⌊ x ⌋ + ⌊ 2 x ⌋ + ⌊ 4 x ⌋ + ⌊ 8 x ⌋ + ⌊ 1 6 x ⌋ + ⌊ 3 2 x ⌋ = 1 2 3 4 5
What is the number of integral values of n such that 1 7 2 9 n satisfies the equation above?
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I'm pointing out that the way the question is phrased, it is clear that the answer is 0. Suppose that there is a real solution x , then x + ϵ is also going to be another real solution, and hence the sum is infinity. Since the answer cannot be infinity, it must be 0.
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Any better idea?
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A slightly better way to ask is "How many integers n are there such that for x = 1 2 8 n , x satisfies the equation ...."
This way, it is less obvious that there is no solutions. If you want to get rid of the power of 2, using x = 1 0 0 0 n or x = 1 2 3 4 5 n will also work
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@Calvin Lin – As the answer won't change, I should edit question directly, right?
@Calvin Lin – Edited! Check it once and delete these comments.
Ugh! I had that exactly, but I am always hesitant to put 0 solutions. Awesome problem, Pranjal!
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Thanks! Its same as tapping "None of the other options" in MCQs!
Hey... I feel while adding fractions it should be like...
f+2f+....+6f Instead of 2f+....6f
One f is missing
Hey i didn't get the 3rd last line where u say f → 1 − Can u please explain...
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We know that f ∈ [ 0 , 1 ) . So to maximize the function, we'll substitute f just less than 1 .
how is the minimum value of the last eqn is 57? ... can u pls explain the method??
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Since f is fractional part, f < 1 . Now substitute f = 0 . 9 9 9 9 9 9 9 9 . . . and you'll get 57.
Did the exact same! Little overrated!
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I kept it at Level 3. It increased to 4 and then 5
A another way to guess : say it satisfies for one n then it would satisfy for another 54 values ( till floor(32x) increases by one) also ( making a total of 55 values) , so the answer could be either 55 if such value exist or zero
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Let integral part of x be I and fractional part be f .
⌊ x ⌋ + ⌊ 2 x ⌋ + ⌊ 4 x ⌋ + ⌊ 8 x ⌋ + ⌊ 1 6 x ⌋ + ⌊ 3 2 x ⌋ = 1 2 3 4 5 ⇒ 6 3 I + ⌊ 2 f ⌋ + ⌊ 4 f ⌋ + ⌊ 8 f ⌋ + ⌊ 1 6 f ⌋ + ⌊ 3 2 f ⌋ = 1 2 3 4 5
1 2 3 4 5 ≡ 6 0 m o d 6 3
⌊ 2 f ⌋ + ⌊ 4 f ⌋ + ⌊ 8 f ⌋ + ⌊ 1 6 f ⌋ + ⌊ 3 2 f ⌋ = 6 0
Note that maximum value of ⌊ 2 f ⌋ + ⌊ 4 f ⌋ + ⌊ 8 f ⌋ + ⌊ 1 6 f ⌋ + ⌊ 3 2 f ⌋ is 5 7 when f → 1 − .
So no solution!
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