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! ❤
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 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.
@@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.
@@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.
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
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?
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! ❤
At the beginning of the video, I assumed the correct answer would be 100. Thank you very much for proving me wrong!
By the same token, the minimum of a1 is -450.
How Pls explain
I did the same thing at begaining, I ust assumed eveyrthing was zero eccept a1, and forgot the neagtives...
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.
a_i can't be negative, otherwise the rateio itself would be negative. The video itself is fine.
@@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.
@@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.
@@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.
Nice. I don't think omission of -500 is an issue because question only ask for largest a1.
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
nice prob
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?
That's the one that yields the maximum value, yes.
There is no assigned value. However, for A1 to be maximum, you must assume all others are 50 each
@@PrimeNewtons Bonus question: Is there a minimum value for a_1+a_2+...+a_100?
Doesnt positive 100 mean that the denominator must also be positive?
Obviously. Where's the problem with that?
yes, but not that all individual terms of the denominator are also positive.
@@mistermudpie Their sum must be, e.g, a_1=-450 and a_2 up to a_100=50, whose sum is 4500.
@@niloneto1608 why are you explaining the obvious?
@@mistermudpie I wasn't explaining it to you per se, but rather for the guy who asked it in first place.