Prime Factorization in Maths
Grade 7 Questions With Detailed Solutions

Practice Questions

1. Which of the following is NOT a prime factorization?
   1. $20 = 2 \times 10$
   2. $14 = 2 \times 7$
   3. $64 = 4^3$
   4. $120 = 2^3 \times 15$
   View Solution

   Answers: a, c, and d are NOT prime factorizations.

   Prime factorization must involve only prime numbers.

   • a) $20 = 2 \times 10$: The number $10$ is composite, not prime.
   • c) $64 = 4^3$: The base number $4$ is composite. (The correct prime factorization is $2^6$).
   • d) $120 = 2^3 \times 15$: The number $15$ is composite. (The correct prime factorization is $2^3 \times 3 \times 5$).
2. What is the prime factorization of the following numbers?
   1. $28$
   2. $32$
   3. $100$
   4. $126$
   5. $900$
   View Solution

   Divide each number successively by prime numbers until the final quotient is prime:

   1. $28 = 2 \times 2 \times 7 = \mathbf{2^2 \times 7}$
   2. $32 = 2 \times 2 \times 2 \times 2 \times 2 = \mathbf{2^5}$
   3. $100 = 2 \times 2 \times 5 \times 5 = \mathbf{2^2 \times 5^2}$
   4. $126 = 2 \times 3 \times 3 \times 7 = \mathbf{2 \times 3^2 \times 7}$
   5. $900 = 2 \times 2 \times 3 \times 3 \times 5 \times 5 = \mathbf{2^2 \times 3^2 \times 5^2}$
3. Two numbers, A and B, are given by:
   $A = 2^3 \times 5^2 \times 11$ and $B = 2^2 \times 5 \times 13$.
   What is the prime factorization of $A \times B$?
   View Solution

   Multiply the two prime factorizations together:

   $A \times B = (2^3 \times 5^2 \times 11) \times (2^2 \times 5 \times 13)$

   Group the bases that are the same and add their exponents:

   $A \times B = (2^3 \times 2^2) \times (5^2 \times 5^1) \times 11 \times 13$

   $A \times B = \mathbf{2^5 \times 5^3 \times 11 \times 13}$

4. Find the prime factorization of $100$ and $70$. Then, use those results to find the prime factorization of $7000$, knowing that $7000 = 100 \times 70$.
   View Solution

   First, find the individual prime factorizations:

   $100 = 2^2 \times 5^2$

   $70 = 2 \times 5 \times 7$

   Use the fact that $7000 = 100 \times 70$ to combine the factors:

   $7000 = (2^2 \times 5^2) \times (2^1 \times 5^1 \times 7^1)$

   Combine like bases by adding exponents:

   $7000 = \mathbf{2^3 \times 5^3 \times 7}$

5. Discovering Patterns in Powers of 10:
   1. Find the prime factorization of $10$.
   2. Use the result in part (a) and the fact that $100 = 10 \times 10$ to find the prime factorization of $100$.
   3. Use the result in part (a) and the fact that $1000 = 10 \times 10 \times 10$ to find the prime factorization of $1000$.
   4. Use the results above to identify a pattern and find the prime factorization of $1,000,000$.
   View Solution

   1. $10 = \mathbf{2 \times 5}$
   2. $100 = 10 \times 10 = 10^2 = (2 \times 5) \times (2 \times 5) = \mathbf{2^2 \times 5^2}$
   3. $1000 = 10 \times 10 \times 10 = 10^3 = (2 \times 5) \times (2 \times 5) \times (2 \times 5) = \mathbf{2^3 \times 5^3}$
   4. Following the pattern established above, $10^n = 2^n \times 5^n$. Therefore, for one million:
      $1,000,000 = 10^6 = (2 \times 5)^6 = \mathbf{2^6 \times 5^6}$

More References and Links

• Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
• High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers
• Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers