A nice Completing the squares problem

[@jasonryan2545]

This was quite simple really, I hadn't expected it to be so straightforward. I have a feeling I would mees up on the negatives as well. Great video, as always, Prime Newtons! ❤

[@wesleydeng71]

By the same token, the minimum of a1 is -450.

[@ShriABCD]

How Pls explain

[@Kero-zc5tc]

⁠​⁠@@ShriABCDbecause when he sqrts 500^2 instead of crossing off the squared sign it should be + or - sqrt(500^2) or a-50= + 500 or a-50=-500 and the second one becomes -450

[@PureHanbali]

At the beginning of the video, I assumed the correct answer would be 100. Thank you very much for proving me wrong!

[@slavinojunepri7648]

Excellent solution

[@dan-florinchereches4892]

Wow this is a nice problem you can solve in your head. If you expand the quadratic and solve you get the solution a1,2=(100+-√10000-4*(P(a2)+P(a3)+...+P(a100)))/2 where P(x)= x^2-100x. This second degree polinomial has a minimum when x=-b/2a=-50 as a quadratic. Because all the terms are subtracted inside the square root logically the maximum value for a1 occurs when P(a2)...P(a100) are miminal as they are subtracted from something. the mimimum value of P(X)=P(-50)=-2500 multiplied by -4 is 10000 so under square root sign we have a sum of 100 numbers each equal to 10000. Then a1,2=(100+-1000)/2 so maximum a1 is 550. Honestly i was expecting to have to figure out an inequality to statt from to solve the problem but this was really straightforward

[@KorlinAng-bs7rh]

Nice. I don't think omission of -500 is an issue because question only ask for largest a1.

[@RR-bs9mr]

I did the same thing at begaining, I ust assumed eveyrthing was zero eccept a1, and forgot the neagtives...

[@Arjunan-d1l]

Great,Sir...Thank You very much...

[@randomjin9392]

A bit sloppy at the end. Following our condition, a₁-50 = ±500, it's just that the maximization of a₁ rules out the negative value. Also, if we're careful till the end, we need to show that maximization of (a₁-50)² leads to maximization of a₁.

[@niloneto1608]

a_i can't be negative, otherwise the rateio itself would be negative. The video itself is fine.

[@randomjin9392]

@@niloneto1608 Wrong. The aₖ cannot be all negative, but a₁ perfectly could. If we have aₖ=50, so that (aₖ-50)² = 0, 2 ≤ k ≤ 100, then a₁ = -550 yields a₁+...+a₁ₒₒ = -550 + 99x50 > 0, so there's no issue. The point is being careful and derive the value from maximization, not because a sign was lost after taking the square root.

[@niloneto1608]

@@randomjin9392 You mean a1=-450 right?(-500+50). And yeah, not all a_i may necessairly be positive, but the overall sum of the a_i must. And the maximization itself is fairly obvious as he has shown. What could be interesting is whether we can minimize that sum.

[@randomjin9392]

@@niloneto1608 Yes, 450, thanks! My thoughts were also: What if we have k not a perfect square? (so, say we have 99 members). Then with the condition of a₁ being an integer, we can't just zero other members. It's a harder problem.

[@flowers42195]

A little bit of lack in the final solution: A1=550 and all the others Ai (i from 2 to 100) = 50

[@kapishkumar4252]

You must try solving INMO papers of india, they are just like USAMO papers , they have a bunch of good problems to do, I recommend you to solve it on this channel

[@AryanKumar-vo1ic]

nice prob

[@hassankhamis77]

I thought we are going to use a sequence because of the notation of the question. From the solution, can I conclude that all values of a’s are 50 except a1 that is 550?

[@niloneto1608]

That's the one that yields the maximum value, yes.

[@PrimeNewtons]

There is no assigned value. However, for A1 to be maximum, you must assume all others are 50 each

[@niloneto1608]

@@PrimeNewtons Bonus question: Is there a minimum value for a_1+a_2+...+a_100?

[@davidgagen9856]

Doesnt positive 100 mean that the denominator must also be positive?

[@niloneto1608]

Obviously. Where's the problem with that?

[@mistermudpie]

yes, but not that all individual terms of the denominator are also positive.

[@niloneto1608]

@@mistermudpie Their sum must be, e.g, a_1=-450 and a_2 up to a_100=50, whose sum is 4500.

[@mistermudpie]

@@niloneto1608 why are you explaining the obvious?

[@niloneto1608]

@@mistermudpie I wasn't explaining it to you per se, but rather for the guy who asked it in first place.