Split 2016

Algebra Level 5

201 { x } + 6 x = 2016 \large 201\{x\}+6 \lfloor x \rfloor=2016

How many non-integral real numbers x x satisfy the above equation?

Notations :


The answer is 33.

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1 solution

Chan Lye Lee
May 14, 2016

Relevant wiki: Floor and Ceiling Functions - Problem Solving

Let x = A + k 201 x = A + \frac{k}{201} where k , A Z k,A \in \mathbb{Z} and 1 k 200 1\le k \le 200 .

Then the equation is equivalent to k + 6 A = 2016 k+6A=2016 . Note that ( k , A ) = ( 6 , 335 ) (k,A)=(6,335) is a solution. Hence ( k , A ) = ( 6 + 6 m , 335 m ) (k,A)=(6+6m, 335-m) , where 1 6 + 6 m 200 1\le 6+6m \le 200 , are solutions. So 0 m 32 0 \le m \le 32 and thus there are 33 \boxed{33} solutions.

Why can't x be an integer? I think you have missed a solution of x=336.

Kushagra Sahni - 5 years, 1 month ago

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Thanks for pointing out. I edited the question.

Chan Lye Lee - 5 years, 1 month ago

I think it's 33 + 1. x = 336 is also a solution? So it must be 34.

Siva Bathula - 5 years, 1 month ago

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You are right. I missed it. The question is edited. Thanks.

Chan Lye Lee - 5 years, 1 month ago

nice question+solution.. +1

Sabhrant Sachan - 5 years, 1 month ago

great ques had to think, solved after 1 failed attempt liked it (+1)

A Former Brilliant Member - 4 years, 9 months ago

Nice solution

Vaibhav Gupta - 2 years, 9 months ago

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