Question

I have a three-digit number. the digits are all different. it is divisble by 5. it has tens digit that is less than one. the ones digit is greater than the hundreds digit. it is also divisible by 3. what number could it be?

Answer 1

The number that satisfies all the given criteria is 105. It is a three-digit number divisible by both 5 and 3, has 0 as its tens digit, and the ones digit (5) is greater than the hundreds digit (1).

Step-by-step explanation:

To find a three-digit number that meets the criteria provided, we need to consider each condition separately and use them to narrow down the possibilities:

1. The number is divisible by 5, so its ones digit must be either 0 or 5.
2. The tens digit is less than one, which means the tens digit must be 0 since it's a digit in a three-digit number and can't be negative.
3. The ones digit is greater than the hundreds digit, which gives us a clue about their relative sizes.
4. The number is also divisible by 3, meaning the sum of the digits must be divisible by 3.

Taking the first two points together, we know the ones digit must be 5, since it cannot be 0 (as the tens digit is already 0). Now we need to find a hundreds digit less than 5 (so that the ones digit is greater) that creates a number whose digits sum to a multiple of 3.

Starting with 1 as the hundreds digit, the sum of the digits would be 1 + 0 + 5 = 6, which is divisible by 3. Therefore, the number could be 105. This number is the only three-digit number that fits all the criteria given.