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

Question

Is the integer n odd?

(1) n is divisible by 3 
(2) 2n is divisible by twice as many positive integers as n

Bunuel · Accepted Answer

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

KarishmaB · Answer

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*

masuhari · Answer

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.

mendelay · Answer

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

IanStewart · Answer

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.

Bunuel · Answer

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.

pgmat · Answer

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

Bunuel · Answer

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.

Sachin9 · Answer

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.

PrashantPonde · Answer

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.

s111 · Answer

Bunuel : Is this applicable only on odd numbers?

Posted from my mobile device

Bunuel · Answer

# 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.

Probus · Answer

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  
eq  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

Kimberly77 · Answer

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

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