find the HCF of 42, 63, and 126 using prime factorization method
Answers
Step-by-step explanation:
Factor each number into its prime factors:
Factor each number into its prime factors:42 = 2*3*7
Factor each number into its prime factors:42 = 2*3*763=3*3*7
Factor each number into its prime factors:42 = 2*3*763=3*3*7126=2*3*3*7
Factor each number into its prime factors:42 = 2*3*763=3*3*7126=2*3*3*7Multiply together the largest common factors to get the HCF/GCD. 2 does not appear in all 3 numbers, so it is discarded. 3 appears in all of them, but only once in all of them, so we only use it once. 7 appears in all of them, again only once. Our HCF/GCD is therefore 3*7 = 21.
Factor each number into its prime factors:42 = 2*3*763=3*3*7126=2*3*3*7Multiply together the largest common factors to get the HCF/GCD. 2 does not appear in all 3 numbers, so it is discarded. 3 appears in all of them, but only once in all of them, so we only use it once. 7 appears in all of them, again only once. Our HCF/GCD is therefore 3*7 = 21.42/21 = 2
Factor each number into its prime factors:42 = 2*3*763=3*3*7126=2*3*3*7Multiply together the largest common factors to get the HCF/GCD. 2 does not appear in all 3 numbers, so it is discarded. 3 appears in all of them, but only once in all of them, so we only use it once. 7 appears in all of them, again only once. Our HCF/GCD is therefore 3*7 = 21.42/21 = 263/21 = 3
Factor each number into its prime factors:42 = 2*3*763=3*3*7126=2*3*3*7Multiply together the largest common factors to get the HCF/GCD. 2 does not appear in all 3 numbers, so it is discarded. 3 appears in all of them, but only once in all of them, so we only use it once. 7 appears in all of them, again only once. Our HCF/GCD is therefore 3*7 = 21.42/21 = 263/21 = 3126/21 = 6
Factor each number into its prime factors:42 = 2*3*763=3*3*7126=2*3*3*7Multiply together the largest common factors to get the HCF/GCD. 2 does not appear in all 3 numbers, so it is discarded. 3 appears in all of them, but only once in all of them, so we only use it once. 7 appears in all of them, again only once. Our HCF/GCD is therefore 3*7 = 21.42/21 = 263/21 = 3126/21 = 6All three numbers are multiples of 21, as expected.