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Is the positive integer n a multiple of 24?

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Is the positive integer n a multiple of 24? [#permalink] New post 22 Feb 2011, 23:35
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Is the positive integer n a multiple of 24?

(1) n is a multiple of 4.
(2) n is a multiple of 6.
[Reveal] Spoiler: OA

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Re: 115. Multiple [#permalink] New post 22 Feb 2011, 23:40
Prime factors of 24: 2^3*3
(1) 4: 2^2; Not sufficient.
(2) 6: 2*3; Not sufficient.

Combining both; minimum factors of n= 2^2*2*3 = 2^3*3 = all factors of 24. Sufficient.

Ans: "C"
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Re: 115. Multiple [#permalink] New post 23 Feb 2011, 00:02
i'd say E..

12 is a multiple of both 4 and 6 but not of 24.
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Re: 115. Multiple [#permalink] New post 23 Feb 2011, 00:19
dreambeliever wrote:
i'd say E..

12 is a multiple of both 4 and 6 but not of 24.


Yes, you are right. I ignored the fact that the 2 in the prime factor of 6 may be the same 2 from the prime factor of 2's in the factors of 12. Thus, n definitely has only two 2's and one 3 as factor, which is 12. thanks.
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Re: 115. Multiple [#permalink] New post 23 Feb 2011, 03:51
fluke wrote:
dreambeliever wrote:
i'd say E..

12 is a multiple of both 4 and 6 but not of 24.


Yes, you are right. I ignored the fact that the 2 in the prime factor of 6 may be the same 2 from the prime factor of 2's in the factors of 12. Thus, n definitely has only two 2's and one 3 as factor, which is 12. thanks.


fluke please make clear in your way such as:

Prime factors of 24: 2^3*3
(1) 4: 2^2; Not sufficient.
(2) 6: 2*3; Not sufficient.

Combining both; minimum factors of n= 2^2*2*3 = 2^3*3 = all factors of 24. Sufficient.

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Re: 115. Multiple [#permalink] New post 23 Feb 2011, 04:49
Baten80 wrote:
fluke wrote:
dreambeliever wrote:
i'd say E..

12 is a multiple of both 4 and 6 but not of 24.


Yes, you are right. I ignored the fact that the 2 in the prime factor of 6 may be the same 2 from the prime factor of 2's in the factors of 12. Thus, n definitely has only two 2's and one 3 as factor, which is 12. thanks.


fluke please make clear in your way such as:

Prime factors of 24: 2^3*3
(1) 4: 2^2; Not sufficient.
(2) 6: 2*3; Not sufficient.

Combining both; minimum factors of n= 2^2*2*3 = 2^3*3 = all factors of 24. Sufficient.



First of all; my answer was wrong; the actual answer is "E"

We need to prove that n is definitely a multiple of 24.
What could be the possible multiples for 24; 24,48,72,96,.....

Prime factors of 24: (2^3*3)
Prime factors of 48: 2*(2^3*3)
Prime factors of 72: 3*(2^3*3)
Prime factors of 96: 4*(2^3*3)

So; we see that if n has at least (2^3*3) as factors; it must be a multiple of 24.

1) n is a multiple of 4.
4 = 2*2 = 2^2. 2^2 contains only two 2's and no 3's. But, we need three 2's and one 3; at least. Thus not sufficient.

2) n is a multiple of 6.
6 = 2*3 = It contains only one 2 and one 3. But, we need three 2's and one 3; at least. Thus not sufficient.

Combining both;
We can't say, as I foolishly did, that n contains all factors "2*2*2*3". It would be wrong. Because the 2 that you see in the second statement may be the same 2 that appeared in statement 1.

e.g.
12: 2*2*3. As you can see.
12 is a factor of 6: 2*3
12 is a factor of 4: 2*2

As you can see the 2 in red color is used by both 6 and 4.

If the two factors were 6,16:
16: 2^4
6: 2*3
It would be sufficient. Just take the maximum power of the each prime factor available for both numbers and check whether the final result is equal or a greater multiple of 24. LCM.

The idea is to find the LCM: if LCM(6,4) >= 24 and LCM(6,4) is a multiple of 24; then n must be a multiple of 24.
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Re: 115. Multiple [#permalink] New post 18 Nov 2011, 23:38
Baten80 wrote:
fluke wrote:
dreambeliever wrote:
i'd say E..

12 is a multiple of both 4 and 6 but not of 24.


Yes, you are right. I ignored the fact that the 2 in the prime factor of 6 may be the same 2 from the prime factor of 2's in the factors of 12. Thus, n definitely has only two 2's and one 3 as factor, which is 12. thanks.


fluke please make clear in your way such as:

Prime factors of 24: 2^3*3
(1) 4: 2^2; Not sufficient.
(2) 6: 2*3; Not sufficient.

Combining both; minimum factors of n= 2^2*2*3 = 2^3*3 = all factors of 24. Sufficient.

If in the same question we were to replace 'multiple' by 'divisible' what the difference??? What exactly happens when something is a multiple of something or when something is a divisbile of something ?
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Re: 115. Multiple [#permalink] New post 04 Feb 2012, 13:50
Expert's post
siddhans wrote:
If in the same question we were to replace 'multiple' by 'divisible' what the difference??? What exactly happens when something is a multiple of something or when something is a divisbile of something ?


There is no difference between saying that 12 is a multiple of 4 and that 12 is divisible by 4.

As for the question.
Is the positive integer n a multiple of 24?

(1) n is a multiple of 4. Not sufficient.
(2) n is a multiple of 6. Not sufficient.

(1)+(2) n is a multiple of both 4 and 6 which means that it's a multiple of least common multiple of 4 and 6, which is 12. So, even taken together statements are not sufficient, since n can be for example 12 as well as 24. Not sufficient.

Answer: E.

Generally if a positive integer n is a multiple of positive integer a and positive integer b, then n is a multiple of LCM(a,b).

Hope it helps.
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Re: Is the positive integer n a multiple of 24? [#permalink] New post 14 Sep 2014, 00:19
aaah I thought you can solve this question in the same way like that one: is-the-integer-x-divisible-by-167497.html

Obviously I overlooked that one of the prime factors of 2 in 6 (S2) is the same as in 4 (S1). Can anyone give me some tips how to notice that immediately?
The other problem asks if x is divisible by 36 and says in S1 that x is divisible by 12 and in S2 that x is divisible by 9. How can I say that the prime factors of 3 in both numbers are not the same?
Re: Is the positive integer n a multiple of 24?   [#permalink] 14 Sep 2014, 00:19
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