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

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Math Expert
Joined: 02 Sep 2009
Posts: 65238
If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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12 Nov 2019, 01:50
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Difficulty:

95% (hard)

Question Stats:

48% (02:42) correct 52% (02:47) wrong based on 184 sessions

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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

Are You Up For the Challenge: 700 Level Questions

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Posts: 8755
Re: If the highest common factor of 2,472, 1,284 and positive integer N is  [#permalink]

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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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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.
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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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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.

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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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

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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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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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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

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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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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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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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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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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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).

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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