GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Apr 2019, 01:55

### 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

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

# 3 positive Divisors

Author Message
TAGS:

### Hide Tags

Manager
Joined: 24 Dec 2009
Posts: 167

### Show Tags

22 Mar 2011, 08:46
1
How many integers from 1 to 900 inclusive have exactly 3 positive divisors?

1. 10
2. 14
3. 15
4. 29
5. 30
Director
Status: -=Given to Fly=-
Joined: 04 Jan 2011
Posts: 799
Location: India
Schools: Haas '18, Kelley '18
GMAT 1: 650 Q44 V37
GMAT 2: 710 Q48 V40
GMAT 3: 750 Q51 V40
GPA: 3.5
WE: Education (Education)

### Show Tags

22 Mar 2011, 11:15
Now, if N = a^p x b^q x c^r ... where a, b, c... are prime numbers and p, q, r are positive integers, then number of factors of N is (p+1)(q+1)(r+1)...

In this case, Number of factors is 3 and two of the three factors of a number are 1 and the number itself. The third factor has to be a prime number.

P+1 = 3 or p = 2

Therefore, N = a^2

N < 900

or a^2 < 900

or a < 30

Since P is a prime number, p can assume 10 values (Take every prime number between 1 and 30 viz. 2, 3, 5, 7, 11, 13, 17, 19, 23 & 29)

Thus Ans = 10
_________________
Manager
Status: GMAT in 4 weeks
Joined: 28 Mar 2010
Posts: 157
GPA: 3.89

### Show Tags

22 Mar 2011, 11:20
1
All prime number have 2 divisors. 1 & the number it self.

So only Square of prime numbers (p) will have 3 divisors.
i.e. 1, p, p*p.

So the total number is count of those prime numbers whos square in the given range.

They are 2 3 5 7 11 13 17 19 23 29.

Total count 10 -- > choice A
_________________
If you liked my post, please consider a Kudos for me. Thanks!
Intern
Joined: 28 Mar 2011
Posts: 2
Schools: Haas

### Show Tags

27 Apr 2011, 07:29
Hi:
What about numbers like 35 - 1*3*5 - whcih also have 3 divisors but aren't squares of a prime no?

Retired Moderator
Joined: 20 Dec 2010
Posts: 1784

### Show Tags

27 Apr 2011, 07:59
1
thebigkats wrote:
Hi:
What about numbers like 35 - 1*3*5 - whcih also have 3 divisors but aren't squares of a prime no?

35=7^1*5^1.
Thus it will have (1+1)*(1+1)=4 Divisors

Check the section: "Finding the Number of Factors of an Integer" in the following link.
math-number-theory-88376.html

35 = 1*35
35 = 7*5

Total 4 divisors: 1,7,5,35.

All positive integers have even number of factors(or divisors) except perfect squares.

1: Perfect Square: 1^2: Factors=1 (Number of factors=1, Odd)
2: Not a perfect Square: Factors=1,2 (Number of factors=2, Even)
6: Not a perfect Square: Factors=1,2,3,6 (Number of factors=4, Even)
9: A perfect Square: 3^2; Factors=1,3,9 (Number of factors=3, Odd)
12: Not a perfect Square: Factors=1,2,3,4,6,12 (Number of factors=6, Even)
16: A perfect Square: 4^2; Factors=1,2,4,8,16 (Number of factors=5, Odd)

Q:
How many integers from 1 to 900 inclusive have exactly 3 positive divisors?

3=odd. We know that the number of factors is odd. Thus, the question is asking us to find the perfect squares as only perfect squares can have odd numbers of factors.

But, that's not all. Reason: We need the perfect squares with only 3 factors.
16: A perfect Square: 4^2; Factors=1,2,4,8,16 (Number of factors=5, Odd but not 3)
We see that 16 has 5 factors; we need only 3 factors. The reason is: the base(4) can be factored further resulting in more number of factors.

However, if we choose just the prime number as base and square them. The base can't be factorized further resulting in exactly 3 factors.

2^2=4; Factors: 1,2,4(Count=3) because 2(base) can't be factored.
3^2=9; Factors: 1,3,9(Count=3) because 3(base) can't be factored.
5^2=25; Factors: 1,5,25(Count=3) because 5(base) can't be factored.
13^2=169. Factors: 1,13,169(Count=3) because 13(base) can't be factored.

Thus, we need to find the number of primes between 1 and 900, such that its square shouldn't exceed 900.

1^2<(Prime Number)^2<900
OR
1^2<(Prime Number)^2<(30)^2
OR
1<Prime Number<30

Numbers that fit this range:
2,3,5,7,11,13,17,19,23,29(Count=10)

Ans: "A"
_________________
Director
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 976

### Show Tags

29 Apr 2011, 23:25
It has to be essentially a perfect square of prime number.

29 ^2 = 30^2 - (30+29) <900
Hence the max value of prime number is 29. Counting from 2 to 29 gives 10 prime numbers. Hence A.
Manager
Status: 700 (q47,v40); AWA 6.0
Joined: 16 Mar 2011
Posts: 80
GMAT 1: 700 Q47 V40

### Show Tags

08 May 2011, 03:43
For each factor of n (apart from 1 and itself), there must be a counter-factor. In other words, if x is a factor of n, so is n/x. So effectively for every new factor you add, you have an additional counter-factor coming in unless both the factor and its counter are equal (which is the case where n is a perfect square). Hope this helps!

Regards
Rahul
_________________
Regards
Rahul
Non-Human User
Joined: 09 Sep 2013
Posts: 10567

### Show Tags

13 Aug 2017, 10:47
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.
_________________
Re: 3 positive Divisors   [#permalink] 13 Aug 2017, 10:47
Display posts from previous: Sort by