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VP
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n' is a natural number. State whether n (n┬▓ - [#permalink]
 25 Nov 2003, 23:15
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'n' is a natural number. State whether n (n┬▓ - 1) is divisible by 24.
(1) 3 divides 'n' completely without leaving any remainder.
(2) 'n' is odd.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Director
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as far as I know natural numbers=positive integers without 0, so from A n could be 3,6,9,and so on. The outcome is 24,210,720, so A is not sufficient, B states n is odd 3,9,15 .. which gives 24,720,3360 which are all divisible by 24 so IMO B is sufficient
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SVP
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agree with B
n(n┬▓-1)=n(n-1)(n+1)=(n-1)n(n+1)
(1) n is div by 3.
consider n=3: 2*3*4=24 is div by 24
consider n=6: 5*6*7 is not
(2) n is odd
n=2k+1
(n-1)n(n+1) is divisible by 6
moreover =2k(2k+1)(2k+2)=8k^3+12k^2+4k divisible by 4
thus divisible by 24
B.
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Manager
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I'd say B too.
(1) Since n is divisible by 3, (n┬▓ - 1) has to be divisible by 8 for the quanitity in question. We also know that n > 3. But not all values of n are such that (n┬▓ - 1) will be evenly divisible by 8. Not suff.
(2) If n is odd then n(n┬▓ - 1) is divisible by 24 for all values of n.
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VP
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good explanation, stolyar. right answer is B
I just substituted values and got the answer... 
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Director
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To stolyar's point,
The product of three consecutive numbers will always be divisible by 24 if the middle number is odd. Because:
one of three consecutive numbers is always divisible by three (for obvious reasons)
every other even number is divisible by four.
The even number that's not divisible by four, when multiplied with the one that is, gives you a multiple of 8.
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close
