Fractional sum

Algebra Level 4

x 3 − ⌊ x ⌋ = 3 , x 6 = ? \large x^3-\lfloor x\rfloor = 3 \qquad,\qquad x^6 = \ ?


The answer is 16.

Atul Shivam by Atul Shivam , 23, India

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3 solutions

Aditya Garg
Oct 3, 2015

[x] will be 1 because cube of 2 will be grater than 8 so, pow(x,3) - 1 = 3; pow(x,3) = 4; so pow(x,6) = 16

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I f x < 1 , x 3 − 0 < 1 , b u t w e h a v e = 3. S o x > 1. F o r 1 ≤ x < 2 , x 3 − 1 = 3 , a n d x 3 = 4. ∴ x 6 = 16. F o r 2 ≤ x < 3 x 3 − 2 = 3 , a n d x 3 = 5. B u t e v e n x = 2 , x 3 = 8 n o t 5. S o x ≥ 2 i s n o t p o s s i b l e . S o x 6 = 4 2 = 16. If~x<1, ~~~x^3-0<1,~ but~we~have~=3.~So~x>1.\\ For~~~1\leq x<2,~x^3-1=3, ~and~x^3=4.~\therefore~x^6=16.\\ For~~~2\leq x <3~x^3-2=3,~and~x^3=5.~But~even~x=2,~x^3=8~not~5.\\ So~~~x \geq 2~is~not~possible.\\ So~~~x^6=4^2=\color{#D61F06}{16}.

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Aaghaz Mahajan
Jun 4, 2018

Well, we can see that x lies between 1 and 2............Taking n as the fractional part of x, we solve and get x equal to the cuberoot of 4........!!!

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