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If n is an integer, is n odd? (1) n/3 is divisible by 3 (2)

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If n is an integer, is n odd? (1) n/3 is divisible by 3 (2)  [#permalink]

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New post 23 Jan 2020, 08:16
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A
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C
D
E

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48% (01:15) correct 52% (01:56) wrong based on 21 sessions

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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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If n is an integer, is n odd? (1) n/3 is divisible by 3 (2)  [#permalink]

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New post Updated on: 24 Jan 2020, 13:05
If n/3=3, n is odd and n/3 is divisible by 3. If n/3=6, n is even and n/3 is divisible by 3. So statement 1 is insufficient.

The only way I can think for statement 2 to be true is prime numbers other than 2.
if n=5 (factors 1, 5); 2n=10 (factors 1, 2, 5, 10)
if n=7 (factors 1, 7); 2n=14 (factors 1, 2, 7, 14)
So, statement 2 alone is sufficient. answer is B.

Originally posted by s1lntz on 24 Jan 2020, 09:39.
Last edited by s1lntz on 24 Jan 2020, 13:05, edited 1 time in total.
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Re: If n is an integer, is n odd? (1) n/3 is divisible by 3 (2)  [#permalink]

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New post 24 Jan 2020, 10:56
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 \(n\)

Posted from my mobile device



Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
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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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Re: If n is an integer, is n odd? (1) n/3 is divisible by 3 (2)   [#permalink] 24 Jan 2020, 10:56
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If n is an integer, is n odd? (1) n/3 is divisible by 3 (2)

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