Finding 3-Digit Numbers With A Digit Product Of 36

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Embark on a mathematical journey to discover the fascinating world of 3-digit numbers. In this exploration, we aim to unravel a specific numerical puzzle: how many 3-digit positive numbers exist where the product of their individual digits equals 36? This seemingly simple question opens the door to a realm of number theory, combinatorics, and logical reasoning. Let's delve into the intricacies of this problem and systematically uncover the solution.

Decoding the Digit Product: A Foundation for Discovery

To effectively tackle this problem, we must first understand the fundamental concept of a digit product. The digit product of a number is simply the result of multiplying all of its digits together. For instance, the digit product of 234 is 2 * 3 * 4 = 24. In our quest, we are seeking 3-digit numbers where this product culminates in the specific value of 36.

Now, let's break down the number 36 into its prime factors. This decomposition will be instrumental in identifying the possible digit combinations that satisfy our condition. The prime factorization of 36 is 2 * 2 * 3 * 3, or 2² * 3². This reveals that the digits of our target numbers must be composed of the factors 2 and 3, along with the possibility of the digit 1, which acts as a neutral element in multiplication.

Considering this prime factorization, we can begin to brainstorm the possible sets of three digits that multiply to 36. We must keep in mind that each digit must be a single-digit number (0 through 9). The systematic exploration of these combinations is the key to solving our puzzle. By carefully examining the factors of 36, we can construct a comprehensive list of digit triplets that meet our criteria.

Unearthing the Digit Combinations: A Systematic Approach

Our objective is to identify all possible combinations of three digits whose product equals 36. We can approach this systematically, ensuring that we don't miss any valid sets. Let's begin by considering the largest possible digit, 9, and work our way down:

  • Starting with 9: If one digit is 9, the other two digits must multiply to 4 (36 / 9 = 4). The possible pairs are (1, 4) and (2, 2). This gives us two sets: (9, 1, 4) and (9, 2, 2).

  • Considering 6: If one digit is 6, the other two digits must multiply to 6 (36 / 6 = 6). The possible pairs are (1, 6) and (2, 3). Since we already have 6 as a digit, the only new set we get is (6, 2, 3).

  • Examining 4: If one digit is 4, the other two digits must multiply to 9 (36 / 4 = 9). The only possible pair is (1, 9), which we already have, and (3,3), which gives us the set (4,3,3).

  • Checking 3: If one digit is 3, the other two digits must multiply to 12 (36 / 3 = 12). The possible pairs are (3, 4), which we already have, and no other combination of single digits will reach 12.

  • Investigating smaller digits: We don't need to consider digits smaller than 3, as the product of the remaining two digits would need to be greater than 12, and we've already explored the possible combinations. So, (2,x,y) would require x*y = 18, and we have covered (2,9,1) already.

Therefore, we have identified the following unique sets of digits that multiply to 36: (9, 1, 4), (9, 2, 2), (6, 2, 3), and (4, 3, 3). These are the fundamental building blocks for constructing our 3-digit numbers.

Constructing the Numbers: Permutations and Possibilities

Now that we have the digit sets, the next step is to determine how many distinct 3-digit numbers can be formed from each set. This involves the concept of permutations, which deals with the arrangement of objects in a specific order.

Let's analyze each set:

  • (9, 1, 4): These three distinct digits can be arranged in 3! (3 factorial) ways, which is 3 * 2 * 1 = 6 permutations. These are 914, 941, 149, 194, 419, and 491.
  • (9, 2, 2): This set has a repeated digit (2), so we need to account for the overcounting. The number of permutations is 3! / 2! = 3. These are 922, 292, and 229.
  • (6, 2, 3): Similar to the first set, these three distinct digits can be arranged in 3! = 6 permutations. These are 623, 632, 236, 263, 326, and 362.
  • (4, 3, 3): This set also has a repeated digit (3), so the number of permutations is 3! / 2! = 3. These are 433, 343, and 334.

By calculating the permutations for each set, we are determining the number of unique 3-digit numbers that can be formed using those specific digits.

The Final Count: Summing Up the Solutions

To arrive at the final answer, we simply need to sum the number of permutations for each of our digit sets. We found 6 permutations for (9, 1, 4), 3 permutations for (9, 2, 2), 6 permutations for (6, 2, 3), and 3 permutations for (4, 3, 3).

Adding these together, we get 6 + 3 + 6 + 3 = 18. Therefore, there are a total of 18 three-digit positive numbers whose digits have a product of 36.

Conclusion: A Journey Through Number Theory

In this exploration, we successfully determined that there are 18 three-digit positive numbers where the product of their digits equals 36. We achieved this by systematically breaking down the problem into smaller parts:

  1. Understanding the concept of digit products.
  2. Prime factorizing 36 to identify possible digit combinations.
  3. Listing all unique sets of digits that multiply to 36.
  4. Calculating the permutations for each set to account for different arrangements.
  5. Summing the permutations to arrive at the final answer.

This problem showcases the beauty of number theory and how seemingly simple questions can lead to fascinating mathematical explorations. By employing logical reasoning, combinatorics, and a systematic approach, we were able to unravel the solution and gain a deeper appreciation for the intricacies of numbers.

This exercise not only provides a solution to a specific mathematical puzzle but also illustrates the power of structured problem-solving in mathematics and beyond. The ability to break down complex problems into smaller, manageable steps is a valuable skill that can be applied in various domains.