Floor Functions

Algebra Level 5

x + 2 x + 4 x + 8 x + 16 x + 32 x = 12345 \large \lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 8x\rfloor+\lfloor 16x\rfloor+\lfloor 32x\rfloor=12345

What is the number of integral values of n n such that n 1729 \dfrac{n}{1729} satisfies the equation above?

Details and assumptions

  • x \lfloor x\rfloor represents the greatest integer function.


The answer is 0.

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2 solutions

Pranjal Jain
Jan 15, 2015

Let integral part of x x be I I and fractional part be f f .

x + 2 x + 4 x + 8 x + 16 x + 32 x = 12345 \lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 4x\rfloor+\lfloor 8x\rfloor+\lfloor 16x\rfloor+\lfloor 32x\rfloor=12345 63 I + 2 f + 4 f + 8 f + 16 f + 32 f = 12345 \Rightarrow 63I+\lfloor 2f\rfloor+\lfloor 4f\rfloor+\lfloor 8f\rfloor+\lfloor 16f\rfloor+\lfloor 32f\rfloor=12345

12345 60 m o d 63 12345\equiv 60\mod 63

2 f + 4 f + 8 f + 16 f + 32 f = 60 \lfloor 2f\rfloor+\lfloor 4f\rfloor+\lfloor 8f\rfloor+\lfloor 16f\rfloor+\lfloor 32f\rfloor=60

Note that maximum value of 2 f + 4 f + 8 f + 16 f + 32 f \lfloor 2f\rfloor+\lfloor 4f\rfloor+\lfloor 8f\rfloor+\lfloor 16f\rfloor+\lfloor 32f\rfloor is 57 57 when f 1 f\rightarrow 1^{-} .

So no solution!

Properties used

  • x + n = n + x \lfloor x+n\rfloor=n+\lfloor x\rfloor if n is integer.

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 x , then x + ϵ x + \epsilon is also going to be another real solution, and hence the sum is infinity. Since the answer cannot be infinity, it must be 0.

Calvin Lin Staff - 6 years, 4 months ago

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Any better idea?

Pranjal Jain - 6 years, 4 months ago

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A slightly better way to ask is "How many integers n n are there such that for x = n 128 x = \frac{ n}{ 128} , x 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 = n 1000 x = \frac{n}{1000} or x = n 12345 x = \frac{n}{12345} will also work

Calvin Lin Staff - 6 years, 4 months ago

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@Calvin Lin As the answer won't change, I should edit question directly, right?

Pranjal Jain - 6 years, 4 months ago

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@Pranjal Jain Yup, go ahead.

Calvin Lin Staff - 6 years, 4 months ago

@Calvin Lin Edited! Check it once and delete these comments.

Pranjal Jain - 6 years, 4 months ago

Ugh! I had that exactly, but I am always hesitant to put 0 solutions. Awesome problem, Pranjal!

Ryan Tamburrino - 6 years, 4 months ago

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Thanks! Its same as tapping "None of the other options" in MCQs!

Pranjal Jain - 6 years, 4 months ago

Hey... I feel while adding fractions it should be like...

f+2f+....+6f Instead of 2f+....6f

One f is missing

Mohit Vaidh - 6 years, 4 months ago

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f \lfloor f\rfloor =0

Pranjal Jain - 6 years, 4 months ago

Hey i didn't get the 3rd last line where u say f 1 f \rightarrow 1^- Can u please explain...

abc xyz - 5 years, 3 months ago

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We know that f [ 0 , 1 ) f\in [0,1) . So to maximize the function, we'll substitute f f just less than 1 1 .

Pranjal Jain - 5 years, 2 months ago

how is the minimum value of the last eqn is 57? ... can u pls explain the method??

Ganesh Ayyappan - 5 years, 5 months ago

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Since f f is fractional part, f < 1 f<1 . Now substitute f = 0.99999999... f=0.99999999... and you'll get 57.

Pranjal Jain - 5 years, 4 months ago

Did the exact same! Little overrated!

Kartik Sharma - 6 years, 4 months ago

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I kept it at Level 3. It increased to 4 and then 5

Pranjal Jain - 6 years, 4 months ago
Mrigank Krishan
Mar 29, 2016

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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