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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