juantheron wrote:

If A = (2^(20) -1) and B = (2^(110) - 1). Then HCF(A,B) =

Use a^2 - b^2 = (a + b)(a - b) to factorize the expressions.

\(A = 2^{20} - 1^{20} = 2^{10*2} - 1^{10*2} = (2^{10})^2 - (1^{10})^2 = (2^{10} + 1^{10})(2^{10} - 1^{10})\)

\(B = 2^{110} - 1^{110} = 2^{10*11} - 1^{10*11} = (2^{10})^{11} - (1^{10})^{11} = (2^{10} - 1^{10})(2^{100} + ....)\)

(Difference of odd powers is divisible by the difference of the numbers e.g. x^3 - y^3 is divisible by x-y)

The highest common factor must be\((2^{10} - 1^{10}) = 2^{10} - 1\)

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