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How many positive integers, from 2 to 100, inclusive, are no

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How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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New post 09 Feb 2011, 17:31
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How many positive integers, from 2 to 100, inclusive, are not divisible by odd integers greater than 1?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 50
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Re: How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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New post 09 Feb 2011, 17:38
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loveparis wrote:
87. How many positive integers, from 2 to 100, inclusive, are not divisible by odd integers greater than 1?
(A) 5
(B) 6
(C) 8
(D) 10
(E) 50


Integer not to have an odd factor greater than 1 should be of the type \(2^n\), for some positive integer \(n\). So we are given that \(2\leq{2^n}\leq{100}\) --> n can take the following values: 1, 2, 3, 4, 5, and 6, which means that the number itself could be 2, 4, 8, 16, 32, or 64.

Answer: B.
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Re: How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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New post 17 Dec 2013, 02:36
The no will be of the form 2^n to achieve this. Since any other form will have odd no in the prime factorization.
Hence we need to find solution of n for equation - 1< 2^n < 101.
2^6 = 64, 2^7 - 128
Hence n can take values from 1 to 6.
Hence Ans- (B) 6
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Re: How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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New post 10 May 2015, 09:24
Can someone please help me with this question, I am truly not understanding it.
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Re: How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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New post 10 May 2015, 11:53
We need to find no of Integers between 2 and 100 ( including 2 & 100) that are not divisible by odd Numbers > 1 i.e by 3,5,7,...

So the numbers need to be of the Form 2^N where N can take values of 1,2,3,4,5,6 i.e. Number of Integers are 6.
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Re: How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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New post 10 May 2015, 16:52
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Hi theandrizzle,

This question is based on a subtle Number Property rule. If you don't immediate recognize the NP rule, then that's okay, but you'll have to do some work to discover that 'pattern' involved....

We're asked to think about the numbers 2 to 100, inclusive. To start, there's NO way that the GMAT would ask us to truly think about each of these numbers individually, so you should be thinking that there's a pattern involved.

Now to the specifics: which of these numbers are NOT divisible by an odd integer that is greater than 1???

Let's start at the first number and work our way "up" until the pattern becomes clear:

2 - this is NOT divisible by any odd integers, so this "fits" what we're looking for...
3 - this IS divisible by an odd integer (3), so it's out
4 - this is NOT divisible by any odd integers, so this "fits"
5 - this IS divisible by an odd integer (5), so it's out
6 - this IS divisible by an odd integer (3), so it's out
7 - this IS divisible by an odd integer (7), so it's out
8 - this is NOT divisible by any odd integers, so this "fits"

Now, looking at the numbers that "fit", we have 2, 4 and 8.....that's 2^1, 2^2 and 2^3....that MUST be the pattern involved, so we can use this against the rest of the question to find the other values that "fit":
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128, but that's outside the range that we were given. Thus, there are 6 values that "fit" what we're looking for.

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Re: How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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New post 10 May 2015, 23:32
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theandrizzle wrote:
Can someone please help me with this question, I am truly not understanding it.


Note that all integers are made up of some combination of prime factors. That is why we can prime factorize them.
e.g. 10 = 2*5
12 = 2*2*3
23 = 23
27 3*3*3
etc

Which are the odd integers greater than 1?
3, 5, 7, 9, 11, 13, 15, 17, ...

Note that, other than 2, this list will contain all prime numbers (since all prime numbers except 2 are odd). So basically, the number that is acceptable should have NO prime factor other than 2. In how many ways can you make a number which is between 2 and 100 using only 2s?
2
2*2 = 4
2*2*2 = 8
2*2*2*2 = 16
2*2*2*2*2 = 32
2*2*2*2*2*2 = 64
Next multiplication by 2 will yield a number greater than 100. So you have 6 such numbers.
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Re: How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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New post 11 May 2015, 08:52
Thank you Rich & Karishma! Very helpful! Boom.
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Re: How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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New post 12 Dec 2015, 11:38
it might be a very difficult question at first look...but it is a really easy one...
i started with finding the total numbers ...100-2+1 = 99 numbers..then tried to find a pattern..but then...it popped in my mind that 2 at any power will not be divisible by odd integers. thus, only the numbers 2, 4, 8, 16, 32, 64 will not be divisible by odd integers. in total, we have 6 numbers.
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Re: How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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New post 24 Dec 2015, 21:45
question is asking for how many numbers between 2 and 100 are even AND do not have any odd number as a factor.
Note that none of the odd numbers (including primes) should be counted because every number is a factor of itself (so are odd numbers and prime numbers),
Essentially question is asking for how many numbers are divisible by 2 and 2 alone.
those are 2^n , with boundaries 2 and 100 ; 2, 4, 8 , 16, 32, 64 .. (not 128) . total # = 6

Answer : B
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Re: How many positive integers, from 2 to 100, inclusive, are no  [#permalink]

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