The numbers 120 and 126, written as the products of their prime factors, are:

- 120 = [tex]2^3 \times 3 \times 5[/tex]
- 126 = [tex]2 \times 3^2 \times 7[/tex]

Find the smallest whole number that is divisible by both 120 and 126.

To find the smallest whole number that is divisible by both 120 and 126, we need to find their common factors. The prime factorization of 120 is 2 x 2 x 2 x 3 x 5, and the prime factorization of 126 is 2 x 3 x 3 x 7.

To find the common factors, we take the highest power of each prime factor that appears in both factorizations, which gives us 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2³ x 3² x 5 x 7. Therefore, the smallest whole number that is divisible by both 120 and 126 is 2³ x 3² x 5 x 7 = 5040.

Learn more about the topic of prime factorization here: brainly.com/question/29775157

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The numbers 120 and 126, written as the products of their prime factors, are:- 120 = [tex]2^3 \times 3 \times 5[/tex]- 126 = [tex]2 \times 3^2 \times 7[/tex]Find the smallest whole number that is divisible by both 120 and 126.

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The numbers 120 and 126, written as the products of their prime factors, are:

- 120 = [tex]2^3 \times 3 \times 5[/tex]
- 126 = [tex]2 \times 3^2 \times 7[/tex]

Find the smallest whole number that is divisible by both 120 and 126.