3. (6 points) If the natural numbers $a$, $b$, $c$ when divided by 14 all leave a remainder of 5, then $a+b+c$ when divided by 14, the remainder obtained is
【Solution】Solution: Let $a=14 x+5, b=14 y+5, c=14 z+5$, so $a+b+c=14(x+y+z)+15$ $[14(x+y+z)+15] \div 14$ $=x+y+z+1 \cdots 1$, therefore the answer is: 1 .