Let $x ∈ R$, then $[\frac{x}{3}]+[\frac{x+1}{3}]+[\frac{x+2}{3}]$, where [.] denotes the greatest integer function, is equal to
Answer & explanation
Correct answer: option 1
Let $x = 3m+k$, where $m ∈ Z$ and $0 ≤ k <1$. Then,
$[\frac{x}{3}]+[\frac{x+1}{3}]+[\frac{x+2}{3}]=m+m+m=3m=[x]$
Let $x=3m+1+k$, where $m ∈ Z$ and $0 ≤k <1$. Then,
$[\frac{x}{3}]+[\frac{x+1}{3}]+[\frac{x+2}{3}]=m+m+m+1=3m+1=[x]$
Let $x = 3m+2$, where $m ∈ Z$ and $0 ≤ k < 1$. Then,
$[\frac{x}{3}]+[\frac{x+1}{3}]+[\frac{x+2}{3}]=m+m+1+m+1=3m+2=[x]$