How many 3-digit positive integers can be added to \(463\) such that no carry occurs when adding the digits?
+2668 A carry occurs when the place values have a sum \(\geq10\).
For the \(1\)'s digit, there are \(7 \) choices \((0,1,2,3,4,5,6)\)
For the \(10\)'s digit, there are \(4\) choices \((0,1,2,3)\)
For the \(100\)'s digit, there are \(5\) choices \((1,2,3,4,5)\). Remember, we can't use \(0\) for the hundred's digit, because then the number wouldn't be 3-digits.
This means that there are \(7 \times 4 \times 5\) numbers. Thus, the answer is \(\color{brown}\boxed {140}\)
