SOLUTION: Find all numbers between 200 and 500 which have exactly 9 factors.

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Question 1182879: Find all numbers between 200 and 500 which have exactly 9 factors.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
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Find all numbers between 200 and 500 which have exactly 9 factors.
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Notice that 9 = ,  and it tells us that the south number N is the square 
of the integer positive number  " n ",  which is the product of two prime numbers.


So, we look into the open interval from   ~ 14.1  to   ~ 22.4


and search there the numbers  " n "  that are the products of two different prime numbers.


We find there ONLY THREE such numbers  15, 21 and 22,  so the  ANSWER  to the problem's question 

are the numbers   = 225,   = 441.  and   = 484.

Solved.


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About the relevant theorem from the number theory on the number of divisors to any integer positive number see the lesson

    - Problems on divisors of a given number

in this site.


This theorem states
    For integer number  N = ,  where p, q, . . . , r are prime divisors  

    and  ,  , . . . ,  are integer exponents  (indexes),  the number of divisors is  .



Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


I will assume, since you are working a problem like this, that you know how to find the number of factors of a given number.

To get exactly 9 factors, the number can be either
(a) -- the 8th power of a single prime p, because 8+1=9; or
(b) -- the product of the squares of two primes p and q, because (2+1)(2+1)=9

For case (a), 2^8=256, which is in the required range.

For case (b), we need the number to be the product of two primes; and the number must be between sqrt(200) and sqrt(500) -- or 15 to 22, inclusive. The numbers that satisfy those conditions are 15=3*5, 21=3*7, and 22=2*11.

ANSWER:
256 = 2^8
225 = 15^2
441 = 21^2
484 = 22^2


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