The set of positive integers less than 50 forming the equivalence class of 6 modulo 9 is given by :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → [6]= {6, 15, 24, 33, 42}
Using conditions mentioned above,
$x≡a(mod\,n)$
$⇒x≡6(mod\,9)$
$⇒x=6+9k$
for $k=0$, $6+9(0)=6$
for $k=1$, $6+9(1)=15$
for $k=2$, $6+9(2)=24$
for $k=3$, $6+9(3)=33$
for $k=4$, $6+9(4)=42$
{6, 15, 24, 33, 42}