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How many numbers between 1 and 1000, inclusive have an odd number of f

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New post 04 Apr 2019, 00:34
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[GMAT math practice question]

How many numbers between 1 and 1000, inclusive have an odd number of factors?

A. 10
B. 25
C. 31
D. 64
E. 128

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New post 04 Apr 2019, 01:02
Only square numbers have an odd number of factors

So the question basically asks how many squares numbers are there within 1-1000

We have 30^2=900 and therefore the answer must be a little over 30

(C) 31
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Re: How many numbers between 1 and 1000, inclusive have an odd number of f  [#permalink]

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New post 07 Apr 2019, 18:28
MathRevolution wrote:
[GMAT math practice question]

How many numbers between 1 and 1000, inclusive have an odd number of factors?

A. 10
B. 25
C. 31
D. 64
E. 128


Only perfect squares have an odd number of factors. Thus, the perfect squares between 1 and 1000, inclusive, are 1^2, 2^2, …, 31^2, so there are a total of 31 numbers.

Answer: C
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Re: How many numbers between 1 and 1000, inclusive have an odd number of f  [#permalink]

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New post 07 Apr 2019, 18:41
>

Recall the property that an integer with an odd number of factors is the perfect square of an integer and only has even exponents in its prime factorization.
If \(n = p^aq^br^c\), then \(n\) has \((a+1)(b+1)(c+1)\) factors, where \(p, q\) and \(r\) are different prime numbers, and \(a, b\) and \(c\) are positive integers. So \(a + 1, b + 1, c + 1\) must be odd numbers in order for \(n\) to have an odd number of factors. This implies that \(a, b\) and \(c\) are even numbers and \(n\) is the perfect square of an integer.
There are \(31\) perfect squares between \(1\) and \(1000\), inclusive, because \(31^2 = 961 < 1000\) and \(32^2 = 1024 > 1000.\)

Therefore, the answer is C.
Answer: C
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Re: How many numbers between 1 and 1000, inclusive have an odd number of f   [#permalink] 07 Apr 2019, 18:41
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