Lesson Problems on divisors of a given number

Problems on divisors of a given number

Problem 1

If the number has exactly 3 factors, it means that the number is the square of a prime number: N = .

Then it has the factors 1 (one), p and .


Indeed, if the number is a prime number, it has only TWO factors: 1 (one) and itself.

If the number is not prime and is not the square of a prime, then it has more than 3 factors.


Therefore, to answer the problem's question, we must take the square of the largest two-digit prime number, which is  = 9409.


Answer. The number under the question is 9409.

Problem 2

1.  For integer number N = ,  where p is a prime number and  is an integer exponent (index), 

    the number of divisors is .


    You can easily check it:  the divisors  are  1, p, , . . . , .



2.  For integer number  N = ,  where p, q, . . . , r are prime divisors and , , . . . ,  are integer exponents (indexes)  

    the number of divisors is  .



3.  From these facts, you can easily obtain the answer.


    Notice that  6 = 2*3.


    It is easy to list those divisors:  1, 2, 4, 3, 6, 12.


Answer.    The smallest positive integer that has exactly  6  divisors is   12 = .

Problem 3

A number, which has exactly 9 factors,  is  EITHER  the 8th degree of a prime number  OR  is
the square of an integer positive number  " n ",  which is the product of two prime numbers.


Regarding the first case, the number can be only   = 256,  which falls into the given interval.


Regarding the second case, we look into the open interval from   ~ 14.1  to   ~ 22.4

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


We find there ONLY THREE such numbers  15, 21 and 22 that produce the squares 

         = 225,   = 441,  and   = 484.


So, the  ANSWER  to the problem's question are the numbers  256,  225,  441  and  484.

Problem 4

In order to facilitate your finding of the two distinct primes, notice that the given numbers 
represent an arithmetic progression with the common difference of 


    2997 - 1998 = 999 = 3*333 = 3*3*111 = 3*3*3*37.


Next, since the number 1998 is divisible by 3 and by 37 (check it directly !),
it implies that the given numbers ALL have 3 and 37 as their prime common divisors,

and DO NOT HAVE ANY OTHER prime common divisors.


Hence, the problem wants you find the sum of 3 and 37, which is 3+37 = 40.


ANSWER.  The sum of these two primes is 40.