Asad wrote:
If \(n\) is an integer, is \(n\) odd?
(1) \(\frac{n}{3}\) is divisible by \(3\)
(2) \(2n\) has twice as many factors as \(n\)
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Since we have 1 variable (\(n\)) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.
Condition 1)
We have an expression \(\frac{n}{3} = 3k\) for some integer \(k\).
Then we have \(n = 3^2k\).
If \(k = 1\), then \(n = 3^2 = 9\), \(n\) is an odd number and the answer is 'yes'.
If \(k = 2\), then \(n = 3^2 \cdot 2 = 18\), \(n\) is an even number and the answer is 'no'.
Since condition 1) does not yield a unique solution, it is not sufficient.
Condition 2)
Assume \(n = p^a \cdot q^b \cdot r^c\) where \(p, q\) and \(r\) are different prime numbers.
Then the number of factors of \(n\) is \((a+1)(b+1)(c+1)\).
If \(p, q\) and \(r\) are odd prime numbers, then \(2n = 2^1 \cdot p^a \cdot q^b \cdot r^c\) and the number of factors of \(2n\) is \((1+1)(a+1)(b+1)(c+1) = 2(a+1)(b+1)(c+1)\) which is twice as many factors as \(n\).
Assume \(p = 2\) which means one of prime factors of \(n\) is \(2\).
Then \(2n = 2p^a \cdot q^b \cdot r^c = 2^{a+1} \cdot q^b \cdot r^c\) and the number of factors of \(2n\) is \((a+2)(b+1)(c+1)\).
We have \((a+2)(b+1)(c+1) = 2(a+1)(b+1)(c+1)\) or \(a+2 = 2a + 2\). Then we have \(a = 0\) and n is an odd number.
Since condition 2) yields a unique solution, it is sufficient.
Therefore, D is the answer.
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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