Final answer:
The second number is determined to be 369 by considering the conditions that it has to be a three times multiple of the first number, both have to be 3-digit numbers, and all digits from 1 to 9 are used once.
Step-by-step explanation:
To solve this problem, we need to establish three 3-digit numbers such that the second is three times the first, and the third is five times the first, using the digits 1-9 exactly once. Let's assume the first number is abc where a, b, and c are distinct digits. Then, the second number would be 3abc and need to also be a 3-digit number. Importantly, since we're dealing with 3-digit numbers, 'a' can't be zero and hence must be at least 1.
Given that all three numbers need to be distinct, we should consider the largest possible 3-digit number made with unique digits - 987. In this scenario, the first number cannot be larger than 329 since 3 times 330 equals 990, which repeats the digit '9' and is out of the range for our second number.
So, we know the first number must be smaller than 330, and the second number must fit within the 900s range. The only number that fulfills this criterion, given that the second number is 3 times the first and we cannot repeat digits, would be starting with the digit '1' for the first number. The only digits left for the second number (after being multiplied by 3) would be 6, 3, and 9 respectively, since they do not repeat any digit and will result in a second number in the 900s range.
Therefore, the first number is 123, the second number, which is three times the first number, is 369, and the third number, which is five times the first number, is 615. So the second number is 369.