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# hcf of two no.

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Joined: 18 Dec 2012
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18 Dec 2012, 19:37
If A = (2^(20) -1) and B = (2^(110) - 1). Then HCF(A,B) =
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Re: hcf of two no.  [#permalink]

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18 Dec 2012, 20:29
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.
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Re: hcf of two no.  [#permalink]

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19 Dec 2012, 00:27
2
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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Re: hcf of two no.  [#permalink]

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19 Dec 2012, 02:57
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.
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Re: hcf of two no.  [#permalink]

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19 Dec 2012, 21:02
Marcab wrote:
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.

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.
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Re: hcf of two no.  [#permalink]

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02 Oct 2017, 21:35
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

VeritasPrepKarishma wrote:
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$$(2^{10} - 1^{10}) = 2^{10} - 1$$
Re: hcf of two no. &nbs [#permalink] 02 Oct 2017, 21:35
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