If the 6-digit number 479xyz is exactly divisible by 7, 11 and 13, then what is the value of \((y+z) \div x\)?
Answer: B
A number that is divisible by 7, 11, and 13 is also divisible by their product, which is \(7 \times 11 \times 13 = 1001\). Any 6-digit number formed by repeating a 3-digit number, i.e., of the form 'abcabc', is divisible by 1001. This is because 'abcabc' = abc × 1001. In our case, the number is 479xyz. For it to be divisible by 1001, it must be of the form 479479. By comparing, we get x=4, y=7, and z=9. Now, we calculate the required value: \((y+z) \div x = (7+9) \div 4 = 16 \div 4 = 4\).