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# hcf of two no.

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18 Dec 2012, 18:37
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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, 19:29
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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18 Dec 2012, 23:27
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18464 [2], given: 237 VP Status: Been a long time guys... Joined: 03 Feb 2011 Posts: 1375 Kudos [?]: 1748 [0], given: 62 Location: United States (NY) Concentration: Finance, Marketing GPA: 3.75 Re: hcf of two no. [#permalink] ### Show Tags 19 Dec 2012, 01:57 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. _________________ Kudos [?]: 1748 [0], given: 62 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7863 Kudos [?]: 18464 [0], given: 237 Location: Pune, India Re: hcf of two no. [#permalink] ### Show Tags 19 Dec 2012, 20:02 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: hcf of two no. [#permalink]

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02 Oct 2017, 20:35
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$$(2^{10} - 1^{10}) = 2^{10} - 1$$

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Re: hcf of two no.   [#permalink] 02 Oct 2017, 20:35
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