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If the highest common factor of 2,472, 1,284 and positive integer N is

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If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 12 Nov 2019, 01:50
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If the highest common factor of 2,472, 1,284 and positive integer N is 12 and the least common multiple of the same three numbers, 2472, 1284 and N, is \(2^3*3^2*5*103*107\), what is the value of N?


(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



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Re: If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 12 Nov 2019, 06:38
Bunuel wrote:
If the highest common factor of 2,472, 1,284 and positive integer N is 12 and the least common multiple of the same three numbers, 2472, 1284 and N, is \(2^3*3^2*5*103*107\), what is the value of N?


(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



Are You Up For the Challenge: 700 Level Questions


\(2472=2*2*2*3*103=2^3*3*103\).
\(1284=2*2*3*107=2^2*3*107\).

LCM=\(2^3*3^2*5*103*107\)
we can check each prime number
2 -- there can be two or 3 2s...\(2^2\) or \(2^3\)
3 -- Surely two 3s as LCM has two 3s but none of 1284 or 2472 have two 3s...\(3^2\)
5 -- Surely one 5..\(5^1\)
103 -- can be none or one of 103...\(103^0\) or \(103^1\)
107 -- can be none or one of 107...\(107^0\) or \(107^1\)

As there are two 3s, A, B and D are out...
N can be any of -
\(2^2*3^2*5\) and any of the combination of 2, 103 or 107 added to it..
for example \(2^3*3^2*5*103*107\) can be the largest value

C

Note : The question should ask for the smallest value of N or it should be ' What can be the value of N?'
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If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 13 Nov 2019, 03:21

Solution



Given

    • HCF of 2,472, 1,284 and positive integer N is 12.
    • LCM of 2,472, 1,284 and positive integer N is 2^3∗3^2∗5∗103∗107.

To find

    • The value of N.

Approach and Working out

    • 2472 = 2^3 * 3 * 103
    • 1284 = 2^2 * 3 * 107
    • HCF (2472, 1284, N) = 12
      o So, N = 2^(2+b) * 3^a * other prime factors where a >= 1 and b>=0

    • LCM (2472, 1284, N) = 2^3∗3^2∗5∗103∗107.
      o So, N can be 2^(2+b) * 3^2 * 5 * 103^c * 107^d where:
         b, c, d can be 0 or 1

Only possible option is option C.

Thus, option C is the correct answer.
Correct Answer: Option C
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Re: If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 15 Nov 2019, 07:01
EgmatQuantExpert wrote:

Solution



Given

    • HCF of 2,472, 1,284 and positive integer N is 12.
    • LCM of 2,472, 1,284 and positive integer N is 2^3∗3^2∗5∗103∗107.

To find

    • The value of N.

Approach and Working out

    • 2472 = 2^3 * 3 * 103
    • 1284 = 2^2 * 3 * 107
    • HCF (2472, 1284, N) = 12
      o So, N = 2^(2+b) * 3^a * other prime factors where a >= 1 and b>=0

    • LCM (2472, 1284, N) = 2^3∗3^2∗5∗103∗107.
      o So, N can be 2^(2+b) * 3^2 * 5 * 103^c * 107^d where:
         b, c, d can be 0 or 1

Only possible option is option C.

Thus, option C is the correct answer.
Correct Answer: Option C




How about this approach ?

HCM X LCM = Product of numbers

2472 X 1284 X N = 12 X 2^3 X 3^2 X 5 X 103 X 107

solving it results in N = 3*5 = 15
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Re: If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 20 Nov 2019, 17:58
Bunuel wrote:
If the highest common factor of 2,472, 1,284 and positive integer N is 12 and the least common multiple of the same three numbers, 2472, 1284 and N, is \(2^3*3^2*5*103*107\), what is the value of N?


(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



Are You Up For the Challenge: 700 Level Questions


Let’s first factor 2,472 and 1,284 into primes:

2,472 = 12 x 206 = 2^2 x 3 x 2 x 103 = 2^3 x 3 x 103

1,284 = 12 x 107 = 2^2 x 3 x 107

We see that that the LCM of 2,472 and 1,284 is 2^3 x 3 x 103 x 107. Notice that the LCM of 2,472 1,284, and N is equal to the LCM of 2^3 x 3 x 103 x 107 and N, which is given to be 2^3 x 3^2 x 5 x 103 x 107. We are also given that the GCF of 2,472 1,284, and N is 12.

Recall that for any two positive integers a and b, we have a x b = LCM(a, b) x GCF(a, b). Therefore, by letting a = 2^3 x 3 x 103 x 107 and b = N, we have:

2^3 x 3 x 103 x 107 x N = 2^3 x 3^2 x 5 x 103 x 107 x 12

N = 3 x 5 x 12

N = 2^2 x 3^2 x 5

Answer: C
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Re: If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 01 Dec 2019, 05:15
chetan2u wrote:
Bunuel wrote:
If the highest common factor of 2,472, 1,284 and positive integer N is 12 and the least common multiple of the same three numbers, 2472, 1284 and N, is \(2^3*3^2*5*103*107\), what is the value of N?


(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



Are You Up For the Challenge: 700 Level Questions


\(2472=2*2*2*3*103=2^3*3*103\).
\(1284=2*2*3*107=2^2*3*107\).

LCM=\(2^3*3^2*5*103*107\)
we can check each prime number
2 -- there can be two or 3 2s...\(2^2\) or \(2^3\)
3 -- Surely two 3s as LCM has two 3s but none of 1284 or 2472 have two 3s...\(3^2\)
5 -- Surely one 5..\(5^1\)
103 -- can be none or one of 103...\(103^0\) or \(103^1\)
107 -- can be none or one of 107...\(107^0\) or \(107^1\)

As there are two 3s, A, B and D are out...
N can be any of -
\(2^2*3^2*5\) and any of the combination of 2, 103 or 107 added to it..
for example \(2^3*3^2*5*103*107\) can be the largest value

C

Note : The question should ask for the smallest value of N or it should be ' What can be the value of N?'



Hi, Can you plz explain why cant we follow the method:

HCF * LCM = Product of numbers?

If we use that, it results in N=15
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Re: If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 01 Dec 2019, 06:46
ScottTargetTestPrep wrote:
Bunuel wrote:
If the highest common factor of 2,472, 1,284 and positive integer N is 12 and the least common multiple of the same three numbers, 2472, 1284 and N, is \(2^3*3^2*5*103*107\), what is the value of N?


(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



Are You Up For the Challenge: 700 Level Questions



Recall that for any two positive integers a and b, we have a x b = LCM(a, b) x GCF(a, b). Therefore, by letting

a = 2^3 x 3 x 103 x 107 and b = N, we have:

2^3 x 3 x 103 x 107 x N = 2^3 x 3^2 x 5 x 103 x 107 x 12

N = 3 x 5 x 12

N = 2^2 x 3^2 x 5

Answer: C


can you explain the bold line?

2472 x 1284 x N = (2^3 x 3 x 103) x (2^2 x 3 x 107) x N

arnt you missing a few terms?
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Re: If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 03 Dec 2019, 19:15
Mansoor50 wrote:

can you explain the bold line?

2472 x 1284 x N = (2^3 x 3 x 103) x (2^2 x 3 x 107) x N

arnt you missing a few terms?


I did not intend to multiply all three numbers together; as a matter of fact, when you have more than two numbers, the formula a x b = LCM(a, b) x GCF(a, b) is no longer valid.

We are taking a = LCM(2472, 1284) and b = N and applying the formula to these numbers.
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Re: If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 05 Dec 2019, 06:17
ScottTargetTestPrep wrote:
Mansoor50 wrote:

can you explain the bold line?

2472 x 1284 x N = (2^3 x 3 x 103) x (2^2 x 3 x 107) x N

arnt you missing a few terms?


I did not intend to multiply all three numbers together; as a matter of fact, when you have more than two numbers, the formula a x b = LCM(a, b) x GCF(a, b) is no longer valid.

We are taking a = LCM(2472, 1284) and b = N and applying the formula to these numbers.


WOOT!......THANKS!!!!!
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If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 22 May 2020, 00:12
1
HCF * LCM = 1284 * 2472 * N
subsequently, N= 3*5 = 15
But, HCF of the three numbers is 12 hence, N must have two 2's as well. So N becomes 4*3*5 . Again LCM of the three numbers has two 3's. Now, each number of the numbers ---1284 and 2472--- has only one 3 , but LCM has two 3's.
Hence, N must have two 3's.


Finally, N has two 2's two 3's and one 5 to get the exact LCM and HCF . C is the correct answer

.......



Dear Moderator,

Could you please remove the comma included in 2,472 and 1,284 as its a bit confusing to read.
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Re: If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 28 May 2020, 16:09
dips1122 wrote:
chetan2u wrote:
Bunuel wrote:
If the highest common factor of 2,472, 1,284 and positive integer N is 12 and the least common multiple of the same three numbers, 2472, 1284 and N, is \(2^3*3^2*5*103*107\), what is the value of N?


(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



Are You Up For the Challenge: 700 Level Questions


\(2472=2*2*2*3*103=2^3*3*103\).
\(1284=2*2*3*107=2^2*3*107\).

LCM=\(2^3*3^2*5*103*107\)
we can check each prime number
2 -- there can be two or 3 2s...\(2^2\) or \(2^3\)
3 -- Surely two 3s as LCM has two 3s but none of 1284 or 2472 have two 3s...\(3^2\)
5 -- Surely one 5..\(5^1\)
103 -- can be none or one of 103...\(103^0\) or \(103^1\)
107 -- can be none or one of 107...\(107^0\) or \(107^1\)

As there are two 3s, A, B and D are out...
N can be any of -
\(2^2*3^2*5\) and any of the combination of 2, 103 or 107 added to it..
for example \(2^3*3^2*5*103*107\) can be the largest value

C

Note : The question should ask for the smallest value of N or it should be ' What can be the value of N?'



Hi, Can you plz explain why cant we follow the method:

HCF * LCM = Product of numbers?

If we use that, it results in N=15



Hi dips1122,

HCF * LCM = Product of numbers?

This formula only applies when there are only two numbers.
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Re: If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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New post 01 Jun 2020, 06:59
Bunuel wrote:
If the highest common factor of 2,472, 1,284 and positive integer N is 12 and the least common multiple of the same three numbers, 2472, 1284 and N, is \(2^3*3^2*5*103*107\), what is the value of N?


(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




Solution:

Let’s prime factorize all the numbers:

2,472 = 6 x 412 = 2 x 3 x 4 x 103 = 2^3 x 3 x 103

1,284 = 12 x 107 = 2^2 x 3 x 107

GCF = 12 = 2^2 x 3

LCM = 2^3 x 3^2 x 5 x 103 x 107

Since the GCF = 2^2 x 3, N also has 2^2 and 3 as factors. Furthermore, since the LCM has an extra factor of 3 and a factor of 5 that 2,472 and 1,284 don’t have, N must have those factors, too. Therefore, N could be 2^2 x 3^2 x 5 (note: N could also be, for example, 2^2 x 3^2 x 5 x 103 or 2^2 x 3^2 x 5 x 107).

Answer: C
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Re: If the highest common factor of 2,472, 1,284 and positive integer N is   [#permalink] 01 Jun 2020, 06:59

If the highest common factor of 2,472, 1,284 and positive integer N is

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