3 positive Divisors

Hi:
What about numbers like 35 - 1*3*5 - whcih also have 3 divisors but aren't squares of a prime no?

please help

35=7^1*5^1.
Thus it will have (1+1)*(1+1)=4 Divisors

Check the section: "Finding the Number of Factors of an Integer" in the following link.
math-number-theory-88376.html

35 = 1*35
35 = 7*5

Total 4 divisors: 1,7,5,35.

All positive integers have even number of factors(or divisors) except perfect squares.

1: Perfect Square: 1^2: Factors=1 (Number of factors=1, Odd)
2: Not a perfect Square: Factors=1,2 (Number of factors=2, Even)
6: Not a perfect Square: Factors=1,2,3,6 (Number of factors=4, Even)
9: A perfect Square: 3^2; Factors=1,3,9 (Number of factors=3, Odd)
12: Not a perfect Square: Factors=1,2,3,4,6,12 (Number of factors=6, Even)
16: A perfect Square: 4^2; Factors=1,2,4,8,16 (Number of factors=5, Odd)

Q:
How many integers from 1 to 900 inclusive have exactly 3 positive divisors?

3=odd. We know that the number of factors is odd. Thus, the question is asking us to find the perfect squares as only perfect squares can have odd numbers of factors.

But, that's not all. Reason: We need the perfect squares with only 3 factors.
16: A perfect Square: 4^2; Factors=1,2,4,8,16 (Number of factors=5, Odd but not 3)
We see that 16 has 5 factors; we need only 3 factors. The reason is: the base(4) can be factored further resulting in more number of factors.

However, if we choose just the prime number as base and square them. The base can't be factorized further resulting in exactly 3 factors.

2^2=4; Factors: 1,2,4(Count=3) because 2(base) can't be factored.
3^2=9; Factors: 1,3,9(Count=3) because 3(base) can't be factored.
5^2=25; Factors: 1,5,25(Count=3) because 5(base) can't be factored.
13^2=169. Factors: 1,13,169(Count=3) because 13(base) can't be factored.

Thus, we need to find the number of primes between 1 and 900, such that its square shouldn't exceed 900.

1^2<(Prime Number)^2<900
OR
1^2<(Prime Number)^2<(30)^2
OR
1<Prime Number<30

Numbers that fit this range:
2,3,5,7,11,13,17,19,23,29(Count=10)

Ans: "A"

It has to be essentially a perfect square of prime number.

29 ^2 = 30^2 - (30+29) <900
Hence the max value of prime number is 29. Counting from 2 to 29 gives 10 prime numbers. Hence A.

For each factor of n (apart from 1 and itself), there must be a counter-factor. In other words, if x is a factor of n, so is n/x. So effectively for every new factor you add, you have an additional counter-factor coming in unless both the factor and its counter are equal (which is the case where n is a perfect square). Hope this helps!

Regards
Rahul

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