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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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23 Jan 2020, 08:16
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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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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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24 Jan 2020, 10:56
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.

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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