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

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Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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05 Jan 2010, 21:35
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Is the integer n odd?

(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n
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Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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06 Jan 2010, 03:56
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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.

For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html
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Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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12 Dec 2010, 04:12
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ajit257 wrote:
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.

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)...

(Check out the post: http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-writing-factors-of-an-ugly-number/ for complete explanation of this formula)

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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14 Mar 2010, 02:28
Bunuel wrote:
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.

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

Nice explanation. Didnt know abt this rule.
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Re: Is the integer n odd  [#permalink]

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10 Dec 2010, 05:42
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ajit257 wrote:
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.
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13 Dec 2010, 05:36
Thanks for the tip.
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Re: Prep Question....is n odd?  [#permalink]

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16 Apr 2012, 00:12
some2none wrote:
Bunuel wrote:
Is the integer n odd

(When even number is doubled, 2n has 1.5 more factors as n.)
Sufficient.

I'm afraid the above statement is incorrect.
Let's take an even number, 4.
The number of factors = 3 (1, 2, 4)
If the statement above is true, then the 2n (4 * 2 = 8) will have 1.5 more factors, that is 3 + 1.5 = 4.5 factors.
Instead the number of factors for 8 = 4 (1, 2, 4, 8)

I tried to find out certain relationship between a even number and its double:

If an even number is
E = 2^x * p^a* q^b* r^c........
2E= 2^(x+1) * p^a* q^b* r^c........

The number of factors that 2E has more than E is

(a+1)*(b+1)* (c+1)*.........

PS: request Bunuel to check whether the observation i derived is correct.

thanx
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Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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04 May 2012, 10:40
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.
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Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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04 May 2012, 10:56
pgmat wrote:
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.
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Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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22 Jan 2013, 23:24
(1)

$$n = 3*I$$

If I is even then n is EVEN...
If I is odd then n is ODD...

INSUFFICIENT!

2)

Let f(n) be the number of factors of n...
Let f(2n) be the number of factors of 2n...

Now let f(n) be some X ==> f(n) = R

The statement says that f(2n) = 2*f(n) = 2*R...

When n is EVEN:

$$2n = 2^(p+1) * {n/2^{p}}$$where p is the number factor of 2 in n...
$$f(2n) = f(2^{p+1}) * f( {n/2^{p}} ) = (p+2) * R/{p+1}$$

let p=1: f(2n) = 3 * R/2
let p=2: f(2n) = 4 * R/3
let p=3: f(2n) = 4 * R/4

When n is ODD:

2n = 2^1 * n
f(2n) = 2 * f(n) = 2*R

SUFFICIENT that n is ODD!

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Re: Prep Question....is n odd?  [#permalink]

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24 Jan 2013, 22:53
Bunuel wrote:
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.

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.

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Re: Prep Question....is n odd?  [#permalink]

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24 Jan 2013, 23:48
2
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Sachin9 wrote:
Bunuel wrote:
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.

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.
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Updated on: 13 May 2013, 09:51
coolpintu wrote:
Is the integer n odd?
(1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n

Statement 1. It states that n is divisible by 3. Doesn't really prove conclusively that n is odd. n could be divisible by 6. Hence not sufficient.

Statement 2. It states that 2n has twice the factors as n. Now, assuming n as odd, 2n would have the below factors {1,n,2n,2...} and n would have {1,n} as its factors. The extra factors appear due to the fact that 2 was multiplied. In case n is not a prime and has odd factors, like 3 and 7 in it, the value will double since 2 is multiplies to individual odd factors leading to double the factors again. For example if n = 3, 2n = 6, the factors double. It doesn't happen in case of n = 4. Once can similarly test the logic for complex numbers like 35.

Hence sufficient.

Hope I my explanation is satisfactory and correct!

Regards,
Arpan
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Originally posted by arpanpatnaik on 13 May 2013, 09:17.
Last edited by arpanpatnaik on 13 May 2013, 09:51, edited 1 time in total.
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13 May 2013, 09:34
St1: N is divisible by 3
N can be 6 (even) or 9 (odd)
Hence Insufficient

St2: 2n is divisible by twice as many divisors as N
if N = 3, divisors = {1,3} that is # divisors = 2
2N = 6 = {1,2,3,6} and divisors = 4
Good solution.

Lets check for an even number also.
N = 6 {1,2,3,6} and #divisors = 4
2N = 12 = {1,2,3,4,6,12} and #divisors = 6
Not a good solution as #divisors for 2N is not twice #divisors for N

Hence sufficient to say that if 2n is divisible by twice as many divisors as N, then N = ODD
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13 May 2013, 22:14
coolpintu wrote:
Is the integer n odd?
(1) n is divisible by 3 (2) 2n is divisible by twice as many positive integers as n

statement 1:- n is divisible by three.If n is 6 it is even and divisible by three .If n is 9 it is divisible by three. So A alone is not sufficient. AD out

statement 2:- number of factors for n = $$a^p * b^q *c^r$$ ....where a,b,c.... are prime numbers and p,q,r.... are positive integers is given by$$(p+1)*(q+1)*(r+1)$$.... Now 2n has twice as many positive integers as n. This is only possible if n is odd.Why??IF n is odd then n doesn't have a 2 .But for 2n n will have a 2 and we will get (1+1) in our formula which doubles the number of factors.But if n is even then we will already have 2 .Lets say the number if 2's is p.The for 2n we will have p+1 2's which doesn't double the number of factors. So n must be odd for this condition to be true.So statement 2 is sufficient.

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07 Jan 2014, 01:11
1. is clearly insufficient. n could be zero for instance and 0 is both even and a multiple of 3.

2. I figured this out pretty quickly, I started off visualizing n as a prime number because prime numbers have 2 factors. Moreover 2 is a prime number itself and has two factors as well, then I tried to visualize the same thing assuming n to be an odd number. Generally n must not be an even number because any even number will cause our "twice as many positive integer" rule to fall.

since we are talking about factors: squares of prime numbers have 3 divisors -the prime number itself, its square, and one.

Sufficient.
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Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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14 Jan 2017, 05:32
Great Official Question.
Here is what i did in this one =>
We are asked if n is odd or not.
We are told that n is an integer.
Statement 1->
n/3=integer.
E.g=> n=3 or n=6 etc.
Hence not sufficient.
Statement 2->
If n is odd => The number of factors of 2n will always be 2 times the number of factors of n.
This is because any odd number will just have odd divisors.It cannot have any even divisor as 2 is not its prime factor.
After we multiple 2 to n => The numb of factors will get doubled as for every odd factor there would be an even factor.

Hence n is odd
Thus-> This statement is sufficient.

Hence B.

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Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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21 May 2018, 06:09
VeritasPrepKarishma wrote:
Now assume the case where n is already even:

$$n = 2^2*3^2*5^3$$ i.e. odd

Just so that my understanding is correct, you meant "i.e. even" and not "i.e.odd". Is my understanding correct?
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Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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21 May 2018, 10:10
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Nived wrote:
VeritasPrepKarishma wrote:
Now assume the case where n is already even:

$$n = 2^2*3^2*5^3$$ i.e. odd

Just so that my understanding is correct, you meant "i.e. even" and not "i.e.odd". Is my understanding correct?

Hello

I think that was a typo. You could actually ignore the phrase "i.e.odd".. and the explanation will make perfect sense.
And yes your understanding is correct - the number is obviously even.
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Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is  [#permalink]

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21 May 2018, 22:10
Nived wrote:
VeritasPrepKarishma wrote:
Now assume the case where n is already even:

$$n = 2^2*3^2*5^3$$ i.e. odd

Just so that my understanding is correct, you meant "i.e. even" and not "i.e.odd". Is my understanding correct?

Yes, even.
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Re: Is the integer n odd? (1) n is divisible by 3. (2) 2n is &nbs [#permalink] 21 May 2018, 22:10

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