Question

What three-digit number can be made from the digits 2, 3, 5, and 7, such that no two digits of the three-digit number are the same and the three-digit number is divisible by each of the digits in it?

Hint:

For problems involving divisibility with multiple constraints, start with the most restrictive rules. Here, the fact that the number cannot be divisible by both 2 and 5 (as it can't end in 0) is the key to splitting the problem into manageable cases.

Answer 1

Correct Answer: 735

Step 1: Understanding the Question:
We are looking for a 3-digit number that meets three criteria:
1. It uses three different digits from the set \{2, 3, 5, 7\}.
2. There are no repeated digits.
3. The number itself must be divisible by each of the three digits it is composed of.
Step 2: Key Formula or Approach:
We can use divisibility rules to narrow down the possibilities.

Divisibility by 2: The number must end in 2.
Divisibility by 5: The number must end in 5.
Divisibility by 3: The sum of its digits must be divisible by 3. A crucial observation is that a number cannot be divisible by both 2 and 5 simultaneously, as it would have to end in 0, and 0 is not an available digit. Therefore, the set of three digits cannot contain both 2 and 5.
Step 3: Detailed Explanation:
This leaves us with two possible combinations of digits:
Case 1: Digits are \{3, 5, 7\}

Divisibility by 3: The sum of the digits is \(3+5+7=15\). Since 15 is divisible by 3, any number formed with these digits will be divisible by 3.
Divisibility by 5: The number must end in 5. This gives us two possibilities: 375 and 735.
Divisibility by 7: We now test these two numbers for divisibility by 7.
\(375 \div 7 = 53\) with a remainder of 4. (Not divisible)
\(735 \div 7 = 105\) with no remainder. (Divisible) So, 735 is a valid solution.
Case 2: Digits are \{2, 3, 7\}

Divisibility by 3: The sum of the digits is \(2+3+7=12\). Since 12 is divisible by 3, any number formed will be divisible by 3.
Divisibility by 2: The number must end in 2. This gives two possibilities: 372 and 732.
Divisibility by 7: We test these two numbers for divisibility by 7.
\(372 \div 7 = 53\) with a remainder of 1. (Not divisible)
\(732 \div 7 = 104\) with a remainder of 4. (Not divisible) There are no solutions in this case.
Step 4: Final Answer:
The only number that satisfies all the given conditions is 735.