Solving the Prime Factorization Problem

The question asks us to find the value of \(ab - cd\) from the equation \(847 \times 385 \times 675 \times 3025 = 3^a \times 5^b \times 7^c \times 11^d\). To solve this, we need to find the prime factorization of the product on the left side of the equation and then compare the exponents with the right side.

Prime Factorization of Each Number

Let's find the prime factors for each number in the product:

• Prime Factorization of 847:

  We start dividing 847 by prime numbers. It's not divisible by 2, 3, or 5.

  \(847 \div 7 = 121\)

  We know that \(121 = 11 \times 11 = 11^2\).

  So, \(847 = 7 \times 11^2\).

• Prime Factorization of 385:

  385 ends in 5, so it's divisible by 5.

  \(385 \div 5 = 77\)

  We know that \(77 = 7 \times 11\).

  So, \(385 = 5 \times 7 \times 11\).

• Prime Factorization of 675:

  675 ends in 5, so it's divisible by 5.

  \(675 \div 5 = 135\)

  135 ends in 5, so it's divisible by 5.

  \(135 \div 5 = 27\)

  We know that \(27 = 3 \times 3 \times 3 = 3^3\).

  So, \(675 = 3^3 \times 5^2\).

• Prime Factorization of 3025:

  3025 ends in 5, so it's divisible by 5.

  \(3025 \div 5 = 605\)

  605 ends in 5, so it's divisible by 5.

  \(605 \div 5 = 121\)

  We know that \(121 = 11 \times 11 = 11^2\).

  So, \(3025 = 5^2 \times 11^2\).

Combining Prime Factors and Exponents

Now, let's multiply the prime factorizations together:

\(847 \times 385 \times 675 \times 3025 = (7 \times 11^2) \times (5 \times 7 \times 11) \times (3^3 \times 5^2) \times (5^2 \times 11^2)\)

To find the exponents for each prime factor in the product, we add the exponents of the same prime factors from each number:

• For prime 3: The only factor of 3 comes from 675, which is \(3^3\). So the exponent of 3 is 3.
• For prime 5: We have \(5^1\) from 385, \(5^2\) from 675, and \(5^2\) from 3025. The total exponent of 5 is \(1 + 2 + 2 = 5\). So we have \(5^5\).
• For prime 7: We have \(7^1\) from 847 and \(7^1\) from 385. The total exponent of 7 is \(1 + 1 = 2\). So we have \(7^2\).
• For prime 11: We have \(11^2\) from 847, \(11^1\) from 385, and \(11^2\) from 3025. The total exponent of 11 is \(2 + 1 + 2 = 5\). So we have \(11^5\).

The prime factorization of the product is \(3^3 \times 5^5 \times 7^2 \times 11^5\).

Identifying the Values of a, b, c, and d

The problem states that the product is equal to \(3^a \times 5^b \times 7^c \times 11^d\).

By comparing our result with the given form, we can identify the values of \(a\), \(b\), \(c\), and \(d\):

• The exponent of 3 is \(a\), so \(a = 3\).
• The exponent of 5 is \(b\), so \(b = 5\).
• The exponent of 7 is \(c\), so \(c = 2\).
• The exponent of 11 is \(d\), so \(d = 5\).

Calculating ab - cd

Now we need to find the value of \(ab - cd\) using the values we found for \(a\), \(b\), \(c\), and \(d\).

• First, calculate \(ab\): \(ab = a \times b = 3 \times 5 = 15\).
• Next, calculate \(cd\): \(cd = c \times d = 2 \times 5 = 10\).
• Finally, calculate \(ab - cd\): \(ab - cd = 15 - 10 = 5\).

The value of \(ab - cd\) is 5.

Revision Table: Key Values

Additional Information: Understanding Prime Factorization

Prime factorization is the process of breaking down a composite number into a product of its prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself (examples: 2, 3, 5, 7, 11, etc.). Every composite number has a unique prime factorization, according to the Fundamental Theorem of Arithmetic.

When you multiply numbers together, their prime factorizations combine. The exponent of each prime factor in the product is the sum of its exponents in the factorizations of the individual numbers. This concept was key to solving this problem, allowing us to find the exponents \(a\), \(b\), \(c\), and \(d\) by combining the exponents of 3, 5, 7, and 11 from the factorizations of 847, 385, 675, and 3025.