Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by

Is the integer n odd ?

(1) n is divisible by 3. Clearly not sufficient. Consider n = 3 and n = 6.

(2) 2n is divisible by twice as many positive integers as n.

TIP:
When odd number n is doubled, 2n has twice as many factors as n.
That's because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2.

Sufficient.

Answer: B.

For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html
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If \(n = a^p*b^q*c^r...\) where a, b, c are distinct prime numbers,
Total number of factors of n = (p+1)(q+1)(r+1)...

Let us assume \(n = 3^2*5^3\) i.e. odd
Total number of factors of n = (2+1)(3+1) = 12

\(2n = 2^1*3^2*5^3\)
Total number of factors of n = (1+1)(2+1)(3+1) = 24 (Twice of 12 obtained before because of additional 2)


Now assume the case where n is already even:

\(n = 2^2*3^2*5^3\) i.e. even
Total number of factors of n = (2+1)(2+1)(3+1) = 36

\(2n = 2^3*3^2*5^3\)
Total number of factors of n = (3+1)(2+1)(3+1) = 48 (More than 36 but not double because 3 is replaced by 4)

This is true for any prime factor. If that prime factor, p, is not in n, then p*n will have double the total number of factors.
If p is already in n, the total number of factors will increase but will never double.


*Edited*
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Statement #1 is not sufficient for reasons mentioned by icandy..so A and D are out as choices

Statement #2 gives an interesting data point. It says 2n has exactly 2xfactors compared to n

Let's take n = 6, then factors are 1, 2, 3 and 6 itself for a total count of 4
Now 2n = 12, so factors are 1, 2, 3, 4, 6, 12 for a total of 6.
Another sample, say n = 8, then factors are 1, 2, 4, 8 (count 4)
2n = 16, then factors are 1, 2, 4, 8, 16 (count 5).

Hmmm ok, doesn't look like even numbers satisfy this. Let's try with a couple of odd number substitutions
Say, n = 5, then factors are 1, 5(count = 2)
2n = 10, factors are 1, 2, 5, 10 (count = 4)...looking good so far

Say n = 9, then factors are 1, 3, 9(count = 3)
2n = 18, factors are 1, 2, 3, 6, 9, 18(count = 6). hmmm good are we ready to call n odd.

Let's hold on and do a prime number test
Say n = 3, then factors are 1, 3(count = 2)
2n = 6, then factors are 1, 2, 3, 6(count = 4). Good let's do the last prime # try

Say n = 2, then factors are 1, 2(count = 2)
2n = 4, then factors are 1, 2, 4 (count = 3)

Seems like #2 is sufficient to answer the question whether n is odd. So pick is B

It'd be a better question, if it was asked whether n is a prime number? That'd have made it a little bit more tricky, by extending us to use both #1 and #2. In which case I think I'd go with C.

There must be a general rule behind this to avoid plugging numbers. Anyone know?

There is a general rule here, which we can arrive at by extending the logic in maliyeci's excellent explanation above. I could explain this abstractly, but it's probably easier to take a specific example - let's use the number 72 = (2^3)(3^2). Now, this number has 12 factors in total, three of which are odd:

1, 3, 3^2

Now, if we multiply each of the numbers above by 2^1, we get three even divisors of 72, and the same will happen if we multiply these numbers by 2^2 or 2^3. So 72 has three odd divisors, and nine even divisors:

1, 3, 3^2
2, 2*3, 2*3^2
2^2, (2^2)*3, (2^2)(3^2)
2^3, (2^3)*3, (2^3)(3^2)

Notice that we have three times as many even divisors as odd divisors because the power on the 2 in the prime factorization of 72 is 3; that guarantees that we have three even divisors for every odd divisor. You could use this logic for any number, of course, from which we have the following general rule:

* The ratio of the number of even divisors of x to the number of odd divisors of x is always equal to the power on the 2 in the prime factorization of x.

So, if the power on the 2 in the prime factorization of x is equal to 1, we have an equal number of odd and even divisors. If the power is greater than 1, we have more even divisors than odd divisors.
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Merging similar topics.

2n is divisible by twice as many positive integers as n, means that the # of factors of 2n is twice the # of factors of n.

For example: # of factors of 3 is two (1, and 3 itself) and the # of factors of 2*3=6 is four (1, 2, 3, and 6 itself), so the # of factors of 6=2n is twice the # of factors of 3=n.

Hope it's clear.
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Bunuel, could you please explain the above TIP with an example. If n=2 and n=15? Thanks.

15 has 4 factors: 1, 3, 5, and 15;
15*2=30 has 4*2=8 factors: 1, 1*2=2, 3, 3*2=6, 5, 5*2=10, 15, and 15*2=30.
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So The following doesn't happen for even numbers?

When odd number n is doubled, 2n has twice as many factors as n.
Thats because odd number has only odd factors and when we multiply n by two we remain all these odd factors as divisors and adding exactly the same number of even divisors, which are odd*2.
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Yes it doesn't follow for even numbers. We can prove it using the formula:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

ODD Number:
Finding the number of all factors of 45: \(45= 3^2*5^1\)
Total number of factors of 45 including 1 and 45 itself is \(((2+1)*(1+1)=3*2=6\) factors.

Finding the number of all factors of 90: \(90=2^1*3^2*5^1\)
Total number of factors of 90 including 1 and 90 itself is \((1+1)*(2+1)*(1+1)=2*3*2=12\) factors.

Note that, the odd numbers have only odd prime factors. If you double it then you introduce factor of 2^1 in the prime factorization and hence you end up multiplying by (1+1) when finding total number of factors, which therefore gets doubled.

EVEN Number:
However, the same is not true for EVEN numbers. They already have prime factorization with 2^x (x>=1) and doubling that EVEN number only increments the exponent of factor 2, but not necessarily doubles the number of factors.
Finding the number of all factors of 12: \(12= 2^2*3^1\)
Total number of factors of 12 including 1 and 12 itself is \(((2+1)*(1+1)=3*2=6\) factors.

Finding the number of all factors of 24: \(24= 2^3*3^1\)
Total number of factors of 24 including 1 and 24 itself is \(((3+1)*(1+1)=4*2=8\) factors.

Bunuel: Is this applicable only on odd numbers?

Posted from my mobile device

# of factors of 2 is two (1, and 2 itself) and the # of factors of 2*3=6 is four (1, 2, 3, and 6 itself), so the # of factors of 6=2n is twice the # of factors of 2=n.
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Hi,

here are my two cents for this question

Let n be expressed in terms of prime factors as \(a^{p}\) \(b^{q}\) \(c^{r}\)

where a,b,c are prime numbers and p q r are postie powers of prime numbers
then total number of factors of n = (p+1)(q+1)(r+1) = X

then 2n if expressed in terms of its prime factors can be

2n=\(a^{p}\) \(b^{q}\) \(c^{r}\)

Now if any of the prime factors of n does not contain 2 as its factor then the total number of factors of 2n would be = (1+1) (p+1)(q+1)(r+1) = 2X

but if any of the prime factors of n contains 2 as its factor then the total number of factors of 2n \(\neq\) twice the number of factors of n


On the same lines we can say that if number of facotrs of a number n =X, multiplying that number 'n' by another prime factor A which is not a prime factor of 'n' the total number of factors of 'An' would be 2X

Let
n= \(2^1 5^1\), total number of factors are 4

then 3n=\(2^1 3^1 5^1\) , total number of factors are 8 which is twice of n.


n= \(2^2\), total number of factors are 3

then 3n=\(2^2 3^1\) , total number of factors are 6 which is twice of n.
then 5n=\(2^2 5^1\) , total number of factors are 6 which is twice of n.

So since from statement B we have that the number of factors of 2n is twice the number of factors of n we can say that 2 is not a prime factor of n. If two is not prime factor of n then n is odd.


Probus
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Great explanation masuhari, not sure I quite understand this last part "#2 is sufficient to answer the question whether n is odd" . Could you expand a bit please? Thanks for your time in advanced.

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