Hi fellow Brilliantians, I found some of these polynomial questions which are quite interesting but hard for me... If you know how to solve these questions, please post your solutions below (or give me a hint). It would be really helpful for me. Thank you!

Let \(a + ar_{1} + ar^2_1+ \ldots \) and \(a + ar_{2} + ar^2_2+ \ldots \) be two different infinite geometric series of positive numbers with the same first term, \(a\). The sum of the first series is \(r_1\), and the sum of the second series is \(r_2\). What is \(r_1 + r_2\)?

The polynomial function \(f\) satisfies \(f(6+x) = f(6-x)\) for all \(x\). Moreover. \(f(x) = 0\) has exactly \(5\) roots. What is the sum of these roots?

(AIME) Amongst all integers \(n\) satisfying \(n+10 \mid n^3 + 100\), which is the largest?

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What is the sum of an infinite geometric series? Write the two equations.

What does the first condition mean? To go further, if \(r\) is a root, what number must also be a root?

Use long division on the polynomials.

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– Happy Melodies · 3 years, 1 month ago

For question \(1\), I have used the method that you have mentioned but didn't see the trick till now! Thanks :) The answer's \(1\) right? For question \(3\), oh my... why didn't I think of it... may I check if answer is \(n= 890\)?Log in to reply

– Daniel Chiu · 3 years, 1 month ago

Yeah, I think those two answers are right, good job!Log in to reply

– Ratnesh Kumar · 3 years ago

1st answer is 1..nd is d 2nd q.ri8Log in to reply

– Happy Melodies · 3 years ago

Yes, first question's answer is 1! :) But I think the 2nd question's answer is 30... (for me at least)Log in to reply

– Happy Melodies · 3 years, 1 month ago

For question 2, if \(r\) is a root, then \(f(r)=0\). So, \(r+12 \text{ or } r-12\)... But I am not sure how is it possible for \(f(x)=0\) to have 5 roots?Log in to reply

\(r\) and \(12-r\) will not be equal unless... – Daniel Chiu · 3 years, 1 month ago

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For 2, I think you may be getting confused about the fact that there are 5 roots. Maybe it would be easier if you tried to solve the problem for when \(f(x) = 0\) has 4 roots. Suppose that \(r_1\) is a root. You can give the given symmetric condition to find another root; what is that root? For example, if \(f(4) = 0\), then what other value \(y\) also definitely gives \(f(y) = 0\), and what is the sum \(4 + y\)?

When you figure that out (and from your comments I think you understand that already), think about how the function could possibly have an odd number of roots. For example, if the function only had one root, what would that root have to be? Do you see what I'm getting at? If so, extend it to the five root case! :) If not, I'd be happy to explain further.

Spoiler Spoiler Spoiler Spoiler Spoiler Spoiler Spoiler:Consider, perhaps, \(f(x) = |x - 6|\) for the one root case, and for the five root case, consider \(f(x) = |(x - 4)(x - 5)(x - 6)(x - 7)(x - 8)|\). – Sotiri Komissopoulos · 3 years, 1 month ago

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– Happy Melodies · 3 years, 1 month ago

Ohhhh i got it! Thanks a lot! :) cause the sum of any 2 roots is just \(6+x+6-x =12\) and the odd root must be \(6\) since \(x=6\) is the line of symmetry. Thus, sum of the roots is just \(12 \cdot 2+6=30\)! Am I right to say that?Log in to reply

– Daniel Chiu · 3 years, 1 month ago

Seems right!Log in to reply

– Happy Melodies · 3 years, 1 month ago

Thank you all for your patience and hints! :) hurray!Log in to reply

About what point or line are the functions \(f(x), f(-x)\) symmetric? About what point or line is the function \(f(x+6), f(-x + 6)\) symmetric then?

What's the remainder when we divide the two polynomials, and how can we use that to help us find a maximum?

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– Happy Melodies · 3 years, 1 month ago

Thanks for your help :) I am not sure where the first question came from.. My teacher gave the question to me... I am done with questions 1 and 3, but really am not sure how to continue with the 5 roots idea.Log in to reply

If you are still stuck on #2 after Daniel's hint, here is another one: \(f(x)\) is symmetric with \(x=6\) as the line of symmetry. Do you see why? – Daniel Liu · 3 years, 1 month ago

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– Happy Melodies · 3 years, 1 month ago

Yes, i see it..Log in to reply

You didn't say anything about #2, so you know, just in case. :) – Daniel Liu · 3 years, 1 month ago

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– Happy Melodies · 3 years, 1 month ago

Oh... No worries, I didn't think it that way! Anyway, thanks for your help :)Log in to reply

How did ya solve the third problem? – Abhimanyu Swami · 3 years, 1 month ago

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may be i did nt get the question right but still see what i got.

a + ar1 + a(r1)^2 … = r1.

this means the sequence must converge to r1. as r1 is the ratio itself so it is finite quantity.

so r1 = a/(1-r1)

so i just got a sequence in head that satisfies it, ¼ + ⅛ + …. = ½ where r1 = ½

so same thing with r2.

r=a/(1-r)

r - r^2 = a.

a is positive (given condition). so r > r^2

and it should be. (r<1 for any sequence to converge).

okay it got many possibilities like

r = ¼ and a = ⅛

r = 1/3 and 2/9.

and many more ..

so how we are able to solve 1st question r1 +r2 ?

and the answer is like : 0 < r1 + r2 < 2 as long as both "r"s are positive. – Soham Zemse · 3 years, 1 month ago

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– Daniel Chiu · 3 years, 1 month ago

So from the first sequence you get \(a=r_1-r_1^2\), and from the second you get \(a=r_2-r_2^2\). What happens when you equate these equations?Log in to reply

– Soham Zemse · 3 years, 1 month ago

ok i got itLog in to reply