instantly recognizable as powers of 2, so even without taking the quarter of a second to work out their ratio, it’s obviously going to be a natural number.

@Bodyknock 2 года назад

It shouldn’t be too surprising that the zeroes 2 and 5 are both solutions since they both correspond to (7n-3) = n² + 7 . 3 and 4 are the ones that it’s kind of unclear without checking.

@stepanvalerivich6999 2 года назад

For the second problem, an alternative straightforward (but not as elegant) solution would be testing small values of n to see 2 and 5 work, then noticing and proving with induction that for n greater than 5, 7n-3 < n^2+7, or the numerator of the fraction bit < denominator and hence the whole expression isn’t a whole number

@user-ik2kd9mb5t 2 года назад

It's obvious that for natural n 7n-3 is positive so (7n-3)/(n^2+7) is positive too

@Happy_Abe 2 года назад

Yes but not always greater than 1 in absolute value

@user-ik2kd9mb5t 2 года назад

@@Happy_Abe this allows us to exclude

@linggamusroji227 2 года назад

@@Happy_Abe but we want it to be an integer, and it is greater than 0, so it is >=1

@Happy_Abe 2 года назад

@@user-ik2kd9mb5t yes that’s true, good point!

@IoT_ 2 года назад

Самое интересное, если n - целое , то оно может быть только натуральным. Так что, можно было вначале утвердить, что n - целое для большего интереса.

@goodplacetostop2973 2 года назад

10:41 Homework 10:46 Good Place To Stop

@Pablo360able 2 года назад

Thanks for including the timestamp for the good place to stop. Otherwise, I might never have been able to stop the video.

@goodplacetostop2973 2 года назад

@@Pablo360able No problem 👍

@jeffreycloete852 2 года назад

Hi Prof Penn just wanted to wish u a healthy and speedy recovery for the hand!!

@Megathescientist 2 года назад

Another way to solve the first problem is the following: 2*(x1x2+...+x2009x1)=999*2=1998 but the LHS can be rewritten as (x1+..+x2009)^2-(x1^2+...+x2009^2)=1998 (x1+...+x2009)^2-2009=1998 (because x1^2=...=x2009^2=1) so (x1+...+x2009)^2=4007, which is impossible because 4007 isn't a perfect square (in fact 4007mod4=-1)

@tonyhaddad1394 2 года назад

Hey michael in the second problem you made it overkiller , you can simply add 7n and subatract it to get n + (3 - 7n)/n^2 +7 then by the absolute inequality you can solve it easly

@mcwulf25 2 года назад

Yes I posted the same. It's more intuitive as we know exactly what we want to extract from the numerator.

@yoav613 2 года назад

10:41 michael penn easiest h.w ever

@mcwulf25 2 года назад

The long division seems less intuitive to me than inserting +7n - 7n into the numerator, to factor out the n. I was going to try and solve this using the resulting quadratic to find conditions where D is a perfect square. But your way is simpler!

@cQunc 2 года назад

That's effectively what the long division does. It just looks different.

@mcwulf25 2 года назад

@@cQunc I suppose it's a kind of reverse engineering of the long division.

@euler30 2 года назад

The first proof was dope😘

@ssttaann11 2 года назад

(n^3+3)/n^2+7=n-(7n-3)/(n^2+7), so the question is: when (7n-3)/(n^2+7)\in Z, if n\in N? Let f(x) = (7x-3)/(x^2+7). But max{f(x):x\geq 0} = 49\sqrt{22}/(176+6\sqrt{22}) < 1.2. Hence, there is only one chance f(x) = 1. Solving: x_1 = 2, x_2 = 5.

@qedmath1729 2 года назад

Alternatively, let (7n-3)/(n^2+7) = k for integer k. We can expand to quadratic in n and quickly find solution with quadratic formula. Your method is good that it takes advantage a useful inequality

@chrisglosser7318 2 года назад

So I let n^3+3 = (n+a) (n^2+7), then you can work out pretty quickly that a = {-1,0,1} and then that a = {0,1} don’t have integer solutions. a=-1 gives n = {2,5) with the ratio being {1,4}

@manucitomx 2 года назад

Thank you, professor!

@mikeschieffer2644 2 года назад

At 6:05 I don't quite understand why we need (7n - 3)/(n^2 + 7) to be an integer when both the numerator and denominator are always positive over the natural numbers?

@thatdude_93 2 года назад

we don't, he just made a mistake

@user-ik2kd9mb5t 2 года назад

It's a difference of two natural numbers n and (n^3 + 3)/(n^2 + 7)

@aioli121 2 года назад

@@thatdude_93 I wouldn’t call that a mistake, just unnecessary additional work. Even if he came to the conclusion accidentally, which I don’t believe he did, his steps at that part are still completely accurate.

@RexxSchneider 2 года назад

@@thatdude_93 It really isn't a mistake. We have found that (n³ + 3)/(n² + 7) = n - (7n - 3)/(n² + 7). That means (7n - 3)/(n² + 7) = n - (n³ + 3)/(n² + 7), but we have supposed that both n and (n³ + 3)/(n² + 7) are natural numbers. The difference between two natural numbers _must_ be an integer, so (7n - 3)/(n² + 7) must be an integer. That is a tighter - and more useful - condition than it being a positive rational number (the quotient of two natural numbers).

@mathboy8188 2 года назад

I'm not sure I understand the question, but we need (7n - 3)/(n^2 + 7) to be an integer because we need [ n - (7n - 3)/(n^2 + 7) ] = (n^3+3)/(n^2+7) to be an integer. (If (n^3+3)/(n^2+7) isn't an integer, then it has no shot at being a positive integer, i.e. it couldn't be a natural number.) The reason he fooled with absolute values, and worried over signs, and mentioned that the numerator is nonzero, is because of this: Suppose you want a/b to be an integer, where a and be are integers. What must be true? (What about a and b is necessary, though not sufficient, for a/b to be an integer?) The instantly reflexive answer is that a >= b. But that's wrong. *If a and b are both positive,* then a/b an integer implies a >=b. However, *if a is negative and b is positive,* then it still could be that a/b is an integer, and yet a < b. For example (-6)/3 = -2 is an integer, and yet -6 < 3. Also, if a = 0, then a/b = 0 is an integer no matter what (nonzero) integer b is, and regardless of whether b is positive or negative. So what's going on is that getting to n^2 + 7

@fdr2275 2 года назад

An easier way for the second problem. Let (n^3+3)/(n^2+7)=k where k is a natural number Then, n^3+3=kn^2+7k (k>=n) => (kn^2 => n^3) =>(kn^2+7k>n^3+3) => Contradiction Therefore, k

@angeloluisrocattojunior3425 2 года назад

Spoiler: 4 doesn't work, 5 works (and the numerator and the denominator are powers of 2).

@ongzz 2 года назад

2:15 ahhhh i'm confused as to why there are 1504 of +1's and 505 of -1's 😳

@endormaster2315 2 года назад

I have the same question, I literally have no idea

@ayushjuvekar 2 года назад

Assume out of the 2009 terms, 'm' of them were +1 and 'n' of them were -1. Then the total number of terms should be m+n=2009, and as the sum equals 999, we have m - n = 999. So we have two equations with two variables which can be solved simultaneously.

@TheKluVerKamp 2 года назад

@@ayushjuvekar and @Michael Penn, that would be a better way to explain, thank you sir!!

@henk7747 2 года назад

1504-505 =999

@ongzz 2 года назад

@@ayushjuvekar ohhh that makes more sense now - thanks!

@hassanalihusseini1717 2 года назад

Nice problems, and nicely explained.

@caffreys1979 2 года назад

As a corollary is it too much to assume that the solutions for the original equation ((n^3)+3)/((n^2)+7) will occur exactly at the roots of the parabola (n^2 - 7n +10). i.e (n-2)(n-5)=0 i.e n = 2,5 or is that an "accident"? ...does it hold in general and can we prove it.?

@jursamaj 2 года назад

Looks like an "accident". If you change the cubic to n^3+4, there don't appear to be any solutions. Conversely, if you change the parabola to n^2+6, its roots are no longer solutions, but n=2 is.

@dubinkaperelmana 2 года назад

Nice job. Thank you sir

@satyamsatapathy1980 2 года назад

both 2 and 5 work

@jimcameron6803 2 года назад

Here's a video from Michael with solutions two There is not one in a hundred questions that he cannot do He's solving problems now from Africa ...

@swapansarkar8881 2 года назад

Fantastic

@LucasDimoveo 2 года назад

Sorry for the dumb question, but what kind of mathematics is this?

@youngmathematician9154 2 года назад

I believe the first one is combinatorics and the second one is number theory.

@shadow-pw4gn 2 года назад

I dont know about the first one but the second one is a number theory problem

@sternmg 2 года назад

The proof of problem (1) is, unfortunately, incorrect, since the identity involving (-1)^505 cannot in fact be expressed by the product expression over x_i the way shown. Alternate proof sketch: Map the problem onto a circular graph with N nodes, and color each NODE by one of 2 colors. Label each graph EDGE by +1 or -1 if its nodes have the same or different color, respectively. The sum expression of the problem equals the sum over all graph edges. If all nodes are colored equal, the edge sum is N. Changing the first node alters BOTH edges, which alters the edge sum by 2 × -2 = -4. Each subsequent node change will change the edge sum by either ±4 or 0. Thus, Σ{Edges} = N - 4k. We would need a sum of 999 = 2009 - 1010, but 1010 is not divisible by 4. Therefore, 999 cannot be expressed as an edge sum. *Edit:* I stand corrected. I could not get Michael's argument into my head because of its odd-numbered exponent for -1. This is because from my graph-based approach, it is clear from the outset that changing a single node either changes the labels (y_i in Alex's notation below) of _both_ its adjoining edges or simply swaps them, meaning x_i * x_(i+1) = -1 can possibly occur only an _even_ number of times.

@aioli121 2 года назад

Michael isn’t saying (-1)^505 equals the product over x_i. It’s the product over each term (x_i * x_(i+1)) where “x_2010” here represents x_1. He’s multiplying every term that appears in the sum in the problem description. It may have helped if he had called each of these terms y_i, where y_i = (x_i * x_(i+1)), so we don’t mistake this result as the product over x_i. We know that since every x_i is 1 or -1, every y_i is also either 1 or -1. He knows how many of those terms y_i are 1 and how many are -1 because of an implicit system of equations, namely A+B=2009 and A*1+B*(-1)=999 where A is the number of y_i equal to 1, and B is the number of y_i equal to -1. This results in A=1504 and B=505. Therefore, if he takes a product over y_i, he knows the result is 1^1504*(-1)^505 which is -1, which he writes (-1)^505. Let me know if any of that doesn’t make sense or seems incorrect, because it seems to me to defend Michael’s explanation.

@sternmg 2 года назад

@@aioli121 Correct. I appended a note to my original comment.

@benjaminvatovez8823 Год назад

@@aioli121 Thank you, I couldn't figure out why (-1)^505=product of ALL y_i and not only 505 of them but it was actually 1^1504 . (-1)^505 , which is trivial.

@playgroundgames3667 2 года назад

There is no negative natural numbers, that's my explanation

@skylardeslypere9909 2 года назад

2:10 Why is this true?

@priyeshajha9820 2 года назад

Look, let the terms (a term like XnXm) be integers, which are either 1 or -1. Given that they are 2009 in number, and their sum is 999, it is pretty clear why. Yk, like, let there be x no. of +1's and y no. of -1's, then: x+y = 2009 x-y = 999 2x = 3008 x = 1504 y = 505 Now, since there are 505 -1's, the product of all the terms should be -1, like, (x1x2)(x2x3)(x3x4).....(x2008x2009)(x2009x1) = -1 But this product is also a square of (x1x2x3....x2009), and hence cannot be a negative number. the contradiction concludes the proof.

@endormaster2315 2 года назад

@@priyeshajha9820 Thank you

@skylardeslypere9909 2 года назад

@@priyeshajha9820 oohh okay thanks! I understood the contradicition, but I just hadn't considered the system of equations {(x+y=2009); (x-y=999)}

@johnny_eth 2 года назад

"from Africa" doesn't really help to nail it down well.

@petereziagor4604 2 года назад

At minute 2:23, Can someone please explain to me how he got that we have 1504 +1. Thank you

@fdr2275 2 года назад

Let n be the number of "+1" Number of "-1" = 2009-n n-(2009-n)=999 2n=999+2009=3008 n=1504

@petereziagor4604 2 года назад

@@fdr2275 thank you very much

@koenth2359 2 года назад

For the first problem I reasoned that there are 2009 products, each being 1 or -1. If there are n positive products and 2009-n negative products, we must have n-(2009-n)=999 => n=1504, and an odd number of negative products. This implies that in the loop there is an odd number of sign changes between two adjacent x's. This implies x1=-x1, so x1=0, which contradicts x1 being -1 or +1. The second problem: (n^3+3)/(n^2+7)=n-(7n-3)/(n^2+7) is integer for integer n, when k(n^2+7)=7n-3. We get kn^2-7n+7k+3=0. This is a quadratic in n with solution n = (7 +/- Sqrt(-28k^2-12k+49))/2k The discriminant itself is a polynomial in k: with zeros k = (3 +/- 4sqrt(22))/14. For the discriminant to be nonnegative, and k integer, we can only have k=-1, k=0, k=1. For k=0 n is undefined, because we have 0 in the denominator. For k=1 we get n=5 or n=2 For k=-1 we get n= (-7+/-sqrt(33))/2 which are no integers. The found integers are > 0, so they are natural numbers. Check n=5: (125+3)/(25+7) = 128/32 = 4 Check n= 2: 11 /11 = 1

@Inovator1911 2 года назад

Alternative solutions: 1. After we explain, that we have 1504 '1' and 505 '-1', lets think, what they mean. '1' means, that x_n have the same digit as x_(n+1) '-1' means, that digit was changed. -> we have 505 changes of digits But, from 1st term to last, we go from x_1 to x_1, which means that we should have even number of digit changing. This is contradiction. 2. Lets write this in other way: n^3+3=(n^2+7)*x, where x is a natural number assume, that x=n n^3+3=(n^2+7)*n n^3+3=n^3+7n. But 7n > 3 for all natural n, then x=0 for having a real solutions. And its quite obvious, when k>=1, then D we have one way: k=1 => x=n-1 Put value of k there: n^3+3=n^3-kn^2+7n-7k n^3+3=n^3-n^2+7n-7 -n^2+7n-10=0 => n=2 and n=5 P.S. Sorry for my bad English)

@adamkhan7234 2 года назад

From the whole of the continent... One collective conscience of nations to provide the question. Other countries get a name check, but this question is from 'Africa'

@mateorios1940 2 года назад

Well, it’s the pan-african math olympiad. You do know what Pan-African means, right?

@adamkhan7234 2 года назад

@@mateorios1940 from that olympiad is not the same thing as 'from Africa' imo

@sjdpfisvrj 2 года назад

@@adamkhan7234 not sure what you expect him to say. If it's from the pan-Africsn math Olympiad, there is no specific African country attached to the problem. Is he supposed to make one up?

@adamkhan7234 2 года назад

@@sjdpfisvrj are you being consciously naive? Other videos have identified the competition the question came from . This one could have too. A question from Pan-African Olympiad (insert year here). Sorry if that blew your mind right there. Literally, like, how is that less obvious than your suggestion of making a country up?

@martinepstein9826 2 года назад

@@adamkhan7234 Your complaint was about the lack of a specific country. "Pan-African Olympiad" still doesn't specify a country so it's not clear why you think this is an improvement.

@makotoniijima862 2 года назад

4 doens't work. However 5 does work; if n = 5, then we get 4 as the answer, thus 5 works

@itismethatguy 2 года назад

Why is there a continent instead of country

@CRGreathouse 2 года назад

Because they come from the Pan-African Mathematical Olympiad, which doesn't come from any one country.

@matniet43 2 года назад

2:51 why is that?

@anantkerur557 2 года назад

Which part exactly is your confusion? The product of all the terms is equal to -1, because 505 of them are equal to -1, and 1504 are equal to 1. Hence the product will equal (-1)^505 * (1)^1504 = -1