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*

2As 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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