Question

Find the HCF of 2^120-1 and 2^50-1​

Answer 1

Given : 2¹²⁰ - 1     and 2⁵⁰ - 1

To Find : HCF

Solution:

2¹²⁰ - 1   =   (2¹⁰)¹² - 1   = (2¹⁰)¹² - 1¹²

2⁵⁰ - 1 = (2¹⁰)⁵ - 1  = (2¹⁰)⁵ - 1⁵

xⁿ - yⁿ  is always divisible by x - y

(2¹⁰)¹² - 1¹² is divisible by

2¹⁰ - 1

(2¹⁰)⁵ - 1⁵ is divisible by

2¹⁰ - 1

Hence 2¹⁰ - 1  is the HCF

2¹⁰ - 1  is the HCF of 2¹²⁰ - 1  and 2⁵⁰ - 1  

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