* Prompted by various changes in my schedules and interests, I decided after much debate to end Weekly Math Challenges at Problem 101, and transition to other projects in TH-cam (such as a course in linear algebra that may come out this summer) or studies in personal life. I express my gratitude and apologies to everyone who submitted a proposal or participated in the challenge in any way.
(Outdated Comment) * I suppose it won't be truly "Weekly" Math Challenges anymore since the problems will be uploaded at a much more flexible (i.e., random) schedule once the fall semester begins. Perhaps we should keep the name just for the sake of tradition, but if you have any suggestions on how to rename the series, please feel free to comment it here. * Edit: Due to a combination of headache/sore throat, busy schedule, and technical difficulties (almost comically, all of my tablet, drawing software, and recording software did not work properly when I first tried to record yesterday), the solutions to Challenges 99 and 100 will be posted sometime in the week of August 11, as the time permits. I sincerely apologize for the unusually long wait between problem videos and their respective solution videos. I do additionally remark that, starting with the 101st challenge, the solutions to some challenge problems will not necessarily be posted, mainly due to time constraints.
@@NoNameAtAll2 I'm not sure if I like the term "periodic." It sounds cleaner to just have it as "Math Challenges," but I'm not sure if that is a necessary change.
Hmm, the English language is rich with synonyms... irregular, sporadic, intermittent (I quite like that one) . You could also just call it The LetsSolveMathsProblems Challenge Series, or just the Maths Problem Challenges. The problem with keeping 'Weekly' is that it might confuse new players who don't know the history. In any case, I'm very pleased that you are keeping it going!
How about something like Secret Math Challenges, because nobody but you knows when they'll be uploaded, and part of the fun is stumbling upon them early by chance and trying to be one of the first 10 to solve them.
There are 1976 First of all, we remark that for any positive integer n, we have floor(n) = n and {n} = 0 while for all n greater than 1 we have floor(n^-3) = 0 and {n^-3} = n^-3 We see that 1 is a problem case, and after some calculations, we find that indeed, f(g(1+1^-3)) < g(f(1+1^-3)) (they are about 1.07 and exactly 2, respectively) We will first see what happens to g(f(n+n^-3)), since its the simplest of the two cases Using what we remarked at the start, we find that f(n+n^-3) = n^2 + n^-9 Using the same kind of logic, we determine that g(n^2 + n^-9) = n + n^-3 So far what we have found is that for any n greater than 1, g(f(n+n^-3)) = n + n^-3 Now we tackle what happens to f(g(n+n^-3)) Again, using what we remarked at the start, we find that g(n+n^-3) = sqrt(n) + n^-1 So, this leads to f(g(n+n^-3)) = floor(sqrt(n) + n^-1)^2 + {sqrt(n) + n^-1}^3 Here there are two possible cases, either n is a square number or it is not If n is a square number, f(g(n+n^-3)) = n + n^-3, which is exactly what we found for g(f(n+n^-3)), there is no strict inequality, so what we're looking for doesn't hold If n is not a square number, f(g(n+n^-3)) = (n-1) + (junk smaller than 1)^3 = n - (small number) In this case, f(g(n+n^-3)) < g(f(n+n^-3)) (since a negative number is always smaller than a positive one) We have thus determined that every single positive integer that isnt a non 1 square number hold the condition! There are 44 square numbers smaller than 2019, but 1 is a square number too, so finaly that gives that the number of solutions is 2019 - 43, which is 1976
@@Артем-х9у9к Yeah I realised shortly after sending the response I had assumed that floor(sqrt(n) + n^-1) = floor(sqrt(n)) for any n, which as we can see with n = 3, is certainly not true There is another mistake in my reasoning (when i go about the n-1 on the second function composition) but it diesnt end up mattering Anyways yeah, I was wrong
g is a left-inverse for f, i.e. g°f=id. However f(g(n+1/n³)) = floor(sqrt(n) +1/n)² + fractionalPart(sqrt(n) +1/n)³, which is equal to the input (n+1/n³) only for those values for which sqrt(n) is integer (and n>1). For all other values one can more or less assume, that floor(sqrt(n) +1/n) < sqrt(n) and therfore f(g(n+1/n³)) < n + 1/n³. The only exception, I've found where sqrt(n) is close enough to it's ceiling, such that floor(sqrt(n)) < floor(sqrt(n) +1/n) is 3, because after that 1/n is too small (however it can not be safely said that holds for every n >3). So the inequality holds for all n apart from 3 and the perfect squares of 2,3,4,.... The answer therefore is |N2019 \ {4,9,16,...} \{3}| = 2019 - floor(sqrt(2019)) = 1975.
frac(n+1/n^3) = 1/n^3. We have f(n+1/n^3) = n^2 + 1/n^9 and g(n+ 1/n^3) = sqrt(n) + 1/n. Then g(f(n+1/n^3)) = n + 1/n^3. f(g(n+1/n^3)) = (floor(sqrt(n))^2 + (frac(sqrt(n)) + 1/n)^3. Whenever n is a square number (remember, n>1), then simplifies to n+1/n^3. So n cannot be a square. Now suppose k^2 < n < (k+1)^2; n = k^2 + r, r < 2k+1. Then f(g(n+1/n^3)) = k^2 + (frac(sqrt(n)) + 1/n)^3. Now 1/n < 1/k^2, and frac(sqrt(k^2+r)) = sqrt(k^2+r)-k = r/(sqrt(k^2+r)+k) < r/(2k) . Then (frac(sqrt(n)) + 1/n)^3 < (1/(2k) + 1/k^2)^3 = ((k+2)/(2k^2))^3. We need ((k+2)/(2k^2))^3 < r + 1/(k^2 +r)^3. One can show this expression increases with r, since k^2+r = n> 1. Thus consider r=1. We can simplify the inequality to (k/2 + 1 + 1/(2k) + 1/k^2)^3 < k^6 + 3k^4 + 3k^2 + 2. Clearly this is true as we increase k. For k = 2, the inequality is true. However, for k = 1, it is false. So we must explicitly check n= 2 and 3. We get it is true for 2, but not 3. In summary, we have the solutions are those n which are not squares, but including 1, and not including 3. These cancel out. The number of squares less than or equal to 2019 is 44. The answer is then 2019-44 = 1975.
For x = n + 1/n^3 it's not hard to check that g(f(x)) = x. So the problem becomes finding integers such that f(g(n+1/n^3)) < n + 1/n^3. We can use the results that (a) floor( sqrt(n) + 1/n ) < n - 1 for all integers n except for perfect squares greater than one* and (b) {x} < 1 for all x to observe that the LHS is < n - 1 + 1 = n for all n =/= m^2. When n = m^2 there is equality. So the problem becomes counting the perfect squares between 1 and 2019, which is 44. So the answer is 2019-44 = 1975. *n = 1 is a special case, but we can check that numerically to find LHS = 1.746... < RHS = 2
The answer is 1975. n=1 must be examined separately, because it is the only positive integer such that n+1/n³ is an integer. This case works. For n>1 clearly floor(n+1/n³)=n {n+1/n³}=1/n³. Therefore we can easily evaluate g(f(n+1/n³))=n+1/n³. f(g(n+1/n³))=f(√(n)+1/n). Since √(n) can have a fractional part we have to distinguish 2 cases. i) n=m²: f(g(n+1/n³))=n+1/n³=g(f(n+1/n³)). Therefore this case does not work. ii) n≠m²: there are 2 subcases: ii-a) {√(n)}+1/n1 √(n)-floor(√(n))+1/n>1 √(n)+1/n>floor(√(n))+1 n+1/n²+2√(n)/n>(floor(√(n))+1)² ≥n+1 2*n√(n)>n² 2>√(n) n
I'm not sure whether the pinned comment means there will still be a solution video for this or not. So I will post mine here just in case. It might be *too* long, since I'll try to be very thorough in my claims. First, we'll prove this lemma: **Lemma:** Let n be a non-perfect-square positive integer. Let k^2 (k positive) be the largest perfect square smaller than n. Then {sqrt(n)} + 1/n < 1, for n > 4. **Proof:** From k^2 < n < (k + 1)^2, we get k < sqrt(n) < k + 1, from which sqrt(n) = k + r, where r is a real number and 0 < r < 1. Again from k^2 < n < (k + 1)^2, we have k^2 < n < k^2 + 2k + 1, from which n = k^2 + m, where m is a positive integer and 1 ≤ m ≤ 2k < 2k + 1. Now: n = k^2 + m => (k + r)^2 = k^2 + m => r^2 + 2kr - m = 0. Solving that equation, we arrive at r = sqrt(k^2 + m) - k (the other solution can be ignored as it is negative). Notice that to maximize the value of r, we must maximize m, so set m = 2k. If the lemma is proven for maximum r, it is proven for all the other, smaller, values of r, given a value of n (r being a shorter notation for {sqrt(n)}). Since 1/k^2 > 1/n, it suffices to prove r + 1/k^2 < 1. Then: sqrt(k^2 + 2k) - k + 1/k^2 < 1 sqrt(k^2 + 2k) < 1 + k - 1/k^2 k^2 + 2k < 1 + k^2 + 1/k^4 + 2k - 2/k - 2/k^2 0 < 1 + 1/k^4 - 2/k - 2/k^2 0 < k^4 - 2k^3 - 2k^2 + 1 k = 1 and k = 2 do not satisfy this inequality, but it doesn't necessarily mean they don't satisfy the lemma; they will be handled separately. For k = 3, the expression is indeed positive, and by studying f(x) = x^4 - 2x^3 - 2x^2 + 1 and its derivatives, we see that for x > 2, the function strictly increases, therefore the inequality is satisifed by all k ≥ 3. Now for k = 2, verify that sqrt(8) - 2 + 1/8 < 1, and since 8 = 2^2 + 2*2, the lemma is valid for n numbers between 2^2 and 3^3. k = 1 doesn't matter for the veracity of the lemma (since it says n > 4), but it will be useful to check either way: sqrt(2) - 1 + 1/2 < 1, but sqrt(3) - 1 + 1/3 > 1. The lemma is proven, QED. (i'll denote [x] as the floor function of x) Now for the *actual* thing, haha. Notice that it holds for n = 1, as it results in 1 + (sqrt(2) - 1) < 2, which is valid as sqrt(2) < 2. Now let's work only with n > 1. That leads to 1/n < 1, which is is very important when computing results with the floor function, since if p is an integer and q is a real number greater than 1, we have that [p + 1/n^q] = p. Assume n = k^2 is a perfect square, then let's verify the inequality's validity: f(g(n + 1/n^3)) < g(f(n + 1/n^3)) f([n + 1/n^3]^(1/2) + {n + 1/n^3}^(1/3)) < g([n + 1/n^3]^2 + {n + 1/n^3}^3 f(n^(1/2) + 1/n) < g(n^2 + 1/n^9) [k + 1/n]^2 + {k + 1/n}^3 < [n^2 + 1/n^9]^(1/2) + {n^2 + 1/n^9}^(1/3) k^2 + 1/n^3 < n + 1/n^3 Which is impossible, therefore if n is a perfect square different than 1, the inequality won't hold. Do realize that g(f(n + 1/n^3)) = n + 1/n^3 irrespective of n being a perfect square, so let us now analyze f(g(n + 1/n^3)) for n non-perfect-square: f(g(n + 1/n^3)) = f([n + 1/n^3]^(1/2) + {n + 1/n^3}^(1/3)) = f(n^(1/2) + 1/n) = [n^(1/2) + 1/n]^2 + {n^(1/2) + 1/n}^3 It is right here that we apply the lemma. For n > 4, and with k^2 being the largest perfect square smaller than n (k positive): k < n^(1/2) + 1/n = k + {n^(1/2)} + 1/n < k + 1, as such [n^(1/2) + 1/n] = k. As we saw before, this is also valid for n = 2, but *not* n = 3; that last one will be handled later. The expression is then equal to: k^2 + ({n^(1/2)} + 1/n)^3 Since {n^(1/2)} + 1/n < 1, k^2 + ({n^(1/2)} + 1/n)^3 < k^2 + 1 ≤ n < n + 1/n^3, thus satisfying the inequality. As for the case n = 3, it is verifiable that it does not satisfy the inequality. Thus, the numbers that *don't* satisfy the inequality are 3 and perfect squares greater than 1. There are 44 perfect squares up to 2019, as 44^2 < 2019 < 45^2. Counting 1 off and counting 3 in, we end up with 44 numbers not satisfying the condition. Therefore, the answer is 2019 - 44 = 1975.
The answer 1975. Let l(x) denote the floor of x, r(x) denote the fractional part of x and s(x) the square root of x. Plugging integer n > 1 into the equation simplifies the inequality to f(s(n) +n^-1) =2k+2 since m and k are integers. Since also l(s(n) +n^-1)=k^2+2k+2>(k+1)^2+q^3 and the proof is complete. Thus the answer is 2019-l(s(2019))=2019-44=1975.
Can you post the problems whenever you are free to do so. As you go along the schedule . Not might be weekly but however it might be. Many people would be glad to hear that.
1976 if n > 1: f(n^1/2 + 1/n) < g(n^2 + 1/n^9) f(n^1/2 + 1/n) < n + 1/n^3 By testing values, we see this is true for 2, 5-8, 10-15, 17-24, and so on---or in other words everything except for perfect squares There are 43 perfect squares up to 2019 (not including 1) so in total there are 2018 - 43 = 1975 for this part. if n = 1: f(g(2)) < g(f(2)) f(√2) < g(4) 1+(√2-1)^3 < ----> This is true so 1 works. The total is 1976 :D
The answer is 1975. Before we simplify the inequality note that any integer > 1, floor(n + 1/(n^3)) = 0 and {n + (1/(n^3))} = n^3 We see that 1 is a problem case. Plugging in 1 in the inequality we get 1 + (0.something)^3 < 2. This shows that 1 is one of the solutions. Simplifying the inequality we get (floor((√n)+(1/n)))^2 + ({(√n)+(1/n)})^3 < n + 1/(n^3) Relise that we can categorise the inequality as 2 cases: Case 1: n is a perfect square Simplifying the inequality we get n + 1/(n^3) < n + 1/(n^3) This contradicts our inequality and thus leads to the point that n cannot be a perfect square(excluding 1) Case 2: n is not a perfect square In this case we first find all the cases that floor((√n)+(1/n)) ≠ floor((√n)). Relise that in order to make this true we need the decimals of √n and 1/n to be big enough such that floor(√n + 1/n) > floor(√n) For the decimals of √n to be largest we need n to be exactly 1 less than some perfect square. By trial and error we find out that n = 3 is the only case that satisfies the condition. Plugging in 3 we get that 4 + something < 3 + 1/27 This contradicts the inequality and thus leading to the point that 3 cannot be one of the solutions. For the other case, we can let n = k^2 + c, where k^2 is the closest perfect square smaller than n. Simplifying the inequality we get k^2 + (0.something)^3 = k^2 + c + 1/(k^2 + c) This proves that all the other integer satisfies the inequality. There are 43 perfect squares(1 excluded) for n ≤ 2019, and we also have to exclude 3 from the numbers Therefore we get the answer of 2019 - 43 - 1 = 1975.
On the RHS. g(f(n+1/n^3)=g(n^2+1/n^6) because n is a integer, similar logic, g(n^2+1/n^6)=n+1/n^3 On the LHS, f(g(n+1/n^3))=f(sqrt(n)+1/n) because n is an integer. First if n is a perfect square, since sqrt(n) is an integer, f(sqrt(n)+1/n)=n+1/n^3, so the inequality doesn't satisfy Since there are 44 perfect squares, right now we have 1975 possible numbers that could work. But, if frac(sqrt(n))+1/n > 1, LHS becomes greater than the RHS because the integer part of LHS is greater than the RHS which means LHS is greater. For frac(sqrt(n))+1/n > 1, frac(sqrt(n)) has to be as close as a perfect square as possible so let n = k^2-1 where k is a positive integer greater than 1. if n = k^2-1, frac(sqrt(n)) = sqrt(k^2-1) - (k-1) since k-1 is the integer part, then the equation becomes sqrt(k^2-1)-(k-1)+1/(k^2-1)>1 isolate the radical, we get (k^2-1)sqrt(k^2-1)>k(k^2-1)-1 squaring both sides since both sides are positive gives us (k^2-1)^3>(k(k^2-1)-1)^2 expanding and isolating 0, we get k^4-2k^3-2k^2+2k+1
ans: 1975, for all n ∈ Z+ gof will always yield the input. for all n = d^2 form where 1 gof only d = 2 satisfies this so, the number of elements for which fog < gof = 2019 - 43 - 1 = 1975
Ans = 1975 Let x=n + 1/n For n=1: x=2 then f(x)=4 then g(f(x))=2 And g(x)=sqrt(2) then f(g(x))=1 + (sqrt(2) - 1)^3 less than 2 so 1 is from our required numbers. For n>1: the integar part will always be n and its integar exponents while the fraction part will be 1/n with its exponents. R.H.S: f(x)=n^2 + 1/n^9 R.H.S=g(f(x))=n + 1/n^3 L.H.S: g(x)=sqrt(n) + 1/n We have two possibilities: i) n is a perfect square. Then L.H.S=f(g(x))=n + 1/n^3=R.H.S so all perfect squares are not of our required numbers. ii) n is not a perfect square. Let s^2 < n < (s+1)^2 Then L.H.S=f(g(x))=s^2 + (sqrt(n)-s+ 1/n)^3 < n < R.H.S Conclusion: We are looking for non-perfect squares from 1 to 2019 Since sqrt(2019)=44.93 Then reqired number=2019-44= 1975
The answer is 1975 We can plug in n=1,2,3,4 and check that the inequality holds for n=1, n=2, but not for n=3 or n=4 For n >=5, we will prove that the inequality holds if and only if n is not a perfect square number. If n is a perfect square number: LHS = RHS = n+1/n^3, therefore the inequality does not hold. If n is not a perfect square number, then we can write n= k^2+q, where k=floor(sqrt(n)) and q is an integer from 1 to 2k, also k>=2 as n >=5. We have RHS = n+1/n^3. LHS = f(sqrt(n)+1/n) . By definition of {x} we have {x} sqrt(n) = sqrt(k^2+q) > sqrt(k^2) = k, so we have to prove that sqrt(n)+1/n < k+1, or sqrt(k^2+q)+1/(k^2+q) < k+1. As the function y= sqrt(x)+ 1/x is increasing in the interval [2,+infinity), we only need to prove this in case q=2k, i.e sqrt(k^2+2k) + 1/(k^2+2k) < k+1, squaring both sides and cancelling the terms give us 1/(k^2+2k)^2 + 2/sqrt(k^2+2k) < 1, which is true for all k>=2. Hence floor(sqrt(n)+1/n) = k Therefore we have LHS < (floor(sqrt(n)+1/n)) ^2 +1. = k^2+1 =5. Therefore the final answer is 1975
The answer is 1975. Let floor[x] denote the floor function on x and let fract[x] = x - floor[x]. Consider first f(g(n + 1/n^3)). By definition, g(n + 1/n^3) = floor[n + 1/n^3]^(1/2) + fract[n + 1/n^3]^(1/3). Since n is an integer, we can simplify this to n^(1/2) + 1/n. Now applying f, we get f(n^(1/2) + 1/n) = floor[n^(1/2) + 1/n]^2 + fract[n^(1/2) + 1/n]^3. Since n is an integer, we can simplify this to S(n) + 1/n^3, where S is defined as the greatest perfect square less than or equal to n. By similar reasoning, we have g(f(n + 1/n^3)) = n + 1/n^3. So, we are looking for n such that S(n) + 1/n^3 < n + 1/n^3 S(n) < n. This is true as long as n isn't a perfect square. The perfect squares between 1 and 2019 inclusive are 1, 2^2, 3^2, ... , 44^2. So, there are 2019 - 44 = 1975 values of n that satisfy the problem, as claimed.
Darn, after reading the other solutions, I now see that I messed up (and got the right answer only by coincidence). It's obvious why my reasoning fails for the special case n=1, but I completely missed how f(n^(1/2)+1/n) needs to be treated more carefully for n
The answer is 1975. First of all, we can start by plug in n+1/n^3 in to both side of the inequality. Because the nature of the function. We would plug in values larger than 1 into the inequality, we can get that the right side of the inequality became g(n^2+1/n^9) because we can ignore the 1/n^3 in the floor function because it is always less than 1 and n in {x} because the functions essentially just asks for the fractional part of x. After simplifying, the left side becomes n+1/n^3, which shows that f(x) and g(x) are inverses of each other in this sense. And then we simplify the right part of the inequality.It first becomes f( sqrt(n)+1/n). For convenience, we are going to call the fractional part of sqrt(n) +1/n, M, and the integral part I. The right side of the inequality becomes I^2+M^3. We realise that it looks a lot like the left side of the inequality. The two sides are equal, if and only if n is a perfect square. The left side of the inequality is always going to be larger than the right side of the inequality because if n is not a perfect square, the floor function is going to make I^2 to be the the closest lower perfect square which would be always lesser than n because the M^2 and 1/n^3 are always less than 1, however this is not going to work if the fraction part of sqrt(n) and 1/n sum to be larger than 1. We can test the extreme case, as the ones that are 1 lower than perfect and see that 3 does not work. As the value goes sqr(n)is going to be arbitrarily close to an integer and 1/n is going to arbitrarily small, we can test the largest extreme case of 1935 as it is one below the largest perfect contained in the set, and it satisfies with inequality. So we can see every perfect square larger than 1 and 3 do not satisfy the inequality and there are 44 perfect squares less than 2019. And then we plug in 1 and see that it satisfies with the inequality. So the final answer is that 2019-43-1=1975
The easy part is that the right side, g(f(n+1/n^3)) is just n + 1/n^3. This is just an inverse (except when n is 1). The left side doesn't work out as nicely. You can reduce to f(sqrt(n) + 1/n) (again, not counting n =1). If n is a perfect square, then that becomes just n + 1/n^3 again. So, the sides are equal when n is a perfect square. If n is not a perfect square, then the question becomes: Is the remainder of sqrt(n), added to 1/n, ever >=1?, for n up to 2019? No, it's always a decimal. Thus, the floor function will always drop it (yay!). So, for all n not a perfect square (and not 1), the left side will be less than the right. There are 43 perfect squares between 1 and 2019 For n = 1, you can just plug it in and see the left side is less then as well, so that's good. So, 43 numbers don't work. 2019 - 43 = 1976
The answer is 1976 By substitution, 1 is one value of n that satisfy the inequality f(g(2)) = f(sqrt2) = 5sqrt2 - 6 and g(f(2)) = g(4) = 2, and 5sqrt2 - 6
Answer: all integers except 3 and perfect squares ( from 1² to 44²). ( [..] stand for the floor function ) Well, if n>1 then [n+1/n³] = n and {n+1/n³]=1/n³ so g°f(n+1/n³) = n+1/n³ ( f and g are inverse functions from this prespective) And f°g(n+1/n³) = f(sqrt(n) + 1/n) but we have 3 cases for f°g: 1- n is a perfect square and in that case f and g are totally inverse functions because [sqrt(n)+1/n]=sqrt(n) and 1/n < 1 so f°g=g°f => perfect squares are not solutions. 2- if n = x²-1 so in order to have [sqrt(n)+1/n] = x we need to find x s.t: sqrt(x²-1)+1/(x²-1)>x so sqrt(n+1) + sqrt(n) > n but we can prove easily that for n>=8 sqrt(n+1)+sqrt(n) < 2sqrt(n+1) < n+1 because 23 hence when n= 3 , f°g>g°f so 3 is excluded as well. 3- the last case left is when n is further from the nearest greatest perfect square meaning that n=x²-a where 2x-1sqrt(x²-a)+sqrt(x²-1) Let a and b be 2 integers >1 we have : (b-1)*(a-1)+1>0 so ab>a+b but a+b>sqrt(a)+sqrt(b) so ab>sqrt(a)+sqrt(b) but this does not apply for x=2 and a =2 so we need to verify this case and we see obviously that sqrt(2)+1/2x f°g>g°f if [sqrt(n)+1/n]=x because x²>x²-1+1/n³ ( 1>1/n³).
The answer is 1976. For.1, the inequality holds. For other n, n+n^-3 has floor equal to n. When computing the LHS and RHS, inequality doesn't hold iff n is a square number. There are 43 square numbers other than 1 up to 2019, so the answer is 2019-43=1976
I got 1975. I plugged these bits of javascript into my console. (press f12 if you're on Google Chrome, and it probably works on other browsers too!) function g(a) {return Math.pow(Math.floor(a), 1/2)+Math.pow(a%1, 1/3)} function f(a) {return Math.pow(Math.floor(a),2) + Math.pow(a%1, 3)} var n = 0; for (var i = 1; i
Let n be positive integers between 1 and 2019 Write m(n) = n + 1/n^3 Write statement f(g(m(n))) < g(f(m(n))) as P(n) (inspired by mathematical induction, though we are definitely not using this method) g(f(m(n))) = g(n^2 + 1/n^9) = n+1/n^3 = m(n) f(g(m(n))) = f(g(2)) = f(sqrt(2)) ~= 1+(0.414)^3 ~= 1.071, if n=1 f(sqrt(n)+1/n) for all n>1 Now we know P(1) satisfies. If n is perfect square larger than 1: f(g(m(n))) = m(n), which does not satisfy P(n) There are floor(sqrt(2019)) = 44 perfect squares for positive integers n, but we have proved P(1) satisfies. There are 44-1 = 43 possible values of n proved not satisfying P(n) Now we want to know if P(n) satisfies when n is not perfect square. From now on, denote any perfect squares larger than 1 as s. As f(g(m(n))) = f(sqrt(n)+1/n) for all n>1 We can test behavior of this function: y = sqrt(x) +1/x - ceil(sqrt(x)) If y(n) > 0 for particular positive integer n that is not perfect square, then P(n) does not satisfy. y(s) = sqrt(s) + 1/s - ceil(sqrt(s)) = sqrt(s) + 1/s - sqrt(s) = 1/s, clearly P(s) does not satisfy. What happens if x is just larger than s? lim(x→s+)y(x) = lim(Δx→0) [sqrt(s+Δx)+1/(s+Δx)-ceil(sqrt(s+Δx))] = sqrt(s) + 1/s - sqrt(s) - 1 = 1/n - 1 = 1 We can still differentiate the continuous part of y function by ignoring ceil part. y'=1/2sqrt(x) - 1/x^2 y" = 2/x^3 - 1/(4x^3/2) Set y' = 0, x=2^k, 1+k/2 = 2k, k=2/3, x=4^(1/3) y"(4^1/3) = 2/4 - 1/4*2 = 3/8 > 0 y' > 0 for all x >= 4^(1/3) ~=1.1587, the continuous part of the function is always increasing for all x >= 4^(1/3) For x>0, which part of y(x) is not continuous? All positive integers x, because of ceil function. That means there is exactly one x-intercept between each 2 perfect squares x (except 1 and 2, as lim(x→1+)y(x) = 0 As we only care about integers 1 4 > 3+1/27, P(3) does not satisfy. y(2) = sqrt(2) +1/2 - 2 ~= -0.0858 < 0 Verify: f(g(m(2))) = f(sqrt(2) + 1/2) ~= f(1.914) < 2+1/8, P(2) satisfies. For all n > 4 that are not perfect squares, as y(s-1) < 0, P(n) satisfies. Finally, number of positive integers n satisfying P(n) = 2019 - 43 -1 = 2019-44+1-1 = 1975 This question is not very difficult by itself, but need a lot of careful investigation, and it is very difficult to fully explain the solution. If I did not fully explain this, please tell me which part I have missed, I'll try to explain more.
Basic concept: Q: Why y(s) is different with lim(x→s+)y(x)? A: Behavior of ceil and floor function: Let i be any integer (do NOT confuse with imaginary number i) ceil(i) = i lim(x→i-)ceil(x) = i lim(x→i+)ceil(x) = i+1 floor(i) = i lim(x→i-)floor(x) = i -1 lim(x→i+)floor(x) = i
I think the answer to be 1975, because n and n² are integers and 1/n³ is a fraction smaller than 1, the right hand side of the equation can be simplified to n + 1/n³. The left hand side simplifies to f(sqrt(n) + 1/n), which can't be calculated easily as sqrt(n) is not necessarily an integer. We can however conclude that the left hand side will never be strictly larger than the right hand side, because 1/n³ is always smaller than 1, so the highest the left hand side can be is equal to the right hand side. This only occurs when n is a perfect square. The right hand side then also simplifies to n + 1/n³. The inequality is therefor valid for all non-perfect squares less than or equal to 2019. The answer is therefore 1975 as there are 44 perfect squares smaller than 2019.
For n=1 the equation however holds (no equality although it is a perfect square). So your solution only randomly coincodes with the actual solution because your two mistake add up to 0
On the RHS. g(f(n+1/n^3)=g(n^2+1/n^6) because n is a integer, similar logic, g(n^2+1/n^6)=n+1/n^3 On the LHS, f(g(n+1/n^3))=f(sqrt(n)+1/n) because n is an integer. First if n is a perfect square, since sqrt(n) is an integer, f(sqrt(n)+1/n)=n+1/n^3, so the inequality doesn't satisfy Since there are 44 perfect squares, right now we have 1975 possible numbers that could work. But, if frac(sqrt(n))+1/n > 1, LHS becomes greater than the RHS because the integer part of LHS is greater than the RHS which means LHS is greater. For frac(sqrt(n))+1/n > 1, frac(sqrt(n)) has to be as close as a perfect square as possible so let n = k^2-1 where k is a positive integer greater than 1. if n = k^2-1, frac(sqrt(n)) = sqrt(k^2-1) - (k-1) since k-1 is the integer part, then the equation becomes sqrt(k^2-1)-(k-1)+1/(k^2-1)>1 isolate the radical, we get (k^2-1)sqrt(k^2-1)>k(k^2-1)-1 squaring both sides since both sides are positive gives us (k^2-1)^3>(k(k^2-1)-1)^2 expanding and isolating 0, we get k^4-2k^3-2k^2+2k+1
* Prompted by various changes in my schedules and interests, I decided after much debate to end Weekly Math Challenges at Problem 101, and transition to other projects in TH-cam (such as a course in linear algebra that may come out this summer) or studies in personal life. I express my gratitude and apologies to everyone who submitted a proposal or participated in the challenge in any way.
(Outdated Comment)
* I suppose it won't be truly "Weekly" Math Challenges anymore since the problems will be uploaded at a much more flexible (i.e., random) schedule once the fall semester begins. Perhaps we should keep the name just for the sake of tradition, but if you have any suggestions on how to rename the series, please feel free to comment it here.
* Edit: Due to a combination of headache/sore throat, busy schedule, and technical difficulties (almost comically, all of my tablet, drawing software, and recording software did not work properly when I first tried to record yesterday), the solutions to Challenges 99 and 100 will be posted sometime in the week of August 11, as the time permits. I sincerely apologize for the unusually long wait between problem videos and their respective solution videos. I do additionally remark that, starting with the 101st challenge, the solutions to some challenge problems will not necessarily be posted, mainly due to time constraints.
rebrand as "periodic math challenges" ?
@@NoNameAtAll2 I'm not sure if I like the term "periodic." It sounds cleaner to just have it as "Math Challenges," but I'm not sure if that is a necessary change.
Hmm, the English language is rich with synonyms... irregular, sporadic, intermittent (I quite like that one) . You could also just call it The LetsSolveMathsProblems Challenge Series, or just the Maths Problem Challenges. The problem with keeping 'Weekly' is that it might confuse new players who don't know the history. In any case, I'm very pleased that you are keeping it going!
@@LetsSolveMathProblems My suggestion is Seasonal Math Challenges =)
How about something like Secret Math Challenges, because nobody but you knows when they'll be uploaded, and part of the fun is stumbling upon them early by chance and trying to be one of the first 10 to solve them.
There are 1976
First of all, we remark that for any positive integer n, we have floor(n) = n and {n} = 0 while for all n greater than 1 we have floor(n^-3) = 0 and {n^-3} = n^-3
We see that 1 is a problem case, and after some calculations, we find that indeed, f(g(1+1^-3)) < g(f(1+1^-3)) (they are about 1.07 and exactly 2, respectively)
We will first see what happens to g(f(n+n^-3)), since its the simplest of the two cases
Using what we remarked at the start, we find that f(n+n^-3) = n^2 + n^-9
Using the same kind of logic, we determine that g(n^2 + n^-9) = n + n^-3
So far what we have found is that for any n greater than 1, g(f(n+n^-3)) = n + n^-3
Now we tackle what happens to f(g(n+n^-3))
Again, using what we remarked at the start, we find that g(n+n^-3) = sqrt(n) + n^-1
So, this leads to f(g(n+n^-3)) = floor(sqrt(n) + n^-1)^2 + {sqrt(n) + n^-1}^3
Here there are two possible cases, either n is a square number or it is not
If n is a square number, f(g(n+n^-3)) = n + n^-3, which is exactly what we found for g(f(n+n^-3)), there is no strict inequality, so what we're looking for doesn't hold
If n is not a square number, f(g(n+n^-3)) = (n-1) + (junk smaller than 1)^3 = n - (small number)
In this case, f(g(n+n^-3)) < g(f(n+n^-3)) (since a negative number is always smaller than a positive one)
We have thus determined that every single positive integer that isnt a non 1 square number hold the condition!
There are 44 square numbers smaller than 2019, but 1 is a square number too, so finaly that gives that the number of solutions is 2019 - 43, which is 1976
how about n=3?
@@Артем-х9у9к Yeah I realised shortly after sending the response
I had assumed that floor(sqrt(n) + n^-1) = floor(sqrt(n)) for any n, which as we can see with n = 3, is certainly not true
There is another mistake in my reasoning (when i go about the n-1 on the second function composition) but it diesnt end up mattering
Anyways yeah, I was wrong
@@taprusthe3rd496 sqrt(n) + n^-1=5. So floor(sqrt(n) + n^-1)=floor(sqrt(n)) for n>=5.
@@Артем-х9у9кwow, that was the missing argue of my proof, thank you. Seems easy sayed like that 😁
g is a left-inverse for f, i.e. g°f=id.
However f(g(n+1/n³)) = floor(sqrt(n) +1/n)² + fractionalPart(sqrt(n) +1/n)³, which is equal to the input (n+1/n³) only for those values for which sqrt(n) is integer (and n>1). For all other values one can more or less assume, that floor(sqrt(n) +1/n) < sqrt(n) and therfore f(g(n+1/n³)) < n + 1/n³. The only exception, I've found where sqrt(n) is close enough to it's ceiling, such that floor(sqrt(n)) < floor(sqrt(n) +1/n) is 3, because after that 1/n is too small (however it can not be safely said that holds for every n >3). So the inequality holds for all n apart from 3 and the perfect squares of 2,3,4,....
The answer therefore is |N2019 \ {4,9,16,...} \{3}| = 2019 - floor(sqrt(2019)) = 1975.
frac(n+1/n^3) = 1/n^3. We have f(n+1/n^3) = n^2 + 1/n^9 and g(n+ 1/n^3) = sqrt(n) + 1/n. Then g(f(n+1/n^3)) = n + 1/n^3. f(g(n+1/n^3)) = (floor(sqrt(n))^2 + (frac(sqrt(n)) + 1/n)^3. Whenever n is a square number (remember, n>1), then simplifies to n+1/n^3. So n cannot be a square. Now suppose k^2 < n < (k+1)^2; n = k^2 + r, r < 2k+1. Then f(g(n+1/n^3)) = k^2 + (frac(sqrt(n)) + 1/n)^3. Now 1/n < 1/k^2, and frac(sqrt(k^2+r)) = sqrt(k^2+r)-k = r/(sqrt(k^2+r)+k) < r/(2k) . Then (frac(sqrt(n)) + 1/n)^3 < (1/(2k) + 1/k^2)^3 = ((k+2)/(2k^2))^3. We need ((k+2)/(2k^2))^3 < r + 1/(k^2 +r)^3. One can show this expression increases with r, since k^2+r = n> 1. Thus consider r=1. We can simplify the inequality to (k/2 + 1 + 1/(2k) + 1/k^2)^3 < k^6 + 3k^4 + 3k^2 + 2. Clearly this is true as we increase k. For k = 2, the inequality is true. However, for k = 1, it is false. So we must explicitly check n= 2 and 3. We get it is true for 2, but not 3. In summary, we have the solutions are those n which are not squares, but including 1, and not including 3. These cancel out. The number of squares less than or equal to 2019 is 44. The answer is then 2019-44 = 1975.
For x = n + 1/n^3 it's not hard to check that g(f(x)) = x. So the problem becomes finding integers such that f(g(n+1/n^3)) < n + 1/n^3. We can use the results that (a) floor( sqrt(n) + 1/n ) < n - 1 for all integers n except for perfect squares greater than one* and (b) {x} < 1 for all x to observe that the LHS is < n - 1 + 1 = n for all n =/= m^2. When n = m^2 there is equality. So the problem becomes counting the perfect squares between 1 and 2019, which is 44. So the answer is 2019-44 = 1975.
*n = 1 is a special case, but we can check that numerically to find LHS = 1.746... < RHS = 2
The answer is 1975.
n=1 must be examined separately, because it is the only positive integer such that n+1/n³ is an integer.
This case works.
For n>1 clearly
floor(n+1/n³)=n
{n+1/n³}=1/n³.
Therefore we can easily evaluate g(f(n+1/n³))=n+1/n³.
f(g(n+1/n³))=f(√(n)+1/n).
Since √(n) can have a fractional part we have to distinguish 2 cases.
i) n=m²: f(g(n+1/n³))=n+1/n³=g(f(n+1/n³)). Therefore this case does not work.
ii) n≠m²: there are 2 subcases:
ii-a) {√(n)}+1/n1
√(n)-floor(√(n))+1/n>1
√(n)+1/n>floor(√(n))+1
n+1/n²+2√(n)/n>(floor(√(n))+1)²
≥n+1
2*n√(n)>n²
2>√(n)
n
Correction: in the last inequality i missed a step:
n+1/n²+2√(n)/n≥n+1
(2n√(n)+1)/n²≥1
since n>1,
2n√(n)+1≥n²
2n√(n)≥n²-1
>n².
...
I'm not sure whether the pinned comment means there will still be a solution video for this or not. So I will post mine here just in case. It might be *too* long, since I'll try to be very thorough in my claims. First, we'll prove this lemma:
**Lemma:** Let n be a non-perfect-square positive integer. Let k^2 (k positive) be the largest perfect square smaller than n. Then {sqrt(n)} + 1/n < 1, for n > 4.
**Proof:** From k^2 < n < (k + 1)^2, we get k < sqrt(n) < k + 1, from which sqrt(n) = k + r, where r is a real number and 0 < r < 1. Again from k^2 < n < (k + 1)^2, we have k^2 < n < k^2 + 2k + 1, from which n = k^2 + m, where m is a positive integer and 1 ≤ m ≤ 2k < 2k + 1. Now:
n = k^2 + m =>
(k + r)^2 = k^2 + m =>
r^2 + 2kr - m = 0.
Solving that equation, we arrive at r = sqrt(k^2 + m) - k (the other solution can be ignored as it is negative). Notice that to maximize the value of r, we must maximize m, so set m = 2k. If the lemma is proven for maximum r, it is proven for all the other, smaller, values of r, given a value of n (r being a shorter notation for {sqrt(n)}). Since 1/k^2 > 1/n, it suffices to prove r + 1/k^2 < 1. Then:
sqrt(k^2 + 2k) - k + 1/k^2 < 1
sqrt(k^2 + 2k) < 1 + k - 1/k^2
k^2 + 2k < 1 + k^2 + 1/k^4 + 2k - 2/k - 2/k^2
0 < 1 + 1/k^4 - 2/k - 2/k^2
0 < k^4 - 2k^3 - 2k^2 + 1
k = 1 and k = 2 do not satisfy this inequality, but it doesn't necessarily mean they don't satisfy the lemma; they will be handled separately. For k = 3, the expression is indeed positive, and by studying f(x) = x^4 - 2x^3 - 2x^2 + 1 and its derivatives, we see that for x > 2, the function strictly increases, therefore the inequality is satisifed by all k ≥ 3. Now for k = 2, verify that sqrt(8) - 2 + 1/8 < 1, and since 8 = 2^2 + 2*2, the lemma is valid for n numbers between 2^2 and 3^3. k = 1 doesn't matter for the veracity of the lemma (since it says n > 4), but it will be useful to check either way: sqrt(2) - 1 + 1/2 < 1, but sqrt(3) - 1 + 1/3 > 1. The lemma is proven, QED.
(i'll denote [x] as the floor function of x) Now for the *actual* thing, haha. Notice that it holds for n = 1, as it results in 1 + (sqrt(2) - 1) < 2, which is valid as sqrt(2) < 2. Now let's work only with n > 1. That leads to 1/n < 1, which is is very important when computing results with the floor function, since if p is an integer and q is a real number greater than 1, we have that [p + 1/n^q] = p. Assume n = k^2 is a perfect square, then let's verify the inequality's validity:
f(g(n + 1/n^3)) < g(f(n + 1/n^3))
f([n + 1/n^3]^(1/2) + {n + 1/n^3}^(1/3)) < g([n + 1/n^3]^2 + {n + 1/n^3}^3
f(n^(1/2) + 1/n) < g(n^2 + 1/n^9)
[k + 1/n]^2 + {k + 1/n}^3 < [n^2 + 1/n^9]^(1/2) + {n^2 + 1/n^9}^(1/3)
k^2 + 1/n^3 < n + 1/n^3
Which is impossible, therefore if n is a perfect square different than 1, the inequality won't hold. Do realize that g(f(n + 1/n^3)) = n + 1/n^3 irrespective of n being a perfect square, so let us now analyze f(g(n + 1/n^3)) for n non-perfect-square:
f(g(n + 1/n^3)) =
f([n + 1/n^3]^(1/2) + {n + 1/n^3}^(1/3)) =
f(n^(1/2) + 1/n) =
[n^(1/2) + 1/n]^2 + {n^(1/2) + 1/n}^3
It is right here that we apply the lemma. For n > 4, and with k^2 being the largest perfect square smaller than n (k positive): k < n^(1/2) + 1/n = k + {n^(1/2)} + 1/n < k + 1, as such [n^(1/2) + 1/n] = k. As we saw before, this is also valid for n = 2, but *not* n = 3; that last one will be handled later. The expression is then equal to:
k^2 + ({n^(1/2)} + 1/n)^3
Since {n^(1/2)} + 1/n < 1, k^2 + ({n^(1/2)} + 1/n)^3 < k^2 + 1 ≤ n < n + 1/n^3, thus satisfying the inequality. As for the case n = 3, it is verifiable that it does not satisfy the inequality.
Thus, the numbers that *don't* satisfy the inequality are 3 and perfect squares greater than 1. There are 44 perfect squares up to 2019, as 44^2 < 2019 < 45^2. Counting 1 off and counting 3 in, we end up with 44 numbers not satisfying the condition. Therefore, the answer is 2019 - 44 = 1975.
The answer 1975.
Let l(x) denote the floor of x, r(x) denote the fractional part of x and s(x) the square root of x.
Plugging integer n > 1 into the equation simplifies the inequality to f(s(n) +n^-1) =2k+2 since m and k are integers. Since also l(s(n) +n^-1)=k^2+2k+2>(k+1)^2+q^3 and the proof is complete.
Thus the answer is 2019-l(s(2019))=2019-44=1975.
Can you post the problems whenever you are free to do so. As you go along the schedule .
Not might be weekly but however it might be. Many people would be glad to hear that.
1976
if n > 1: f(n^1/2 + 1/n) < g(n^2 + 1/n^9)
f(n^1/2 + 1/n) < n + 1/n^3
By testing values, we see this is true for 2, 5-8, 10-15, 17-24, and so on---or in other words everything except for perfect squares
There are 43 perfect squares up to 2019 (not including 1) so in total there are 2018 - 43 = 1975 for this part.
if n = 1: f(g(2)) < g(f(2))
f(√2) < g(4)
1+(√2-1)^3 < ----> This is true so 1 works.
The total is 1976 :D
whoops typo, I meant 1+(√2-1)^3 < 2
Challenge 101 should be an easy one... oh wait
The answer is 1975.
Before we simplify the inequality note that any integer > 1, floor(n + 1/(n^3)) = 0 and {n + (1/(n^3))} = n^3
We see that 1 is a problem case. Plugging in 1 in the inequality we get 1 + (0.something)^3 < 2.
This shows that 1 is one of the solutions.
Simplifying the inequality we get (floor((√n)+(1/n)))^2 + ({(√n)+(1/n)})^3 < n + 1/(n^3)
Relise that we can categorise the inequality as 2 cases:
Case 1: n is a perfect square
Simplifying the inequality we get n + 1/(n^3) < n + 1/(n^3)
This contradicts our inequality and thus leads to the point that n cannot be a perfect square(excluding 1)
Case 2: n is not a perfect square
In this case we first find all the cases that floor((√n)+(1/n)) ≠ floor((√n)).
Relise that in order to make this true we need the decimals of √n and 1/n to be big enough such that floor(√n + 1/n) > floor(√n)
For the decimals of √n to be largest we need n to be exactly 1 less than some perfect square.
By trial and error we find out that n = 3 is the only case that satisfies the condition.
Plugging in 3 we get that 4 + something < 3 + 1/27
This contradicts the inequality and thus leading to the point that 3 cannot be one of the solutions.
For the other case, we can let n = k^2 + c, where k^2 is the closest perfect square smaller than n.
Simplifying the inequality we get k^2 + (0.something)^3 = k^2 + c + 1/(k^2 + c)
This proves that all the other integer satisfies the inequality.
There are 43 perfect squares(1 excluded) for n ≤ 2019, and we also have to exclude 3 from the numbers
Therefore we get the answer of 2019 - 43 - 1 = 1975.
On the RHS. g(f(n+1/n^3)=g(n^2+1/n^6) because n is a integer, similar logic, g(n^2+1/n^6)=n+1/n^3
On the LHS, f(g(n+1/n^3))=f(sqrt(n)+1/n) because n is an integer.
First if n is a perfect square, since sqrt(n) is an integer, f(sqrt(n)+1/n)=n+1/n^3, so the inequality doesn't satisfy
Since there are 44 perfect squares, right now we have 1975 possible numbers that could work.
But, if frac(sqrt(n))+1/n > 1, LHS becomes greater than the RHS because the integer part of LHS is greater than the RHS which means LHS is greater.
For frac(sqrt(n))+1/n > 1, frac(sqrt(n)) has to be as close as a perfect square as possible so let n = k^2-1 where k is a positive integer greater than 1.
if n = k^2-1, frac(sqrt(n)) = sqrt(k^2-1) - (k-1) since k-1 is the integer part,
then the equation becomes sqrt(k^2-1)-(k-1)+1/(k^2-1)>1
isolate the radical, we get (k^2-1)sqrt(k^2-1)>k(k^2-1)-1
squaring both sides since both sides are positive gives us (k^2-1)^3>(k(k^2-1)-1)^2
expanding and isolating 0, we get k^4-2k^3-2k^2+2k+1
ans: 1975,
for all n ∈ Z+ gof will always yield the input.
for all n = d^2 form where 1 gof
only d = 2 satisfies this so,
the number of elements for which fog < gof = 2019 - 43 - 1 = 1975
Ans = 1975
Let x=n + 1/n
For n=1: x=2 then f(x)=4 then g(f(x))=2
And g(x)=sqrt(2)
then f(g(x))=1 + (sqrt(2) - 1)^3 less than 2 so 1 is from our required numbers.
For n>1: the integar part will always be n and its integar exponents while the fraction part will be 1/n with its exponents.
R.H.S:
f(x)=n^2 + 1/n^9
R.H.S=g(f(x))=n + 1/n^3
L.H.S:
g(x)=sqrt(n) + 1/n
We have two possibilities:
i) n is a perfect square. Then L.H.S=f(g(x))=n + 1/n^3=R.H.S so all perfect squares are not of our required numbers.
ii) n is not a perfect square.
Let s^2 < n < (s+1)^2
Then L.H.S=f(g(x))=s^2 + (sqrt(n)-s+ 1/n)^3 < n < R.H.S
Conclusion: We are looking for non-perfect squares from 1 to 2019
Since sqrt(2019)=44.93
Then reqired number=2019-44= 1975
The answer is 1975
We can plug in n=1,2,3,4 and check that the inequality holds for n=1, n=2, but not for n=3 or n=4
For n >=5, we will prove that the inequality holds if and only if n is not a perfect square number.
If n is a perfect square number: LHS = RHS = n+1/n^3, therefore the inequality does not hold.
If n is not a perfect square number, then we can write n= k^2+q, where k=floor(sqrt(n)) and q is an integer from 1 to 2k, also k>=2 as n >=5. We have RHS = n+1/n^3. LHS = f(sqrt(n)+1/n) .
By definition of {x} we have {x} sqrt(n) = sqrt(k^2+q) > sqrt(k^2) = k, so we have to prove that sqrt(n)+1/n < k+1, or sqrt(k^2+q)+1/(k^2+q) < k+1. As the function y= sqrt(x)+ 1/x is increasing in the interval [2,+infinity), we only need to prove this in case q=2k, i.e sqrt(k^2+2k) + 1/(k^2+2k) < k+1, squaring both sides and cancelling the terms give us 1/(k^2+2k)^2 + 2/sqrt(k^2+2k) < 1, which is true for all k>=2. Hence floor(sqrt(n)+1/n) = k
Therefore we have LHS < (floor(sqrt(n)+1/n)) ^2 +1. = k^2+1 =5. Therefore the final answer is 1975
The answer is 1975.
Let floor[x] denote the floor function on x and let fract[x] = x - floor[x].
Consider first f(g(n + 1/n^3)). By definition,
g(n + 1/n^3) = floor[n + 1/n^3]^(1/2) + fract[n + 1/n^3]^(1/3).
Since n is an integer, we can simplify this to n^(1/2) + 1/n. Now applying f, we get
f(n^(1/2) + 1/n) = floor[n^(1/2) + 1/n]^2 + fract[n^(1/2) + 1/n]^3.
Since n is an integer, we can simplify this to S(n) + 1/n^3, where S is defined as the greatest perfect square less than or equal to n.
By similar reasoning, we have g(f(n + 1/n^3)) = n + 1/n^3. So, we are looking for n such that
S(n) + 1/n^3 < n + 1/n^3 S(n) < n.
This is true as long as n isn't a perfect square. The perfect squares between 1 and 2019 inclusive are 1, 2^2, 3^2, ... , 44^2. So, there are 2019 - 44 = 1975 values of n that satisfy the problem, as claimed.
Darn, after reading the other solutions, I now see that I messed up (and got the right answer only by coincidence). It's obvious why my reasoning fails for the special case n=1, but I completely missed how f(n^(1/2)+1/n) needs to be treated more carefully for n
The answer is 1975. First of all, we can start by plug in n+1/n^3 in to both side of the inequality. Because the nature of the function. We would plug in values larger than 1 into the inequality, we can get that the right side of the inequality became g(n^2+1/n^9) because we can ignore the 1/n^3 in the floor function because it is always less than 1 and n in {x} because the functions essentially just asks for the fractional part of x. After simplifying, the left side becomes n+1/n^3, which shows that f(x) and g(x) are inverses of each other in this sense. And then we simplify the right part of the inequality.It first becomes f( sqrt(n)+1/n). For convenience, we are going to call the fractional part of sqrt(n) +1/n, M, and the integral part I. The right side of the inequality becomes I^2+M^3. We realise that it looks a lot like the left side of the inequality. The two sides are equal, if and only if n is a perfect square. The left side of the inequality is always going to be larger than the right side of the inequality because if n is not a perfect square, the floor function is going to make I^2 to be the the closest lower perfect square which would be always lesser than n because the M^2 and 1/n^3 are always less than 1, however this is not going to work if the fraction part of sqrt(n) and 1/n sum to be larger than 1. We can test the extreme case, as the ones that are 1 lower than perfect and see that 3 does not work. As the value goes sqr(n)is going to be arbitrarily close to an integer and 1/n is going to arbitrarily small, we can test the largest extreme case of 1935 as it is one below the largest perfect contained in the set, and it satisfies with inequality. So we can see every perfect square larger than 1 and 3 do not satisfy the inequality and there are 44 perfect squares less than 2019. And then we plug in 1 and see that it satisfies with the inequality. So the final answer is that 2019-43-1=1975
The easy part is that the right side, g(f(n+1/n^3)) is just n + 1/n^3. This is just an inverse (except when n is 1).
The left side doesn't work out as nicely. You can reduce to f(sqrt(n) + 1/n) (again, not counting n =1).
If n is a perfect square, then that becomes just n + 1/n^3 again. So, the sides are equal when n is a perfect square.
If n is not a perfect square, then the question becomes: Is the remainder of sqrt(n), added to 1/n, ever >=1?, for n up to 2019?
No, it's always a decimal. Thus, the floor function will always drop it (yay!).
So, for all n not a perfect square (and not 1), the left side will be less than the right.
There are 43 perfect squares between 1 and 2019
For n = 1, you can just plug it in and see the left side is less then as well, so that's good. So, 43 numbers don't work.
2019 - 43 = 1976
The answer is 1976
By substitution, 1 is one value of n that satisfy the inequality
f(g(2)) = f(sqrt2) = 5sqrt2 - 6 and g(f(2)) = g(4) = 2, and 5sqrt2 - 6
Answer: all integers except 3 and perfect squares ( from 1² to 44²).
( [..] stand for the floor function )
Well, if n>1 then [n+1/n³] = n and {n+1/n³]=1/n³ so g°f(n+1/n³) = n+1/n³ ( f and g are inverse functions from this prespective)
And
f°g(n+1/n³) = f(sqrt(n) + 1/n)
but we have 3 cases for f°g:
1- n is a perfect square and in that case f and g are totally inverse functions because [sqrt(n)+1/n]=sqrt(n) and 1/n < 1 so f°g=g°f => perfect squares are not solutions.
2- if n = x²-1 so in order to have [sqrt(n)+1/n] = x we need to find x s.t: sqrt(x²-1)+1/(x²-1)>x so
sqrt(n+1) + sqrt(n) > n but we can prove easily that for n>=8 sqrt(n+1)+sqrt(n) < 2sqrt(n+1) < n+1 because 23 hence when n= 3 , f°g>g°f so 3 is excluded as well.
3- the last case left is when n is further from the nearest greatest perfect square meaning that n=x²-a where 2x-1sqrt(x²-a)+sqrt(x²-1)
Let a and b be 2 integers >1 we have :
(b-1)*(a-1)+1>0 so ab>a+b but a+b>sqrt(a)+sqrt(b) so ab>sqrt(a)+sqrt(b) but this does not apply for x=2 and a =2 so we need to verify this case and we see obviously that sqrt(2)+1/2x
f°g>g°f if [sqrt(n)+1/n]=x because x²>x²-1+1/n³ ( 1>1/n³).
Sorry 😅 I thought we should be mentionning the numbers and not the number of numbers LOL anyways the answer is 1974.
Sir plz give combinatorics problem using recursive relationship
The answer is 1976. For.1, the inequality holds. For other n, n+n^-3 has floor equal to n. When computing the LHS and RHS, inequality doesn't hold iff n is a square number. There are 43 square numbers other than 1 up to 2019, so the answer is 2019-43=1976
I got 1975. I plugged these bits of javascript into my console. (press f12 if you're on Google Chrome, and it probably works on other browsers too!)
function g(a) {return Math.pow(Math.floor(a), 1/2)+Math.pow(a%1, 1/3)}
function f(a) {return Math.pow(Math.floor(a),2) + Math.pow(a%1, 3)}
var n = 0; for (var i = 1; i
Please put email & last Date for answers. Can you organise Playlist of Weekly Challenge from Old to New( 1,2,3 0 ... .....)
Let n be positive integers between 1 and 2019
Write m(n) = n + 1/n^3
Write statement f(g(m(n))) < g(f(m(n))) as P(n) (inspired by mathematical induction, though we are definitely not using this method)
g(f(m(n))) = g(n^2 + 1/n^9) = n+1/n^3 = m(n)
f(g(m(n))) =
f(g(2)) = f(sqrt(2)) ~= 1+(0.414)^3 ~= 1.071, if n=1
f(sqrt(n)+1/n) for all n>1
Now we know P(1) satisfies.
If n is perfect square larger than 1: f(g(m(n))) = m(n), which does not satisfy P(n)
There are floor(sqrt(2019)) = 44 perfect squares for positive integers n, but we have proved P(1) satisfies.
There are 44-1 = 43 possible values of n proved not satisfying P(n)
Now we want to know if P(n) satisfies when n is not perfect square.
From now on, denote any perfect squares larger than 1 as s.
As f(g(m(n))) = f(sqrt(n)+1/n) for all n>1
We can test behavior of this function:
y = sqrt(x) +1/x - ceil(sqrt(x))
If y(n) > 0 for particular positive integer n that is not perfect square, then P(n) does not satisfy.
y(s) = sqrt(s) + 1/s - ceil(sqrt(s)) = sqrt(s) + 1/s - sqrt(s) = 1/s, clearly P(s) does not satisfy.
What happens if x is just larger than s?
lim(x→s+)y(x)
= lim(Δx→0) [sqrt(s+Δx)+1/(s+Δx)-ceil(sqrt(s+Δx))]
= sqrt(s) + 1/s - sqrt(s) - 1
= 1/n - 1 = 1
We can still differentiate the continuous part of y function by ignoring ceil part.
y'=1/2sqrt(x) - 1/x^2
y" = 2/x^3 - 1/(4x^3/2)
Set y' = 0, x=2^k, 1+k/2 = 2k, k=2/3, x=4^(1/3)
y"(4^1/3) = 2/4 - 1/4*2 = 3/8 > 0
y' > 0 for all x >= 4^(1/3) ~=1.1587, the continuous part of the function is always increasing for all x >= 4^(1/3)
For x>0, which part of y(x) is not continuous? All positive integers x, because of ceil function.
That means there is exactly one x-intercept between each 2 perfect squares x (except 1 and 2, as lim(x→1+)y(x) = 0
As we only care about integers 1 4 > 3+1/27, P(3) does not satisfy.
y(2) = sqrt(2) +1/2 - 2 ~= -0.0858 < 0
Verify: f(g(m(2))) = f(sqrt(2) + 1/2) ~= f(1.914) < 2+1/8, P(2) satisfies.
For all n > 4 that are not perfect squares, as y(s-1) < 0, P(n) satisfies.
Finally, number of positive integers n satisfying P(n) = 2019 - 43 -1 = 2019-44+1-1 = 1975
This question is not very difficult by itself, but need a lot of careful investigation, and it is very difficult to fully explain the solution.
If I did not fully explain this, please tell me which part I have missed, I'll try to explain more.
Basic concept:
Q: Why y(s) is different with lim(x→s+)y(x)?
A: Behavior of ceil and floor function: Let i be any integer (do NOT confuse with imaginary number i)
ceil(i) = i
lim(x→i-)ceil(x) = i
lim(x→i+)ceil(x) = i+1
floor(i) = i
lim(x→i-)floor(x) = i
-1
lim(x→i+)floor(x) = i
What type of question is this???? What grade level is this???
i am really love they way u a lucturing. Mrr. Prof
I have a long way to go before I can be called a professor. =)
@@LetsSolveMathProblems i think it is on time now! haha
@@LetsSolveMathProblems I know one day you will. Math is fun as a game.
I think the answer to be 1975,
because n and n² are integers and 1/n³ is a fraction smaller than 1, the right hand side of the equation can be simplified to n + 1/n³.
The left hand side simplifies to f(sqrt(n) + 1/n), which can't be calculated easily as sqrt(n) is not necessarily an integer. We can however conclude that the left hand side will never be strictly larger than the right hand side, because 1/n³ is always smaller than 1, so the highest the left hand side can be is equal to the right hand side. This only occurs when n is a perfect square. The right hand side then also simplifies to n + 1/n³. The inequality is therefor valid for all non-perfect squares less than or equal to 2019. The answer is therefore 1975 as there are 44 perfect squares smaller than 2019.
You cant assume that, because floor (sqrt(n) +1/n) is not always equal to floor(sqrt(n)). For n=3 it is actually larger
For n=1 the equation however holds (no equality although it is a perfect square).
So your solution only randomly coincodes with the actual solution because your two mistake add up to 0
1976
On the RHS. g(f(n+1/n^3)=g(n^2+1/n^6) because n is a integer, similar logic, g(n^2+1/n^6)=n+1/n^3
On the LHS, f(g(n+1/n^3))=f(sqrt(n)+1/n) because n is an integer.
First if n is a perfect square, since sqrt(n) is an integer, f(sqrt(n)+1/n)=n+1/n^3, so the inequality doesn't satisfy
Since there are 44 perfect squares, right now we have 1975 possible numbers that could work.
But, if frac(sqrt(n))+1/n > 1, LHS becomes greater than the RHS because the integer part of LHS is greater than the RHS which means LHS is greater.
For frac(sqrt(n))+1/n > 1, frac(sqrt(n)) has to be as close as a perfect square as possible so let n = k^2-1 where k is a positive integer greater than 1.
if n = k^2-1, frac(sqrt(n)) = sqrt(k^2-1) - (k-1) since k-1 is the integer part,
then the equation becomes sqrt(k^2-1)-(k-1)+1/(k^2-1)>1
isolate the radical, we get (k^2-1)sqrt(k^2-1)>k(k^2-1)-1
squaring both sides since both sides are positive gives us (k^2-1)^3>(k(k^2-1)-1)^2
expanding and isolating 0, we get k^4-2k^3-2k^2+2k+1