How many numbers between 1 and 100, inclusive, have exactly 5 positive factors?

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Re: How many numbers between 1 and 100, inclusive, have exactly 5 positive factors?

by Brent@GMATPrepNow » Thu Mar 26, 2020 5:33 am

BTGModeratorVI wrote: ↑
Wed Mar 25, 2020 6:28 am
How many numbers between 1 and 100, inclusive, have exactly 5 positive factors?

A. 0
B. 1
C. 2
D. 9
E. 10

Answer: C
Source: Veritas Prep

First note that most positive integers have an EVEN number of positive factors.
Only the SQUARES of integers have an ODD number of positive factors.
So, we need only consider the following squares of integers: 1, 4, 9, 16, 25, 36, . . . . 81, 100

Most of these squares don't have 5 positive factors. Let's check...
Factors of 1: 1 NO
Factors of 4: 1, 2, 4 NO
Factors of 9: 1, 3, 9 NO
Factors of 16: 1, 2, 4, 8, 16 PERFECT!
Factors of 25: 1, 5, 25 NO
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 NO
Factors of 49: 1, 7, 49 NO
Factors of 64: 1, 2, 4, 8, 16, 32, 64 NO
Factors of 81: 1, 3, 9, 27, 81 PERFECT!
Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100 NO

Answer: C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Re: How many numbers between 1 and 100, inclusive, have exactly 5 positive factors?

by Scott@TargetTestPrep » Fri Mar 27, 2020 6:20 am

BTGModeratorVI wrote: ↑
Wed Mar 25, 2020 6:28 am
How many numbers between 1 and 100, inclusive, have exactly 5 positive factors?

A. 0
B. 1
C. 2
D. 9
E. 10

Answer: C
Source: Veritas Prep

We should recall the rule we can use to determine the total number of factors:

For a positive integer n (where n > 1),

i) if the prime factorization of n is p^a (where p is a prime), then the total number of factors of n is equal to a + 1.
ii) if the prime factorization of n is p^a * q^b (where p and q are distinct primes), then the total number of factors of n is equal to (a + 1)(b + 1).
(Note: The concept can be extended to the prime factorization of n when n has 3 or more distinct prime factors.)

Since 5 is a prime number, the number(s) we seek can’t have more than 1 prime factor. For example, if it has two distinct prime factors, then, according to the rule, the total number of factors is equal to (a + 1)(b + 1). But (a + 1)(b + 1) can’t be equal to 5 since 5 is a prime number..

Thus, we see that the number must have only 1 prime factor and is of the form p^a. Since, according to the rule, the total number of factors is equal to a + 1, we can set a + 1 = 5 and obtain a = 4. Now let’s check some values of p (keep in mind that p is a prime):

If p = 2, then p^4 = 2^4 = 16 (which is between 1 and 100)
If p = 3, then p^4 = 3^4 = 81 (which is between 1 and 100)
If p = 5, then p^4 = 5^4 = 625 (which is more than 100)

Thus, there are only two integers between 1 and 100 (6 and 81) that have exactly 5 positive factors.

Answer: C

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