Two cases are possible:

\(x^1y^1=xy=odd<100\), where \(x\) and \(y\) are odd prime numbers, then # of factors will be \((1+1)(1+1)=4\);
If x=3, then y can be: 5, 7, 11, 13, 17, 19, 23, 29, 31 - 9 numbers.
If x=5, then y can be: 7, 11, 13, 17, 19 - 5 numbers.
If x=7, then y can be: 11, 13 - 2 numbers.

OR:
\(x^3=odd<100\), where \(x\) is odd prime number, then # of factors will be \((3+1)=4\).
x can be only 3 - 1 number.

Total \(9+5+2+1=17\).

Answer: D.

P.S. Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Hope it helps.
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this is only possible when factors are 1,a,b,c

where c is the number itself and c= a*b

when a = 3, b can be 5 7 9 11 13 17 19 23 29 31, total 10
when a = 5, b can be 7,11,13,17,19 , total 5
when a = 7, b can be 11 total 1

so total = 16 but its not there in the options.

PS: 9 does not have 4 factors...1,3,9 total 3 factors
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Thanks bunnel I missed 7,13 combination. I need to curb such mistakes. 99% of the time I miss the question because of such mistakes....
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I had this question in 800score's first test. QA is 15. Does this mean that they are wrong?

Integers less than 100 with exactly 4 odd factors but no even factors:

1. 15=3*5 --> factors 1, 3, 5, 15;
2. 21=3*7 --> factors 1, 3, 7, 21;
3. 27=3^3 --> factors 1, 3, 9, 27;
4. 33=3*11 --> factors 1, 3, 11, 33;
5. 35=5*7 --> factors 1, 5, 7, 35;
6. 39=3*13 --> factors 1, 3, 13, 39;
7. 51=3*17 --> factors 1, 3, 17, 51;
8. 55=5*11 --> factors 1, 5, 11, 55;
9. 57=3*19 --> factors 1, 3, 19, 57;
10. 65=5*13 --> factors 1, 5, 13, 65;
11. 69=3*23 --> factors 1, 3, 23, 69;
12. 77=7*11 --> factors 1, 7, 11, 77;
13. 85=5*17 --> factors 1, 5, 17, 85;
14. 87=3*29 --> factors 1, 3, 29, 87;
15. 91=7*13 --> factors 1, 7, 13, 91;
16. 93=3*31 --> factors 1, 3, 31, 93;
17. 95=5*19 --> factors 1, 5, 19, 95.

OA must be wrong.
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Hey Bunuel,



I did not understand, why the number '3' in the first line was not counted. If '3' is also counted then the total numbers will be 10 for the first line, 6 for the second and 3 for the 3rs line. Please explain.

Also, is there any other easy way to get to the solution? Thinking and working out the solution in the manner that you showed seems very daunting to me.....it was like a genius solution. It might never occur to me to solve it this way on test day.......Or is it a matter of practice.

If x=3, then y can be: 5, 7, 11, 13, 17, 19, 23, 29, 31 - 9 numbers --> we are counting # of \(xy\) (not \(x\) and \(y\) separately) --> \(3*5\), \(3*7\), \(3*11\), ... so 9 numbers.



First you should realize that the numbers we are looking for can be either of a type \(xy\) or \(x^3\) (where \(x\) and \(y\) are distinct primes), next count them: should take less than 2 minutes. You are right it's matter of practice.
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Thanks so mcuh Bunuel. Makes sense now.

Do we by any chance have a collection of Number Properties problems (posted in the forum) that we can practice?

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DS number properties questions: search.php?search_id=tag&tag_id=38

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Hello,
the solution is the extension of the property that every number is a factor of 1 and itself right ?

Cheers
Mustu

Hi Bunnel,

how many integers less than 100 have exactly 4 odd [highlight]prime[/highlight] factors but no even
factors?

Since if we take 99 we have

99 - factors - 1 3 33 99 11

this is also having 4 odd factors but no even factors

99 = 33 x 3 => 11 x 3 x 3 => 3 factors + 2 factors( 1 and number itself) = > Hence according to you there are 5 factors and not 4. You shouldnt count 3 twice.

Regards
Mustu

Actually 99 has 6 factors
1
99
9
11
3
33

Please see Bunuel's solution above.

Kind of perfect explanation Kudos to your approach bunuel

Dear Fluke
assiduo is asking Odd [highlight]Prime[/highlight]

can we still keep 9 and 33 as odd [highlight]PRIME[/highlight] factors (as both of them are not prime)

please clarify

Hi Bunnel

What you mean by following

OR:
x^3=odd<100, where x is odd prime number, then # of factors will be (3+1)=4.
x can be only 3 - 1 number.

Total 9+5+2+1=17.

(odd prime)^3 to be less than 100, (odd prime) can only be 3.
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OA is wrong. Answer should be D.
Please do not post questions with wrong answers to avoid any confusion.

Edited the OA. Thank you.

Similar question to practice: how-many-of-the-factors-of-72-are-divisible-by-100708.html
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