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16 Jul 2010, 12:26
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D
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Difficulty:
5%
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Question Stats:
85% (00:47) correct
15%
(01:01)
wrong
based on 79
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Is x(x + 1)(x + 2) divisible by 4 (where x is an integer)?
(1) x is even
(2) x is 2
(1) x is even
(2) x is 2
16 Jul 2010, 12:33
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This question is form Manhattan guide. I couldn't understand the reasoning for A is being sufficient.
Manhattan said that if x is even then we have 2, 3, 2+2 so basically two prime factor with 2 in it and hence divisible by 4.
But how about x=0. zero is also even and in the above case the product is zero. Oh so basically 0 is divisible by 4 and hence having x as even satisfy the condition??? am I getting this correct??
By the way the answer in above should be D.
Manhattan said that if x is even then we have 2, 3, 2+2 so basically two prime factor with 2 in it and hence divisible by 4.
But how about x=0. zero is also even and in the above case the product is zero. Oh so basically 0 is divisible by 4 and hence having x as even satisfy the condition??? am I getting this correct??
By the way the answer in above should be D.
16 Jul 2010, 12:45
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paam0101
Zero is even and divisible by every integer (except zero itself, as division by zero is undefined).
Integer \(a\) is divisible by integer \(b\) means that \(\frac{a}{b}=integer\) (so remainder is zero).
Now, if \(a=0\) --> \(\frac{a}{b}=\frac{0}{b}=0=integer\), that's why zero divisible by every integer (except zero itself, as division by zero is undefined).
Hope it helps.
