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The HCF of 2472, 1284 and a third number ‘N’ is 12. If their LCM

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The HCF of 2472, 1284 and a third number ‘N’ is 12. If their LCM [#permalink]

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The HCF of 2472, 1284 and a third number ‘N’ is 12. If their LCM is 2^3 × 3^2 × 5 × 103 × 107,
then the number ‘N’ is:

(A) 2^2 × 3^2 × 7
(B) 2^2 × 3^3 × 103
(C) 2^2 × 3^2 × 5
(D) 2^2 × 3 × 5
(E) None of these
[Reveal] Spoiler: OA

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Re: The HCF of 2472, 1284 and a third number ‘N’ is 12. If their LCM [#permalink]

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New post 23 Jan 2017, 12:08
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saswata4s wrote:
The HCF of 2472, 1284 and a third number ‘N’ is 12. If their LCM is 2^3 × 3^2 × 5 × 103 × 107,
then the number ‘N’ is:

(A) 2^2 × 3^2 × 7
(B) 2^2 × 3^3 × 103
(C) 2^2 × 3^2 × 5
(D) 2^2 × 3 × 5
(E) None of these



\(LCM (2472, 1284, N) = 2^3 × 3^2 × 5 × 103 × 107\)

\(2472 = 2^3*3*103\)

\(1284 = 2^2*3*107\)

In order to produce required LCM our N should have 3^2 and 5, because none of the 2472 and 1284 have them.

Only option that fits is C.

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The HCF of 2472, 1284 and a third number ‘N’ is 12. If their LCM [#permalink]

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New post 03 Oct 2017, 05:26
HCF = 12 = 2^2 x 3
LCM = 2^3 × 3^2 × 5 × 103 × 107

2472 = 2^3 x 3 x 103
1284 = 2^2 x 3 x 107

Since we know, Product of LCM and HCF = Product of Numbers

(2^2 x 3) x (2^3 x 3^2 x 5 x 103 x 107) = (2^3 x 3 x 103) x ( 2^2 x 3 x 107) x N

2^5 x 3^3 x 5 = 2^5 x 3^2 x N

3 x 5 = N

So N is 3*5 and answer is E (None of the Above)

Edited: And, I realised that below formula is applicable only when there are two numbers. If there are more than two numbers then their HCF should be 1(they should be prime numbers).
Product of LCM and HCF = Product of Numbers

Hence above solution is incorrect.

Last edited by ramanbajwa2003 on 22 Oct 2017, 08:48, edited 1 time in total.

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Re: The HCF of 2472, 1284 and a third number ‘N’ is 12. If their LCM [#permalink]

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New post 21 Oct 2017, 04:33
ramanbajwa2003 wrote:
HCF = 12 = 2^2 x 3
LCM = 2^3 × 3^2 × 5 × 103 × 107

2472 = 2^3 x 3 x 103
1284 = 2^2 x 3 x 107

Since we know, Product of LCM and HCF = Product of Numbers

(2^2 x 3) x (2^3 x 3^2 x 5 x 103 x 107) = (2^3 x 3 x 103) x ( 2^2 x 3 x 107) x N

2^5 x 3^3 x 5 = 2^5 x 3^2 x N

3 x 5 = N

So N is 3*5 and answer is E (None of the Above)


Hi Raman,

If N is 15, then how can it have HCF as 12?

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Re: The HCF of 2472, 1284 and a third number ‘N’ is 12. If their LCM [#permalink]

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New post 21 Oct 2017, 06:21
ramanbajwa2003 wrote:
HCF = 12 = 2^2 x 3
LCM = 2^3 × 3^2 × 5 × 103 × 107

2472 = 2^3 x 3 x 103
1284 = 2^2 x 3 x 107

Since we know, Product of LCM and HCF = Product of Numbers

(2^2 x 3) x (2^3 x 3^2 x 5 x 103 x 107) = (2^3 x 3 x 103) x ( 2^2 x 3 x 107) x N

2^5 x 3^3 x 5 = 2^5 x 3^2 x N

3 x 5 = N

So N is 3*5 and answer is E (None of the Above)


We cannot use this logic as:

Product of two numbers = Product of their H.C.F. and L.C.M.

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Re: The HCF of 2472, 1284 and a third number ‘N’ is 12. If their LCM   [#permalink] 21 Oct 2017, 06:21
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