This question is based on the five three-digit numbers given below. (Left) 471…

Concept: Two whole numbers that have the same count of digits are compared place by place from the left: the hundreds digit decides first, and only when the hundreds digits are equal does the tens digit decide, and after that the units digit. The position of a digit — first, second or third — is fixed by the place it occupies counting from the left, and does not depend on how large that digit is.

Application:

1. The five numbers are 471, 496, 687, 492 and 678, and their hundreds digits are 4, 4, 6, 4 and 6.

2. A hundreds digit of 6 outranks a hundreds digit of 4, so the highest number is one of 687 and 678, while the lowest number is one of 471, 496 and 492.

3. Between 687 and 678 the hundreds digits are equal, so the tens digits decide: 8 is greater than 7, hence the highest number is 687.

4. Among 471, 496 and 492 the hundreds digits are equal as well, and their tens digits are 7, 9 and 9, so the lowest number is 471.

5. The second digit of the highest number 687 is 8, and the second digit of the lowest number 471 is 7.

6. Working from left to right as the note requires, 8 + 7 = 15.

Cross-check: Placing all five numbers in ascending order gives 471 < 492 < 496 < 678 < 687, which confirms 471 at the bottom and 687 at the top. The table below lists the three digits of each number.

Reading the second-digit column for 687 and for 471 gives 8 and 7, whose sum is 15.