# Divisibility

• Mar 1st 2010, 10:13 AM
novice
Divisibility
Since it is commonly known that if an integer \$\displaystyle x\$ is not divisible by 3, then \$\displaystyle x = 3q + 1\$ or \$\displaystyle x = 3q + 2\$, does it mean that if an integer \$\displaystyle y\$ is not divisible by 11, then \$\displaystyle y = 11q + 1\$ or \$\displaystyle y = 11q+2\$ or \$\displaystyle y=11q + 3 \$or …or \$\displaystyle y = 11q+10 \$ for some integer \$\displaystyle q\$?
• Mar 1st 2010, 11:55 AM
hatsoff
Quote:

Originally Posted by novice
Since it is commonly known that if an integer \$\displaystyle x\$ is not divisible by 3, then \$\displaystyle x = 3q + 1\$ or \$\displaystyle x = 3q + 2\$, does it mean that if an integer \$\displaystyle y\$ is not divisible by 11, then \$\displaystyle y = 11q + 1\$ or \$\displaystyle y = 11q+2\$ or \$\displaystyle y=11q + 3 \$or …or \$\displaystyle y = 11q+10 \$ for some integer \$\displaystyle q\$?

Yes. You may read about the division algorithm to understand why this must be the case.