C'mon, More logs?
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have length
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, where n is a positive integer from 1-10. Find the sum of all the
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radii.
The answer is 1023.
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1 solution
The text of the problem says that n is between 1 and 10. That means: 1) the value "10" is included! 2) the value "0" is excluded. Thus, the solution should be 2046: 2^(10+1) -1 -2^0 = 2048 - 2 = 2046 The text is unclear!
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You mean to say 2 0 4 8 − 1 ⟹ 2 0 4 7 .
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n=0 is also an excluded value. If it were included, then the formula you used would be correct. Here, however, the minimum value for n is 1. Therefore, we also need to subtract 2^0 from the sum.
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@Petru Lupsac – No, it's included! When you apply the totient function to a power of two, it divides it by two. So I'm counting 2 / 2 + 4 / 2 + 8 / 2 + 1 6 / 2 . . . 1 0 2 4 / 2 ⟹ 1 + 2 + 4 + 8 . . . 5 1 2 ⟹ 1 0 2 3 .
@David Lee It is hard to understand what you mean in this question. Can you try rewriting it so that it can be understood? At least mention that ϕ refers to the Euler's Totient function.
It is not clear what "Logarithmic Log" means. If you are trying to define that term, it helps to add the word define.
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Yeah. If I were a moderator, I would be able to fix the poor wording and italicization, as well as the definition of terms. Hint hint. @Peter Taylor @Silas Hundt @Suyeon Khim @Daniel Hirschberg .
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I realize you are a young man, and that this is the Internet, but your attitude, repetition, and methods do not help your cause. That said, the decision is not mine, I'm mostly an observer. I speak mostly from the point of view of a one time "really smart kid" who didn't understand why people didn't just do what he said, who is now a much older kid is it a reasonable win rate.
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@Daniel Hirschberg – Ugg... I already spoke with Peter about this, and I've told him that although I may come off as a bratty, demanding, obnoxious, kid, if you could hear my tone behind the screen called "The Internet", it's a respectful, (partially contained) excited tone! It is for this reason that I almost ALWAYS end my sentences with :D. I would get excited if somebody told me my problem was cool. In short, I'm just a really cheery dude who likes socializing. Of course, I'm OCD as well as (not medically) obsessive, which is why I spend hours and hours on Brilliant, and little flaws make me want to pull my hair out. I really mean the best for this website when I make my snotty suggestions and point out glitches and stuff. Sometimes things don't work out for me, though, like not making it to the "Who to follow" page, and stuff like that. Anyways, I'm sorry for my attitude, repetition, and methods. :D
When we apply Euler's totient function to powers of two, we actually divide whatever power of two that we're being asked to find by two! The reason being that the only factors of powers of two are multiples of two, and exactly half of the numbers less than 2 n are divisible by two. Thus we are asked to find the sum 1 + 2 + 4 + 8 . . . 5 1 2 . We could just hack this out, but there's a really cool way to do this that's really useful for speed math. For a sum 2 0 + 2 1 + 2 2 + 2 3 . . . 2 n , the answer is simply 2 n + 1 − 1 . You're welcome to experiment around with it some more, but we can apply this to our desired sum. Thus, the answer is 2 9 + 1 − 1 ⟹ 2 1 0 − 1 ⟹ 1 0 2 4 − 1 = 1 0 2 3 . And we're finished! Great problem David! :D