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by GMATGuruNY » Wed Aug 06, 2014 2:56 am
sapuna wrote:How many positive integers less than 100 have exactly 4 odd factors but no even factors ?

1) 13

2) 14

3) 15

4) 16

5) 17
Case 1: The product of any PAIR OF DISTINCT PRIME NUMBERS will have exactly 4 odd factors.
To illustrate:
If x=3 and y=5, then xy = 15.
The factors of 15 are 1, 3, 5, and 15, for a total of 4 odd factors.

Options for xy that are less than 100:
3*5, 3*7, 3*11, 3*13, 3*17, 3*19, 3*23, 3*29, 3*31
5*7, 5*11, 5*13, 5*17, 5*19
7*11, 7*13.
Total options = 16.

Case 2: The cube of an odd prime number will have 4 odd factors
Options less than 100:
3³ = 27, which has as factors 1, 3, 9, and 27.
Total options = 1.

Total = Case 1 + Case 2 = 16+1 = 17.

The correct answer is E.
Last edited by GMATGuruNY on Wed Aug 06, 2014 11:34 am, edited 2 times in total.
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by sapuna » Wed Aug 06, 2014 3:32 am
I think you made a small mistake here mate as 3x37 > 100.


3 can be paired with 8 numbers - 5, 7 , 11 , 13, 17, 23, 29 31. inclusive. The next prime 37 x 3 will be more over 100.

5 can be paired with 4 numbers - 7 , 11 , 13 and 17 .

7 with 2 numbers - 11 and 13.

This is a total of 14 numbers. The correct answer given is 15. Maybe 3^3 = 27. 27 can be factored into 1 , 3, , 3 and 3. And this is the 15 number. Is this possible ?
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by GMATGuruNY » Wed Aug 06, 2014 4:47 am
sapuna wrote:I think you made a small mistake here mate as 3x37 > 100.
Good catch.
Please check my amended post above.
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by sapuna » Wed Aug 06, 2014 8:12 am
GMATGuruNY wrote:
sapuna wrote:I think you made a small mistake here mate as 3x37 > 100.
Good catch.
Please check my amended post above.
Ah, I tihnk together we managed to solve it. I forgot 19 :D So , the answer that the program says is correct actually is wrong , correct :D Meaning our answer is right ?
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by Brent@GMATPrepNow » Wed Aug 06, 2014 8:12 am
This one doesn't smell very "GMAT-y". What's the source?

I ask because the GMAT typically doesn't have questions that can be solved by only brute force (e.g., listing all of the possible pairs of primes). I'm not saying that real GMAT questions can never be solved by brute force; I'm saying that it's unlikely that the ONLY valid approach to a GMAT question involves brute force.

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by sapuna » Wed Aug 06, 2014 8:26 am
It`s a program called 800 score.

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by Matt@VeritasPrep » Wed Aug 06, 2014 4:18 pm
Not to defend 800score -- I'm not a fan -- but I think this one is fair enough.

Building on Mitch's approach, we have two cases.

Case 1: Odd Cube > 1
There's only one in range (27), so we have one of these.

Case 2: Product of two odd primes, with the product less than 100
Here I'd take each odd prime and look at possibilities.

If the smaller odd prime is 3, we could multiply it by any other prime from 5 to 31.
If the smaller odd prime is 5, we could multiply it by any other prime from 7 to 19.
If the smaller odd prime is 7, we could multiply it by any other prime from 11 to 13.

Counting primes, the first case gives us 9 possibilities, the second case gives us 4 possibilities, and the third case gives us 2 possibilities. This is a total of 9+5+2 = 16.

Combining the two cases gives us 16 + 1, or 17.

This sort of prime counting seems in bounds to me, and actually rather clever (I love the odd cube, that's a real monkeywrench) ... though the GMAT does play Calvinball with us sometimes!
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by GMATinsight » Wed Aug 06, 2014 8:56 pm
sapuna wrote:How many positive integers less than 100 have exactly 4 odd factors but no even factors ?

1) 13

2) 14

3) 15

4) 16

5) 17
If Number = A^p x B^q x C^r .... and so on
Where A, B, C are Distinct Prime numbers and p, q, r... are their respective powers
Then, No. of Factors of a Number = (p+1) x (q+1) x (r+1) .... and so on

[Please note that each (p+1,(q+1), (r+1).. must be an integer greater than 1]

For No. of factors to be 4 the only break up of (p+1)x(q+1)..numbers possible is
i.e. EITHER (p+1) x (q+1) = 4 = 2 x 2 [i.e. product of 2 prime numbers]
OR (p+1) = 4 [i.e. perfect cube with power multiple of 3]

CASE 1: Options for xy that are less than 100 (Like Mitch has shown)
3*5, 3*7, 3*11, 3*13, 3*17, 3*19, 3*23, 3*29, 3*31
5*7, 5*11, 5*13, 5*17, 5*19
7*11, 7*13.
Total options = 16

Case 2: perfect cube with power multiple of 3, 3³ = 27

Answer: Option E
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