Just Some Fractional Parts
{ x } + { x 1 } = 2 0 1 6 1
How many positive rational solutions are there to the equation above?
Notation : { ⋅ } denotes the fractional part function .
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1 solution
What do you exactly say the solutions are sir.?
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Can you be a bit more specific ? (And no need for sir :))
@Rishabh Tiwari The only solutions are x = 2 0 1 6 , 2 0 1 6 1 .
If you'd like more explanation from @UTKARSH DWIVEDI , you should mention which part of the solution is confusing to you. E.g., where did you follow until, and which steps became unclear?
Plz explain it further.
@UTKARSH DWIVEDI can you explain why the HCF was 1?
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Hi, Anik, I think you mean - "Why H C F ( [ b ( a y + b ) + a ] , ( a y + b ) ) = 1 ?". Well, I already mentioned a a y + b is in its simplest form so a y + b and a are coprime and a < a y + b so a y + b in any case can't divide [ b ( a y + b ) + a ] .
Relevant wiki: Floor and Ceiling Functions - Problem Solving
First, we must acknowledge that if a positive rational number x satisfies this so should x 1 . And so we proceed as :
As x is positive and rational, suppose : x = a a y + b ; 0 ≤ b < a where x > 1 and is in its simplest form.
{ a a y + b } + { a y + b a } = 2 0 1 6 1 ⇒ a b + a y + b a = 2 0 1 6 1
⇒ 2 0 1 6 [ b ( a y + b ) + a ] = a ( a y + b )
If b = 0 , clearly, a y + b does not divide [ b ( a y + b ) + a ] ⇒ ( a y + b ) ∣ 2 0 1 6 because H C F ( [ b ( a y + b ) + a ] , ( a y + b ) ) = 1 (based on assumptions)
& as a does not divide b ( a y + b ) so it does not divide [ b ( a y + b ) + a ] ⇒ a ∣ 2 0 1 6
So H C F ( a , ( a y + b ) ) = 1 → Not Possible.
Hence , b = 0 then x = 2 0 1 6 or its reciprocal x = 2 0 1 6 1 (in which case x < 1 )