Solving the Prime Number Product Problem

The question asks us to find the largest among four prime numbers arranged in ascending order. We are given the product of the first three numbers and the product of the last three numbers.

Setting Up the Problem

Let the four prime numbers in ascending order be \(p_1, p_2, p_3,\) and \(p_4\), such that \(p_1 < p_2 < p_3 < p_4\).

We are given two pieces of information:

• Product of the first three prime numbers: \(p_1 \times p_2 \times p_3 = 255\)
• Product of the last three prime numbers: \(p_2 \times p_3 \times p_4 = 1955\)

Our goal is to find the value of \(p_4\), the largest prime number.

Finding the First Three Prime Numbers

We know that \(p_1, p_2,\) and \(p_3\) are prime numbers and their product is 255. To find these numbers, we need to find the prime factorization of 255.

Let's find the prime factors of 255:

• 255 ends in 5, so it is divisible by 5. \(255 \div 5 = 51\).
• Now consider 51. The sum of its digits is \(5+1=6\), which is divisible by 3. So, 51 is divisible by 3. \(51 \div 3 = 17\).
• 17 is a prime number.

So, the prime factorization of 255 is \(3 \times 5 \times 17\).

Since \(p_1, p_2,\) and \(p_3\) are prime numbers in ascending order whose product is 255, they must be the prime factors of 255 arranged in ascending order.

Therefore, \(p_1 = 3\), \(p_2 = 5\), and \(p_3 = 17\). These are indeed prime numbers and are in ascending order \(3 < 5 < 17\).

Finding the Largest Prime Number

Now we use the second piece of information: the product of the last three prime numbers is 1955.

\(p_2 \times p_3 \times p_4 = 1955\)

Substitute the values we found for \(p_2\) and \(p_3\):

\(5 \times 17 \times p_4 = 1955\)

\(85 \times p_4 = 1955\)

To find \(p_4\), we need to divide 1955 by 85:

\(p_4 = \frac{1955}{85}\)

Let's perform the division:

\(1955 \div 85\)

We can try multiplying 85 by numbers to get close to 1955.

• \(85 \times 10 = 850\)
• \(85 \times 20 = 1700\)
• The remainder is \(1955 - 1700 = 255\).
• Now divide 255 by 85. We found earlier that \(255 = 3 \times 85\).
• So, \(1955 = 85 \times 20 + 85 \times 3 = 85 \times (20 + 3) = 85 \times 23\).

Therefore, \(p_4 = 23\).

Verifying the Solution

The four prime numbers in ascending order are \(p_1 = 3, p_2 = 5, p_3 = 17,\) and \(p_4 = 23\). Let's check if they meet the conditions:

• Are they prime? Yes: 3, 5, 17, 23 are all prime numbers.
• Are they in ascending order? Yes: \(3 < 5 < 17 < 23\).
• Is the product of the first three 255? \(3 \times 5 \times 17 = 15 \times 17 = 255\). Yes.
• Is the product of the last three 1955? \(5 \times 17 \times 23 = 85 \times 23 = 1955\). Yes.

All conditions are satisfied. The largest prime number is \(p_4 = 23\).

Summary of Prime Numbers Found

Revision Table: Key Concepts

Additional Information: Working with Prime Factors

Problems involving products of numbers, especially prime numbers, often rely on the unique prime factorization theorem. This theorem states that every integer greater than 1 can be represented uniquely as a product of prime numbers, up to the order of the factors.

In this prime number problem, the fact that the numbers are prime and in ascending order is crucial. When we found the prime factors of 255 (\(3, 5, 17\)), the ascending order immediately told us which prime factor corresponds to \(p_1, p_2,\) and \(p_3\).

If the numbers were not required to be in ascending order, there would be multiple possibilities for \(p_1, p_2,\) and \(p_3\) using the factors 3, 5, and 17. However, the ascending order constraint simplifies the problem significantly, allowing us to uniquely identify the first three prime numbers.

Similarly, the second equation \(p_2 \times p_3 \times p_4 = 1955\) allowed us to find \(p_4\) because we had already identified \(p_2\) and \(p_3\). Dividing the second product by the first product (after removing common terms) is another way to approach this type of problem:

\[ \frac{p_2 \times p_3 \times p_4}{p_1 \times p_2 \times p_3} = \frac{1955}{255} \]

Simplifying the left side gives \(\frac{p_4}{p_1}\). So, \(\frac{p_4}{p_1} = \frac{1955}{255}\).

We can simplify the fraction \(\frac{1955}{255}\) by dividing both numerator and denominator by their greatest common divisor. Notice that \(255 = 5 \times 51\) and \(1955 = 5 \times 391\). Their common factor is 5.

\[ \frac{1955}{255} = \frac{5 \times 391}{5 \times 51} = \frac{391}{51} \]

We know \(p_1 = 3\). So, \(\frac{p_4}{3} = \frac{391}{51}\). This means \(p_4 = 3 \times \frac{391}{51}\). Since \(51 = 3 \times 17\), we have:

\[ p_4 = 3 \times \frac{391}{3 \times 17} = \frac{391}{17} \]

Now we divide 391 by 17:

• \(17 \times 10 = 170\)
• \(17 \times 20 = 340\)
• Remainder is \(391 - 340 = 51\).
• \(17 \times 3 = 51\).
• So, \(391 = 17 \times 20 + 17 \times 3 = 17 \times (20 + 3) = 17 \times 23\).

Thus, \(p_4 = 23\). This method gives the same result and highlights the relationship between the two products and the outer prime numbers \(p_1\) and \(p_4\).