Question

find the HCF of the number in in each of the the following using the prime factorization method (1)106,159,371 (2)272,425 (3)144,252,630 (4)1197,5320,4289 very urgent send answer fast who send me answer first i join in brianlist answer please send me answer fast please solve step by step​

Answer 1

Step-by-step explanation:

(1)106,159,371

The factors of 106 = 2 * 53

The factors of 159 = 3 * 53

The factors of 371 = 7 * 53

Common factors = 53

So, HCF = 53.

(2)272, 425

The factors of 272 = 2 * 2 * 2 * 2 * 17 = 2⁴ * 17

The factors of 425 = 5 * 5 * 17 = 5² * 17

Common factors = 17

So, HCF = 17

(3)144, 252 ,630

The factors of 144 = 2 * 2 * 2 * 2 * 3 * 3 = 2⁴ * 3²

The factors of 252 = 2 * 2 * 3 * 3 * 7 = 2² * 3² * 7

The factors of 630 = 2 * 3 * 3 * 5 * 7 = 2 * 3² * 5 * 7

Common factors = 2 * 3 * 3 = 18

HCF = 18

(4)1197, 5320, 4289

The factors of 1197 =   3 * 3 * 7 * 19 =   3² * 7 * 19

The factors of 5320 = 2 * 2 *  2  * 5  * 7  * 19 = 2³ * 5  *   7

The factors of 4289 = 4289

No common factors

HCF =  1