Last visit was: 06 Oct 2024, 19:55 It is currently 06 Oct 2024, 19:55
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
User avatar
Joined: 02 Dec 2007
Posts: 259
Own Kudos [?]: 2142 [113]
Given Kudos: 6
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 95949
Own Kudos [?]: 665803 [81]
Given Kudos: 87512
Send PM
Tutor
Joined: 16 Oct 2010
Posts: 15344
Own Kudos [?]: 68587 [44]
Given Kudos: 443
Location: Pune, India
Send PM
General Discussion
User avatar
Joined: 14 Sep 2003
Posts: 24
Own Kudos [?]: 44 [7]
Given Kudos: 0
Location: california
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
4
Kudos
3
Bookmarks
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.
User avatar
Joined: 21 Jul 2009
Posts: 79
Own Kudos [?]: 591 [1]
Given Kudos: 23
Location: New York, NY
 Q45  V42
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
1
Kudos
There must be a general rule behind this to avoid plugging numbers. Anyone know?
GMAT Tutor
Joined: 24 Jun 2008
Posts: 4127
Own Kudos [?]: 9681 [3]
Given Kudos: 97
 Q51  V47
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
2
Kudos
1
Bookmarks
Expert Reply
mendelay
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.
Math Expert
Joined: 02 Sep 2009
Posts: 95949
Own Kudos [?]: 665803 [6]
Given Kudos: 87512
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
5
Kudos
1
Bookmarks
Expert Reply
ajit257
Is the integer n odd?
(1) n is divisible by 3.
(2) 2n is divisible by twice as many positive integers as n.


What do you mean by the second statement.

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.
avatar
Joined: 26 Jul 2010
Posts: 22
Own Kudos [?]: 1077 [0]
Given Kudos: 8
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
Quote:
(2) TIP: 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.


Bunuel, could you please explain the above TIP with an example. If n=2 and n=15? Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 95949
Own Kudos [?]: 665803 [0]
Given Kudos: 87512
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
Expert Reply
pgmat
Quote:
(2) TIP: 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.


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.
User avatar
Joined: 22 Jul 2012
Status:Gonna rock this time!!!
Posts: 356
Own Kudos [?]: 169 [1]
Given Kudos: 562
Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE:Information Technology (Computer Software)
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
1
Kudos
Bunuel
Is the integer n odd
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

(1) 3 or 6. Clearly not sufficient.

(2) TIP:
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.

Sufficient.

Answer: B.

For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html

P.S. You can attach the screenshot of a question directly to the post so that everyone will see it.

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.
User avatar
Joined: 27 Jun 2012
Posts: 323
Own Kudos [?]: 2527 [4]
Given Kudos: 185
Concentration: Strategy, Finance
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
3
Kudos
1
Bookmarks
Sachin9
Bunuel
Is the integer n odd
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

(1) 3 or 6. Clearly not sufficient.

(2) TIP:
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.

Sufficient.

Answer: B.

For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html

P.S. You can attach the screenshot of a question directly to the post so that everyone will see it.

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.

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.
Joined: 29 Apr 2018
Posts: 10
Own Kudos [?]: 7 [0]
Given Kudos: 42
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
Bunuel
ajit257
Is the integer n odd?
(1) n is divisible by 3.
(2) 2n is divisible by twice as many positive integers as n.


What do you mean by the second statement.

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.

Bunuel: Is this applicable only on odd numbers?

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 95949
Own Kudos [?]: 665803 [0]
Given Kudos: 87512
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
Expert Reply
s111
Bunuel
ajit257
Is the integer n odd?
(1) n is divisible by 3.
(2) 2n is divisible by twice as many positive integers as n.


What do you mean by the second statement.

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.

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.
Joined: 10 Apr 2018
Posts: 183
Own Kudos [?]: 464 [0]
Given Kudos: 115
Location: United States (NC)
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
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
Joined: 16 Nov 2021
Posts: 467
Own Kudos [?]: 29 [0]
Given Kudos: 5901
Location: United Kingdom
GMAT 1: 450 Q42 V34
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
masuhari
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.


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.
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 35131
Own Kudos [?]: 891 [0]
Given Kudos: 0
Send PM
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: Is the integer n odd? (1) n is divisible by 3 (2) 2n is divisible by [#permalink]
Moderator:
Math Expert
95949 posts