700+ question

[Night reader]

In the set of integers from 0 to 700, how many possible values have more than 3 positive divisors

A. 540
B. 566
C. 610
D. 640
E. 680

[anshumishra]

In the set of integers from 0 to 700, how many possible values have more than 3 positive divisors

A. 540
B. 580
C. 610
D. 640
E. 680

Lets calculate the nos. which have less than 4 positive divisors.

1 positive divisor : 1

2 positive divisors : 125 (prime numbers less than 700. I am sure we don't need to remember this, so either there could be a better approach to solve this problem or in real exam the number would be far less than 700. I used the prime table from here : https://www.tutorvista.com/math/prime-numbers-list-all)

3 positive divisors : 9 (All p^2 such that p is prime and p^2 < 700)


Total = 135

Any other number will be a multiple of at least one power of two primes : p^n*q^m; n,m > =1 , which will have (1+m)*(1+n) >= 4 positive divisors

Hence, number of numbers in the set having more than 3 positive divisors = 701 - 135 = 566.

Answer is B

[diebeatsthegmat]

In the set of integers from 0 to 700, how many possible values have more than 3 positive divisors

A. 540
B. 580
C. 610
D. 640
E. 680

i dont believe this is a GMAT question because it kinda takes time to solve and find the answer ,

[Night reader]

In the set of integers from 0 to 700, how many possible values have more than 3 positive divisors

A. 540
B. 580
C. 610
D. 640
E. 680

I am posting my approach here, although my answer doesn't match with the options , I am sure people will find what I have missed :

Lets calculate the nos. which have less than 4 positive divisors.

1 positive divisor : 1

2 positive divisors : 125 (prime numbers less than 700. I am sure we don't need to remember this, so either there could be a better approach to solve this problem or in real exam the number would be far less than 700. I used the prime table from here : https://www.tutorvista.com/math/prime-numbers-list-all)

3 positive divisors : 9 (All p^2 such that p is prime and p^2 < 700)


Total = 135

Any other number will be a multiple of at least one power of two primes : p^n*q^m; n,m > =1 , which will have (1+m)*(1+n) >= 4 positive divisors

Hence, number of numbers in the set having more than 3 positive divisors = 700 - 135 = 565.

Please let me know if I have missed anything or there is some flaw in my approach

Anshu excellent work!

{0... 700} makes 701 integers => 701 - 1 integer for '0' - 1 integer for '1' - 9 integers for 'powers of two of primes' - 125 integers for 'primes' = 701-136 = 565
and I have to correct the choice B => makes B correct answer with 565.

thanks Anshu for great elaboration on this problem.

[anshumishra]

In the set of integers from 0 to 700, how many possible values have more than 3 positive divisors

A. 540
B. 580
C. 610
D. 640
E. 680

I am posting my approach here, although my answer doesn't match with the options , I am sure people will find what I have missed :

Lets calculate the nos. which have less than 4 positive divisors.

1 positive divisor : 1

2 positive divisors : 125 (prime numbers less than 700. I am sure we don't need to remember this, so either there could be a better approach to solve this problem or in real exam the number would be far less than 700. I used the prime table from here : https://www.tutorvista.com/math/prime-numbers-list-all)

3 positive divisors : 9 (All p^2 such that p is prime and p^2 < 700)


Total = 135

Any other number will be a multiple of at least one power of two primes : p^n*q^m; n,m > =1 , which will have (1+m)*(1+n) >= 4 positive divisors

Hence, number of numbers in the set having more than 3 positive divisors = 700 - 135 = 565.

Please let me know if I have missed anything or there is some flaw in my approach

Anshu excellent work!

{0... 700} makes 701 integers => 701 - 1 integer for '0' - 1 integer for '1' - 9 integers for 'powers of two of primes' - 125 integers for 'primes' = 701-136 = 565
and I have to correct the choice B => makes B correct answer with 565.

thanks Anshu for great elaboration on this problem.

Cool ! I think 0 has infinite divisors (as it is divisible by any number). If that is right, then the answer should be 701-135 = 566.
What do you say?

[Night reader]

this came to my mind too. I was discounting '0' mistakenly, as it has endless positive divisors (factors) always.

[anshumishra]

this came to my mind too. I was discounting '0' mistakenly, as it has endless positive divisors (factors) always.

Great, Night reader ! Nice question !!!
I have edited my first post accordingly.