How many positive three digit integers have exactly 7 factor

Question

How many positive three digit integers have exactly 7 factors?

(A) none 
(B) one 
(C) two 
(D) three 
(E) more than three

Reinfrank2011 · Accepted Answer

Theory: 
The total number of factors of integer n with prime factorization  n=a^x*b^y*c^z  is equal to  (x+1)(y+1)(z+1) ; this includes 1 and  n . In other words, to find the total number of factors of integer  n , make the prime factorization of  n  and then take the product of (each exponent +1).

Back to the question: 
Assume we have a generic integer  n  with prime factorization  n=a^x*b^y*c^z

We are told that the total number of factors is 7; this implies that  (x+1)(y+1)(z+1) = 7

Now, when we inspect that equation, we see that 7 is a PRIME number, which means it's only factors are "one and itself". Thus, the only way to have the product equal 7 is for one of  x, y, z  to equal 6 while the other two are zero; that way we get the required  (6+1)(0+1)(0+1)=7*1*1=7 .

The key inference is thus that our generic integer  n  has only one prime factor whose exponent is 6.

Now, if  n=a^6 , we just need to see how many prime numbers  a  fit the given conditions:  100   ≤   a^6   ≤    999

2 ^6= 64  --> no good; only two digits
  3 ^6 = 27^2=729  --> good; three digits
  5 ^6 = 625*5*5=3125*5  --> no good; four digits and counting

Thus, there is only ONE. Answer B.

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v1rok · Answer

(A) None.

This was actually pretty easy once I realized that exactly 7 factors can happen only if these seven factors are prime numbers.

So I started writing down 7 smallest prime numbers:

2,3,5,7,11,13,19

If you start multiplying them the product pretty quickly becomes larger than 3-digit number.

sgrover · Answer

You are missing 1 and the no. itself.. so,for that we need to consider 5 prime #s by ur approach, which yields the same answer though..

kevincan · Answer

Hint: What is the smallest positive integer with EXACTLY 7 factors?

sharadGmat · Answer

Answer would still be A.. 
1*2*3*5*7*11*13 =30030 , isn't it ?

kevincan · Answer

Another clue: If an integer has exactly 7 factors, how many  prime  factors must it have?

15=5*3 has 4 factors , 5^0*3^0 , 5^0*3^1, 5^1*3^0 and 5^1,5^1

kevincan · Answer

Right! 15 has 4 factors but only 2 prime factors.

sharadGmat · Answer

Anyone has a say on this ? I am not sure what's correct.. but somehow used to thinking that 7 factors mean minimum 7 prime factors where they don't repeat..

kevincan · Answer

1000 has two prime factors, but 16 factors!

acfuture · Answer

I think I found one... but this question took me way tooooooo much time. I just could not forget about this problem and it was bothering me...

I think the question is unique in the sense that it wants exactly 7 factors. Which means that it has to be some form of a perfect square.
ie. 13*13 = 169, this has 3 factors = 1, 13, 169 

So my first guess was 25*25 = 625, but I was only able to find the following factors; 1,5,25,125,625.

but how about 27*27 = 729
I was able to find 7 factors for 729 
1x729
3x249
9x81
27x27

therefore: 1, 3, 9, 27, 81, 249, 729

in advance, if I missed a factor in 729...or if I totally mis-understood the question disregard this message... 

kevin, any more help?

kevincan · Answer

If p1 and p2 are two different prime numbers, how many factors does
p1*p2 have? 4=2*2

What about p1*p2^2 ? 2*3=6 

and p1^n1*p2^n2 if n1 and n2 are postive integers? (n1+1)(n2+1)

So, if an integer has seven (a prime number) factors, it must be that...

jimwild · Answer

7 factors, means the number is a perfect square, also means that 7 is a prime so our number have power like (x+1) = 7
so number has power of 6

2pow6= 64 not 3 digits
3pow6= 27 x 27 = cool
5pow6 = not 3 digit integer

so only one number

Pls give me kudos so i can unlock the free cat

PKay · Answer

Property of prime nos- Any prime no. has only 2 factors-1 and itself.
Eg. For X-a prime no., factors are 1 & x
When you multiply the prime no with itself i.e, square the number, it has three factors
x^2--factors are 1, x, x^2-3 factors.
When you cube it, it has 4 factors-1,x^, x^2 & x^3.
for x^n, we have n+1 factors.

In the question, n+1=7=> n=6. and the requite no ishould be in the form x^6.
Now try with the first prime-2
2^6=64- Not a 3 digit no.
3^6=729-YES. It's a 3 digit no.
5^6=625*625 Big no....No need to check further.
 Hence, there is one and only one three-digit no (729) which has 7 factors.

iammohakjain · Answer

I used logic to find this answer.

Prob:  Need all 3 digit numbers that exactly have 7 factors.
 Soln:  For any number to have an odd number of factors, the number has to be a perfect square.   Hence I limited my set starting from 10^2 to 31^2 i.e. {10^2, 11^2, 12^2, 13^2.......... 31^2}. 
  - Why? Because I need to evaluate 3 digit numbers that are perfect squares

Step1 . You can safely remove all the prime numbers from the above set ( {Set} - {11,13,17,19,23,29,31})
 ---since all prime numbers would exactly have 3 factors eg: 11^2 --> 1, 11, 121

Step2 . You can safely remove all numbers from the above set  which have 2 or more prime factors  ({Set} - {10,12,14,15,16,18,20,21,22,24,26,28,30})
 ---since all numbers having 2 or more prime factors would  lead to the 3 digit number having more than 7 factors  
eg: 22^2--> 1, 2,4,11,22,44,121,242,484. 
eg: 15^2--> 1, 3,5,9,15,25,45,75,225.

Step3 . You are left with 25, 27 - 
25 ^2- 5 factors
27^2- 7 factors

Archit3110 · Answer

key point to know is that only square of prime numbers have odd number of factors 
in this case (x^2)^3 ; x^6 ; so x has to be prime number which would be 3 as only 3^6 is a 3 digit number ...
option B

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How many positive three digit integers have exactly 7 factor

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