hcf of two no.

Question

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

Marcab · Answer

Hii welcome to GMATCLUB.
Please see the rules before posting.

In this question, the answer must be 3.
Take two numbers for easy calculation: 2^6 and 2^8.
So 2^6 - 1=63, 2^8 - 1=255.
Hcf is 3.
Hence IMO for the two numbers in question, the hcf will be 3.

KarishmaB · Answer

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

Marcab · Answer

Hi Karishma.
Is there any alternate way to do this question?
I tried picking up a smaller number but failed as evitable by your explanation.

KarishmaB · Answer

There is no reason that the HCF of two small numbers will be the same as the HCF of two larger numbers. The question is meant to test your application of algebraic identities.

mdacosta · Answer

do you mind writing out the "..." part in B please? I understood the rest of your explanation perfect, just got lost there. Many thanks!

Michael

hcf of two no.

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