This is a problem from 101 problems in Algebra for International Mathematics Olympiad Training for the USA team.
Пікірлер: 101
@jay_sensz2 күн бұрын
I don't think the conclusion is quite properly argued. If we choose (a,b,c,d) = (1,1,1,5), then we get 8+e = 5e, implying e=2. But that violates the assumption that a≤b≤c≤d≤e. Rigorously, you would have to check all five candidates for (a,b,c,d) and check if there's an integer solution for e such that d≤e. The possible solutions (a,b,c,d,e) seem to be (1,1,1,2,5), (1,1,1,3,3), and (1,1,2,2,2), so the answer being 5 still stands.
@AnantGoswami22 күн бұрын
Same here, it doesn't work for max {a,b,c,d,e} = 4. Oddly enough, it works for 3, and then 5.
@LearnmoreMoyo-q1o2 күн бұрын
But the key word is the maximum possible which key word to note is "possible" so 5 is the solution
@LearnmoreMoyo-q1o2 күн бұрын
All these satisfy the equation but they are ruled out coz 5 is the max possible number in all those solution sets
@jay_sensz2 күн бұрын
@@LearnmoreMoyo-q1o Of course 5 is the solution. What I'm saying is that you can't have d=5 with a≤b≤c≤d≤e.
@cret859Күн бұрын
@@jay_sensz You right! A last step is missing. By chance in the first case (a,b,c,d)=(1,1,1,2) since 1+1+1+2+5=1*1*1*2*5=10 the value e=5 is a valide maximum. But, we may have check this. Perhaps we also may have check that the other cases (1,1,1,3) (1,1,1,4) and (1,1,2,2) produce a solution less than 5 or at least no valide solution. Unless we have to demonstrate the general case where whatever the length of the set, if a+b+c+...+x+y+z=a*b*c*...x*y*z then the maximum corresponds to the number of elements because (a,b,c,...,x,y)=(1,1,...,1,2) always produces a valid maximum z because of the sum and the product simultaneously make 2z. And when we increase any single element of (a,b,c,...,x,y) the product or the sum increase, so the new solution, when it exist, is less than the z of the first one.
@misterj.a912 күн бұрын
I don't understand the conclusion. I get that you've proven (considering your generalization) that the maximum value for d (a, b and c included) is 5 but if translated in the original comparison it just means (for me) that 5 5 + max{e} = 2.max{e} so max{e} is equal to 5. Still a valid conclusion I guess.
@ChristopherBittiКүн бұрын
Brute force solution: Say a + b + c + d + e = abcde and without loss of generality, say e is the max of {a, b, c, d, e} Now note that abcde = a + b + c + d + e e = 3. This is another possible value of the max. Case 4: abcd = 4 We have two subcases here. First, we can have one term being 4 and all the other terms being 1, then we get 7 + e = 4e => 3e = 7. There are no solutions in this case. The next subcase is two terms being 2 and the other two being 1. In this case we get 6 + e = 4e => e = 2. Case 5: abcd = 5 In this case one of the terms is 5 and the other three are 1, so we get 8 + e = 5e => e = 2. This is not valid, though, as e = 2 is smaller than the term that is 5, contradicting the definition of e as the max. Thus, we have gathered that e can be 2, 3, or 5. So, 5 is the answer.
@MikeGz92Күн бұрын
I think that is the right way to solve. Without checking all the possibilities, how you say that 5 is a valid solution? Perhaps e=5 wouldn't satisfy original equation and so you have to try e=4. I think that inequality only gives you candidates, setting a boundary, but then you have to try and check if they work
@sanamite2 күн бұрын
I understand the approach, but I don't understand how we showed that e can't be greater than 5 (I know it can't be)
@rufusjasko2 күн бұрын
You can solve for e in each of the cases to show that 5 is the maximum. But this wasn't shown in the video.
@glorrin2 күн бұрын
I agree, the video is missing something
@sanamite2 күн бұрын
@@rufusjasko Oh right you just use the given equation to do that, thanks
@sanamite2 күн бұрын
@@rufusjasko what do you think of my solution in the comments?
@konraddapper77642 күн бұрын
He did Not Proof it There are two flaws First e is only bounded from below by his Argumente. Second nothing.garentees you hat a solution exsists for a given abcd touple or at all for that matter
@antosandrasКүн бұрын
Although 5 is correct, the finishing argument is a total nonsense, as others commented. The solution (a,b,c,d,e)=(1,1,1,2,5) must be found, and also that the other possible 4-tupples for (a,b,c,d) cannot yield greater e. I am disappointed.
@brian554xxКүн бұрын
You have argued that e can't be > 5, but you haven't shown an example that fits the formula with e = 5. You're a few steps short of { 1, 1, 1, 2, 5 }, which would be conclusive.
@kinshuksinghania4289Күн бұрын
Although 5 is the correct answer but I don’t agree with how you deduce 5 being the answer at the end.
@NotAvailable-b6d2 күн бұрын
You solved for the maximum of your inequalities, but 5 is not an element in the set that solves A+B+C+D+E = A*B*C*D*E .. Testing 4 as Max element fails to solve the equality as well .. Testing 3 as Max element, I found 1,1,1,3,3 solves the equality, thus the Max element of the set is 3 ..
@sanamite2 күн бұрын
1,1,1,2,5 works
@icetruckthrilla2 күн бұрын
The last couple of minutes of the video left me scratching my head. You’re right that 1,1,1,2,5 works. But we’re trying to do this set of numbers in increasing order.
@LearnmoreMoyo-q1o2 күн бұрын
Valid
@LITHICKROSHANMS-gw2lx2 күн бұрын
Given equation a+b+c+d+e=abcde....(1) Solution:- Taking the given equation a+b+c+d+e=abcde Modifying the equation (((10)^(log(a)))+((10)^(log(b)))+((10)^(log(c)))+((10)^(log(d)))+((10)^(log(e))))=((10)^(log(abcde))) (((10)^(log(a)))+((10)^(log(b)))+((10)^(log(c)))+((10)^(log(d)))+((10)^(log(e))))=((10)^((log(a))+(log(b))+(log(c))+(log(d))+(log(e)))) Taking the logarithmic function on both sides log((((10)^(log(a)))+((10)^(log(b)))+((10)^(log(c)))+((10)^(log(d)))+((10)^(log(e)))))=log(((10)^((log(a))+(log(b))+(log(c))+(log(d))+(log(e))))) log((((10)^(log(a)))+((10)^(log(b)))+((10)^(log(c)))+((10)^(log(d)))+((10)^(log(e)))))=log((10)^(log(abcde))) log((((10)^(log(a)))+((10)^(log(b)))+((10)^(log(c)))+((10)^(log(d)))+((10)^(log(e)))))=log(abcde) log(a+b+c+d+e)=log(abcde) log(a+b+c+d+e)=log(a)+log(b)+log(c)+log(d)+log(e) [Since, a+b+c+d+e=abcde] log(abcde)=log(a)+log(b)+log(c)+log(d)+log(e)....(2) The solutions 1) {a,b,c,d,e}→{1,1,2,2,2} Now substitute this result in the equation (2) as we get log((1)(1)(2)(2)(2))=log(1)+log(1)+log(2)+log(2)+log(2) log(8)=3(log(2)) This is true but it is one of the solution We can check this solution either it is maximum or minimum 0.90308998699=0.90308998699→minimum value of the solution 2) {a,b,c,d,e}→{1,1,1,3,3} log((1)(1)(1)(3)(3))=log(1)+log(1)+log(1)+log(3)+log(3) log(9)=2(log(3)) 0.95424250943=0.95424250943→slight minimum value of the solution 3) {a,b,c,d,e}→{1,1,1,2,5} log((1)(1)(1)(2)(5))=log(1)+log(1)+log(1)+log(2)+log(5) log(10)=log(2)+log(5) 1=1→maximum value of the solution Hence, the maximum value of the solution {a,b,c,d,e}→{1,1,1,2,5}
@Grecks7523 сағат бұрын
W.l.o.g. assume that a
@kereric_cКүн бұрын
(a,b,c,d)=(1,1,1,2) then e=5 (a,b,c,d)=(1,1,1,3) then e=3 (a,b,c,d)=(1,1,1,4) then e=7/6 valid (a,b,c,d)=(1,1,1,5) then e=2 e
@sanamite2 күн бұрын
I have a straightforward solution : assuming e is the maximum of the set : e = (a+b+c+d)/(abcd-1) abcd >= 2 d(abcd-1)/d(d) >= d(a+b+c+d)/d(d) abc >= 1 True (the growth of abcd-1 is greater or equal than that of a+b+c+d here, so we want the closest values to {1,1,1,1} as possible while respecting abcd >= 2) so say d=2 and a=b=c=1 e = (1+1+1+2)/1=5
@Dr_piFrogКүн бұрын
Only solutions when sum=product=10, 9, 8. When all permutations of the three solution sets are considered there are 40.
@assiya30232 күн бұрын
شكرا أستاذ فعلا الإنسان يتعلم من ولادته لمماته ، أنا أستاذة ( في مجال غير الرياضيات) منذ ما يزيد عن التلاتين سنة وأتعلم وأستمتع بالفيدوات التعليمية وأتابع قناتك دائما . فشكرا مرة أخرى
@RyanLewis-Johnson-wq6xs2 күн бұрын
Find the maximum possible value of max {a,b,c,d,e} if a+b+c+d+e=abcde e
@maherom111 сағат бұрын
Sir I suggest this two équations for next vidéo thanks: 1) 2^x + 3^(x^2) = 6 2) 2^x * 3^(x^2) = 6
@SALogics2 күн бұрын
Nice problem with nice solution! ❤❤
@majora413 сағат бұрын
The strategy I came up with is a bit weird but I think it works, unless I've made some error or unjustified leap in logic. I considered how many of a,b,c,d,e are >= 2. Obviously zero won't work because 5 ≠ 1. And we can also see that one won't work either since that would require a + 4 = a and that's impossible. If exactly two of a,b,c,d,e are >= 2 then we have a + b + 3 = ab which implies b = (a+3)/(a-1). In order for a and b to both be natural numbers it must be the case that a = 2 and that corresponds to b = 5. So in this case we have max(a,b,c,d,e) = 5. If exactly three of a,b,c,d,e are >= 2 then we have a + b + c + 2 = abc. The smallest possible state would be 2 + 2 + 2 + 1 + 1 = 2^3 which is actually true. From there we can see that if we increase any of the values by some positive integer k, that would increase the right-hand side by 4k > k and invalidate the equality. This does tell us that max(a,b,c,d,e) can be 2, but that's uninteresting for our purposes. If exactly four of a,b,c,d,e are >= 2 the smallest possible state is 2 + 2 + 2 + 2 + 1 < 2^4, and playing the same "game" of increasing the left-hand side by k increases the right-hand side by 8k, only growing the gulf between them and leaving no possible way for them to be equal. And by an almost identical argument we can rule out all of a,b,c,d,e being >= 2 Therefore we can see that the maximum possible value of max(a,b,c,d,e) is 5.
@golddddusКүн бұрын
You failed to prove the concrete existence of the existence of the number 5. It is a possible solution of the maximum. Well done. A check was needed. As far as I have calculated, the only possibility with e=5 in natural numbers is 1+1+1+2+5 = 1*1*1*2*5. Never stop learning. I was born in 1950 and I'm still learning. So I'm alive.😎
@markjohansen6048Күн бұрын
Wait, no. You showed that e >=5, not
@RomanOrekhovКүн бұрын
Notice that (a,b,c,d) can't be (1,1,1,1) since then e+4=e. => d >= 2 Now consider abcde >= 1*1*1*2*e=2e, so -abcde
@MinhLe-se6bs19 сағат бұрын
Hi prime Newtons ! Your teaching style is awesome! I hope there are plenty of math teachers like you in USA, unluckily not many like you as I wish. Your handwriting and your steps to the end are almost perfect! It reminds me when I was in Vietnam before 1975, most math teachers were like you.
@spacer999Күн бұрын
Max(a,b,c,d) does not imply max e=5. You have to rearrange the initial equation to e = (a+b+c+d)/(abcd-1), and try all 5 possible sets of (a,b,c,d) to see which one gives the largest e. In fact it is (1,1,1,2) that produces the max e which is 5.
@simoncashewКүн бұрын
Although you got the correct answer, the way is wrong. It starts that a+b+c+d+e has to be strictly less than 5e or else you get a=b=c=d=e which you also mentioned at the beginning is impossible. With that you get abcd < 5 which rules out (1,1,1,5). Your quadruples also just satisfy abcd < 5, but not a+b+c+d+e = abcde.
@ahsgdf115 минут бұрын
Sorry, not correct. Let m=abcde. We can easily see that m = 5^(5/4) is a "greater maximum" than the number 5 as given in the video. Proof: let all five quantities be equal, then we have m = abcde = a^5, but we must have 5 a = a^5 , hence 5 = a^4 -> a = 5^(1/4) and m = 5^(5/4) > 5 QED. Remark: I don't know if it is the "true" maximum the problem was looking for.
@satakesatoshi2 күн бұрын
Is it the solution from the original textbook? Clearly some sets of {a,b,c,d} mentioned above (e.g. 1,1,1,5) are not valid to the original equation a+b+c+d+e=abcde as there will be no natural number fitting the value of e
@tenminuteretreat807Күн бұрын
I just found your channel. Wow! I looked at the list of videos available, and I'm so excited! I see I'm going to be having some fun! Tons of cool problems.
@davidgagen9856Күн бұрын
(a,b,c,d) = (1,1,1,5) is not a valid solution. (a,b,c,d,e) = (1,1,1,2,5) is the correct order. Of course 5 remains the max value.
@MYeganeh100Күн бұрын
I live thousands of miles away from USA. I am an 82 years retired man, who loves Maths. I must admit that you are a wonderful teacher. God bless you. From far away. Moh.
@beholdmaverick7121Күн бұрын
I think the conclusion looks vague largely because of the wordiness of it. "if a+b+c+d+e = abcde" is a hypothetical presumption and on that premise we are asked to find the maximum value in the set of natural numbers. But the deduction is almost useless because none of any values can ever prove the hypothetical condition to be true. Even if max value is considered any number the aforesaid presumption of equality never holds true. So in a way not much value in this proof that I see.
@e6a42 күн бұрын
Can you show please how to compare W(W(1)) and (W(1))^2 without calculator?
@hodanassef4057Күн бұрын
(1,1,2,2,2) sum = product =8
@artandataКүн бұрын
ok. it's a solution but in this case e=2 which is not a maximum for e since {1,1,1,2,5} is also a solution and in this case e=5 which is grater that e=2.
@hodanassef4057Күн бұрын
@@artandata yes you are right, but I ve guessed it from the first trial. My respect . From Egypt.
@artandataКүн бұрын
@@hodanassef4057 my regards to you from an Argentine living in Brazil.
@dhairyaakbariКүн бұрын
Your videos aids me a lot in solving different questions and preparing for my upcoming exams. Thank you sir❤❤
@LearnmoreMoyo-q1o2 күн бұрын
This problem reminds of another Find the letters ABCD such that ABCD X 4 = DCBA
@RyanLewis-Johnson-wq6xs2 күн бұрын
1
@aljawad2 күн бұрын
When I saw the thumbnail, I thought you were referencing a perfect number and its factors!
@glorrin2 күн бұрын
Great video, but I am not convinced by the conclusion to me all we have shown is max{e} >= 5 I fail to see what we have shown about the max of e, all we know is the max of a b c d is 5, e is bigger tha a b c and d, so how can we know that e cannot be 6 and above ?
@randirbox2 күн бұрын
I keep amused how many that questions could be finished with a small py script.
@vandanagarg39062 күн бұрын
First , your videos are really knowledgeable for a maths lover looks like you go fall in another world
@maxborn7400Күн бұрын
really enjoyed this one, very interesting problem
@zactastic4k955Күн бұрын
But 5 is prime and a,b,c,d are natural numbers so that can’t be true
@jamesharmon49942 күн бұрын
Without demonstrating that a+b+c+d+e equals abcde, you cannot verify the given condition.
@jamesharmon4994Күн бұрын
For 10:15 option 1, the sum is 5, and the product is 2. this means e=5. 10:25 the sum is 6, the product is 3. There is no e for which the the sum and product are equal. This holds true for all following options. Therefore, the values must be 1,1,1,2,5 in order for sum and product to be equal. From this, the max is 5. If you cannot construct a set that meets the given condition, the answer to the question is "the empty set"
@manes8008Сағат бұрын
what if they are all equal to 0 than you cant say a + b + c + d > e
@PureExile26 минут бұрын
0:21 His definition of natural numbers doesn't include 0.
@_samin2566Күн бұрын
Shouldn't the condition be less than 5 e cause if we consider less than Or equal to 5e then if it is equal to 5 e abcde =e^5 which is not possible for any natural number
@nasancakКүн бұрын
It is possible when a=b=c=d=e=1.
@Ivan-fc9tp4fh4d2 күн бұрын
I do not understand.
@rogerkearns80942 күн бұрын
Yes, I get that. Still, does the given equation have any solution at all, in N? If it does not, then I cannot see that the question is meaningful. [Edit] If it does have a solution, then that should be demonstrated, don't you think?
@glorrin2 күн бұрын
it has 3 different solutions, if you include all the permutations it has quite a lot of solutions. You can find them by using all the cases for a b c d and you would find 1,1,1,2,5 => 10/10 1,1,1,3,3 => 9/9 1,1,2,2,2 => 8/8 I'll let you find all the permutation I am not bored enough for that
@sanamite2 күн бұрын
We just found a solution in the natural numbers
@rogerkearns80942 күн бұрын
@@sanamite OK. Maybe I missed it.
@ramunasstulga82642 күн бұрын
How would you solve a^a+b^b+c^c+d^d=abcd ?
@LITHICKROSHANMS-gw2lx2 күн бұрын
Given equation a^a+b^b+c^c+d^d=abcd Solution:- a=b=c=d=n So,the equation becomes nⁿ+nⁿ+nⁿ+nⁿ=n⁴ 4(nⁿ)=n⁴ (2²)(nⁿ)=n⁴ (nⁿ)=((n⁴)/(2²)) (n)ⁿ=((n²)/(2))² Equating the power and base So, The case:-1[In base] n=n²/2 2n=n² n²-2n=0 Completing the square root (n-1)²=1 (n-1)=±√1 (n-1)=1&(n-1)=-1 n=2& n≠0 The case:-2[In power] n=2 Therefore, The case 1= The case 2=n=2 It means a=b=c=d=n a=b=c=d=2
@rainerzufall42Күн бұрын
ERROR: You are calculating conditions for the solution without proving the existence of this solution! If you calculate the maximum of all the solutions, but that's an empty set, you won't get the correct result! Example: max { n € IN | n = 0.5 } is not 0.5, as expected, it is just not defined... In this instance, with the tuple (a, b, c, d, e) = (1, 1, 1, 2, 5), you have max { a, b, c, d, e } = e = 5, but max { a, b, c, d } = 2. Why is that? Because for (a, b, c, d) = (1, 1, 1, 5), there exists no such solution! Therefore: max { e € IN AND e >= max { 1, 1, 1, 5 } | (1, 1, 1, 5, e) is a solution, i.e. 1+1+1+5+e =1*1*1*5*e } is undefined.
@tenminuteretreat807Күн бұрын
You don't know what you're talking about.
@holyshit9222 күн бұрын
So {a,b,c,d,e} = {1,1,1,2,5} Only this option satisfies equation
@Illenom2 күн бұрын
{1,1,1,3,3} also satisfies the equation, but has a lower max.
@sanamite2 күн бұрын
1,1,2,2,2
@LearnmoreMoyo-q1o2 күн бұрын
Hence 3 is not a max at all@@Illenom
@sanamite2 күн бұрын
@@LearnmoreMoyo-q1o it satisfies the equation, contradicting holyshit922's comment.
@holyshit9222 күн бұрын
@@sanamite I checked these which he wrote Note that i used word option which is the same that he used and since at least one of his options satisfies equation 5 is indeed the solution
@RyanLewis-Johnson-wq6xs2 күн бұрын
Since 5 is the max of a,b,c,d and a≤b≤c≤d≤e, max{e}=5
@NicolaVozza-w5m2 күн бұрын
(a+b+c+d) +e = (abcd) e and for (1112) a + b +c +d = 5 abcd = 2 and so 5 + e = 2e and so e = 5 for (1113) (1114) (1115) (1122) e < d