Problem Solving

I know this is on old post, but still I have one doubt...
What will make us stop and mark 6 as the answer when we know that another option 8 is still there?
I mean how can we be certain by plugging in that 6 it is?

GMATGuruNY wrote:

koby_gen wrote:If n=8p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8

We can plug in a value for p.
Let p = 3.
Then n = 8*3 = 24.
Even factors of 24 are 2, 4, 6, 8, 12, 24 = 6 even factors.

The correct answer is D.

n = 8p
or
n = 2^3 * p^1 (the factorised form)
It has a total of (3 + 1)*(1 + 1) = 8 positive factors, and two of them - 1 and p - are not even.
So 8 - 2 =6 are the number of positive even factors.

(D) is the answer.

Illustration:
the number 'n' has the following 8 positive factors:
1, 2, 4, 8, 2p, 4p, 8p, p.

Six of these factors - 2, 4, 8, 2p, 4p, 8p - are even.
So 6 is the answer.

Note: This is value for all values of n. Please avoid the plugging method. This is more convincing.

n = 2^3*p
Total divisors = (3+1)*(1+1) = 8
Out of which 2 are odd (1 and p)
So even divisors = 8-2 = 6

jordan23 wrote:I know this is on old post, but still I have one doubt...
What will make us stop and mark 6 as the answer when we know that another option 8 is still there?
I mean how can we be certain by plugging in that 6 it is?

GMATGuruNY wrote:

koby_gen wrote:If n=8p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8

We can plug in a value for p.
Let p = 3.
Then n = 8*3 = 24.
Even factors of 24 are 2, 4, 6, 8, 12, 24 = 6 even factors.

The correct answer is D.

Only one answer choice can be correct.
If plugging in a different value for p could yield a different number of even factors, then there would be no correct answer.

Oh yea...I thought the question asked us to find the MAXIMUM number of positive even divisors of N.
If that was the case, then how to proceed? How many values to plug in and check?
Or a different approach is needed then?

GMATGuruNY wrote:

jordan23 wrote:I know this is on old post, but still I have one doubt...
What will make us stop and mark 6 as the answer when we know that another option 8 is still there?
I mean how can we be certain by plugging in that 6 it is?

GMATGuruNY wrote:

koby_gen wrote:If n=8p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8

We can plug in a value for p.
Let p = 3.
Then n = 8*3 = 24.
Even factors of 24 are 2, 4, 6, 8, 12, 24 = 6 even factors.

The correct answer is D.

Only one answer choice can be correct.
If plugging in a different value for p could yield a different number of even factors, then there would be no correct answer.

I plugged in values.

That's easier !

Answer is D = 6

(D) 6

a = (2^3)*3

divisors are, (1),(3),2,6,4,12,8,24

Quick method:

Substitute p =3
n =24
number of divisors = 2,4,6,8,12,24

Hence 6 is correct option.

D it is, best way to solve this is to substitute a value for n.

8 has 4 positive divisors: 1 2 4 8
p has 2 positive divisors: 1 & p
Hence, n has 8 positive divisors: 1 2 4 8 & 1p 2p 4p 8p(n)
Choose the even divisors >>> D(6)
This is my method!

If n = 8 p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8

Explanation:
n=8p
p= 3,5,7,11, and so on (p>2)
So, n= 24,40,56, and so on
Even divisors:
24: 2,4,6,8,12,24
40: 2,4,8,10,40
56: 2,4,8,28,56

Answer Choice D is correct with six even factors.

simplest method to solve is to take some prime numbers greater than 2

lets say 3 & 5

8 * 3 = 24
2,4,6,8,12,24 = 6 factors

8*5 = 40
2,4,8,10,20,40 = 6 factors

Answer is D = 6

anshumishra wrote:

nipunkathuria wrote:Would like to add something over here:

N=x^p1*y^q1
where x and y are the prime factors of number N
the number of factors of the number N would be (p1+1)(q1+1)

But in our case the restriction has been that of Even factors.
Hence there will not be any role played by the p(for n=8*p) ...
hence we have 2^3=8 factors - (2^0*p and 2^0 * 1)



hence 6 factors

Right, The part in bold is also the number of odd factors, as I have already posted in a post above :
No. of even factors= All factors - odd factors= (3+1)*(1+1) - (1+1) = 8 - 2 = 6.

Anshu - Can you please guide how to get these numbers 3,1,.....

No. of even factors= All factors - odd factors= (3+1)*(1+1) - (1+1) = 8 - 2 = 6.

nikhilgmat31 wrote: Anshu - Can you please guide how to get these numbers 3,1,.....

No. of even factors= All factors - odd factors= (3+1)*(1+1) - (1+1) = 8 - 2 = 6.

Suppose we have a number expressed as a product of its primes. Let's call this number

2ˣ * pʸ * qᶻ

where p and q are distinct primes and x, y, and z are positive integers. (It could obviously have other distinct prime factors, but once you see how to do this you can see how to generalize to other cases.)

We know that this number has (x + 1)*(y + 1)*(z + 1) factors.

We also know that it has (y + 1)*(z + 1) ODD factors.

So it has (x + 1)*(y + 1)*(z + 1) - (y + 1)*(z + 1) EVEN factors, or (y + 1)*(z + 1)*(x - 1) EVEN factors.

For instance, take the number 2² * 3³ * 5¹. It has (2+1)*(3+1)*(1+1), or 24 factors. (3+1)*(1+1), or 8 of them are odd, so the other 16 are even.

Matt@VeritasPrep wrote:

nikhilgmat31 wrote: Anshu - Can you please guide how to get these numbers 3,1,.....

No. of even factors= All factors - odd factors= (3+1)*(1+1) - (1+1) = 8 - 2 = 6.

Suppose we have a number expressed as a product of its primes. Let's call this number

2ˣ * pʸ * qᶻ

where p and q are distinct primes and x, y, and z are positive integers. (It could obviously have other distinct prime factors, but once you see how to do this you can see how to generalize to other cases.)

We know that this number has (x + 1)*(y + 1)*(z + 1) factors.

We also know that it has (y + 1)*(z + 1) ODD factors.

So it has (x + 1)*(y + 1)*(z + 1) - (y + 1)*(z + 1) EVEN factors, or (y + 1)*(z + 1)*(x - 1) EVEN factors.

For instance, take the number 2² * 3³ * 5¹. It has (2+1)*(3+1)*(1+1), or 24 factors. (3+1)*(1+1), or 8 of them are odd, so the other 16 are even.

Thanks, I remember the formula to find the number of distinct factors as

= 2^3 * p^1
so number of factors will be (3+1) (1+1) = 8

but even factors = 8-2 = 6 Answer D