The maximum possible maximum

[@ChristopherBitti]

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.

[@MikeGz92]

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

[@jay_sensz]

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.

[@AnantGoswami2]

Same here, it doesn't work for max {a,b,c,d,e} = 4. Oddly enough, it works for 3, and then 5.

[@LearnmoreMoyo-q1o]

But the key word is the maximum possible which key word to note is "possible" so 5 is the solution

[@LearnmoreMoyo-q1o]

All these satisfy the equation but they are ruled out coz 5 is the max possible number in all those solution sets

[@jay_sensz]

@@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.

[@cret859]

@@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.

[@assiya3023]

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

[@johnconrardy8486]

i don;t know what to say, i watch your videos every day and it make me happy, thank you. you are one of the best teacher out there.

[@bhaveerathod2373]

How do you always blow my brain, I look at the question and think no way this is even possible, I don’t even know where to start, then I watch your explanation and it makes SO much sense, you are a genius man!! This is a brilliant way you solved this!

[@Grecks75]

W.l.o.g. assume that a

[@dhairyaakbari]

Your videos aids me a lot in solving different questions and preparing for my upcoming exams. Thank you sir❤❤

[@nopenotatall397]

I just saw your squeeze thm video and it was so informative. I clicked over to your channel and I'm so happy to see that you're still uploading!! Will definitely be following you :) thanks sm

[@tenminuteretreat807]

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.

[@misterj.a91]

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.

[@sanamite]

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)

[@rufusjasko]

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.

[@glorrin]

I agree, the video is missing something

[@sanamite]

@@rufusjasko Oh right you just use the given equation to do that, thanks

[@sanamite]

@@rufusjasko what do you think of my solution in the comments?

[@konraddapper7764]

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

[@brian554xx]

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.

[@kereric_c]

(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

[@albajasadur2694]

This checking step is straight forward and necessary.

[@MellyhedSuccessfulTips]

Hey coulnd't find any videos about Thales theorem could you make one, you are an amazing teacher and I would love to learn about this and it's crucial for my final exams.

[@antosandras]

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.

[@cmilkau]

11:35 This argument is not valid. First, even if we knew there is a solution with d = 5, e could still be larger than that. (For example, a=b=c=1, d=2, e=5, is a solution with d

[@sanamite]

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

[@SALogics]

Nice problem with nice solution! ❤❤

[@majora4]

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.

[@JavedAlam24]

I like your solution best. It doesn't change your solution, however you forgot the case where a=3 and b=3 for when two of the values are >= 2. This is a valid solution and it works with your equations, which proves their validity. Good job!

[@vandanagarg3906]

First , your videos are really knowledgeable for a maths lover looks like you go fall in another world

[@maxborn7400]

really enjoyed this one, very interesting problem

[@golddddus]

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.😎

[@Dr_piFrog]

Only solutions when sum=product=10, 9, 8. When all permutations of the three solution sets are considered there are 40.

[@simoncashew]

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.

[@derekschmidt5705]

Since e is dropped out of the trial-and-error phase, wouldn't it be more prudent to excluse it from the inequality chain, leaving d = max(a,b,c,d)? Then e could just be the balancing value, whatever it happens to be. Also, i know you finally got to it at the end, but as one of the setup steps, it might have been helpful to render the natural-language question in math notation and say that for each valid solution, e_n = f(a_n,b_n,c_n,d_n) = g(a_n,b_n,c_n,d_n). And what is max(e)?

[@RomanOrekhov]

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

[@alessandrokaroui1095]

can we have some linear algebra or abstract algebra exercises? i think it would be great for students who study pure mathematics

[@RyanLewis-Johnson-wq6xs]

Find the maximum possible value of max {a,b,c,d,e} if a+b+c+d+e=abcde e

[@NotAvailable-b6d]

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 ..

[@sanamite]

1,1,1,2,5 works

[@icetruckthrilla]

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-q1o]

Valid

[@klausao]

If all numbers are equal and then you can write the sum as 5 * n, where n could be a, b, c, d or e. So, choosing 5 * n as the biggest value possible of the equation (the sum is always less than the product), we choose to have that 5n >= a*b*c*d*e where n is in the set of {a, b, c, d, e}. Now, dividing both sides by n (one of the integers) we have that the product of 4 remaining integers must be less than 5 and because 5 is prime the product of the rest of 4 numbers is 4 or less. Conclusion, 5 is the greatest number of the set {a, b , c, d, e} as the product of the rest of 4 is less or equal to 4. Keep up the good work!

[@kinshuksinghania4289]

Although 5 is the correct answer but I don’t agree with how you deduce 5 being the answer at the end.

[@maherom1]

Sir I suggest this two équations for next vidéo thanks: 1) 2^x + 3^(x^2) = 6 2) 2^x * 3^(x^2) = 6

[@spacer999]

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.

[@Aryansbestfriend]

Hey I have a question regarding oblique asymptotes I have a very hard question calculator allowed but it hasn’t helped me all that much

[@johncirillo9544]

Eyeballing it, I had (5, 2, ,1, 1, 1) which yields a sum and product of 10. Therefore the maximum term is 5 and the maximum sum/product is 10.

[@gavintillman1884]

The final conclusion seems plucked from the air. Having enumerated the different cases for (a,b,c,d) I think you need to consider each of these cases and deduce what implies for e when you equate a+b+c+d+e = abcde. Several of those cases have no integer solution, but if those that do, the greatest value for e is indeed 5: 1+1+1+2+5 = 1.1.1.2.5 = 10.

[@hodanassef4057]

(1,1,2,2,2) sum = product =8

[@artandata]

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.

[@hodanassef4057]

@@artandata yes you are right, but I ve guessed it from the first trial. My respect . From Egypt.

[@artandata]

@@hodanassef4057 my regards to you from an Argentine living in Brazil.

[@aljawad]

When I saw the thumbnail, I thought you were referencing a perfect number and its factors!

[@randirbox]

I keep amused how many that questions could be finished with a small py script.

[@talcual2138]

And a general solution for a set of n natural numbers would be n? Prove that, in another of your wonderful videos!

[@glorrin]

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 ?

[@RyanLewis-Johnson-wq6xs]

1

[@MYeganeh100]

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.

[@LITHICKROSHANMS-gw2lx]

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}

[@MinhLe-se6bs]

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.

[@satakesatoshi]

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

[@e6a4]

Can you show please how to compare W(W(1)) and (W(1))^2 without calculator?

[@markjohansen6048]

Wait, no. You showed that e >=5, not

[@JavedAlam24]

Did he even show that e >= 5? I thought he just showed that abcd

[@shennyboi110]

Hi! I think this proof is a few steps short of rigorous! We have not yet proved that there is a practical case where e=5, only that e=5 is the theoretical maximum

[@JavedAlam24]

He actually didn't even prove that e is the theoretical maximum, only that abcd

[@LearnmoreMoyo-q1o]

This problem reminds of another Find the letters ABCD such that ABCD X 4 = DCBA

[@jamesharmon4994]

Without demonstrating that a+b+c+d+e equals abcde, you cannot verify the given condition.

[@jamesharmon4994]

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"

[@PeymanMoradi-gv1ik]

perfect

[@manes8008]

what if they are all equal to 0 than you cant say a + b + c + d > e

[@PureExile]

0:21 His definition of natural numbers doesn't include 0.

[@beholdmaverick7121]

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.

[@ramunasstulga8264]

How would you solve a^a+b^b+c^c+d^d=abcd ?

[@LITHICKROSHANMS-gw2lx]

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

[@_samin2566]

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

[@nasancak]

It is possible when a=b=c=d=e=1.

[@davidgagen9856]

(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.

[@rogerkearns8094]

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?

[@glorrin]

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

[@sanamite]

We just found a solution in the natural numbers

[@rogerkearns8094]

@@sanamite OK. Maybe I missed it.

[@absolutezero9874]

Yup Ignore No reply 👍🏼👍🏼

[@paulwatson746]

The conclusion does not follow, you have just stated it, there is no logical link. You need to prove that it cannot be >5.

[@PrimeNewtons]

Working on it.

[@JavedAlam24]

​​@@PrimeNewtonsI suggest you look at majora4's answer in the comments

[@Ivan-fc9tp4fh4d]

I do not understand.

[@vikasrajput2112]

I love your hat😊

[@rainerzufall42]

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.

[@tenminuteretreat807]

You don't know what you're talking about.

[@NicolaVozza-w5m]

(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

[@RyanLewis-Johnson-wq6xs]

Since 5 is the max of a,b,c,d and a≤b≤c≤d≤e, max{e}=5

[@holyshit922]

So {a,b,c,d,e} = {1,1,1,2,5} Only this option satisfies equation

[@Illenom]

{1,1,1,3,3} also satisfies the equation, but has a lower max.

[@sanamite]

1,1,2,2,2

[@LearnmoreMoyo-q1o]

Hence 3 is not a max at all​@@Illenom

[@sanamite]

@@LearnmoreMoyo-q1o it satisfies the equation, contradicting holyshit922's comment.

[@holyshit922]

@@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

[@zactastic4k955]

But 5 is prime and a,b,c,d are natural numbers so that can’t be true

[@shalinisharma3651]

A + b + c + d + e is never = abcde if they r natural numbers

[@PrimeNewtons]

1+1+1+2+5 = 1×1×1×2×5

[@PrimeNewtons]

Mathematics is brutal

[@PrimeNewtons]

1+1+2+2+2 = 1×1×2×2×2