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

Question

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

Bunuel · Accepted Answer

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.

EMPOWERgmatRichC · Answer

Hi,

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.

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

KarishmaB · Answer

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.

messi16 · Answer

Quick little shortcut. Numbers without any odd divisor greater than or equal to 1 are just powers of 2. List powers of 2 up until 100. 2^7 = 128, thus our answer must be 2^6. Thus, 6 is the answer.

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

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