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Manager  Joined: 14 Jul 2010
Posts: 67
How many integers less than 100 have exactly 4 odd factors b  [#permalink]

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Question Stats: 31% (03:04) correct 69% (02:48) wrong based on 323 sessions

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How many integers less than 100 have exactly 4 odd factors but no even factors?

A. 11
B. 13
C. 15
D. 17
E. 19

Originally posted by efet on 17 Jul 2010, 11:53.
Last edited by Bunuel on 26 Apr 2014, 09:35, edited 1 time in total.
Edited the OA.
Math Expert V
Joined: 02 Sep 2009
Posts: 60647
Re: integer with exactly 4 odd factors  [#permalink]

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10
12
efet wrote:
how many integers less than 100 have exactly 4 odd factors but no even factors?

a. 11
b. 13
c. 15
d. 17
e. 19

--

Is 9 with exactly 4 odd factors

gurpreetsingh wrote:
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

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 $$9+5+2+1=17$$.

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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Re: integer with exactly 4 odd factors  [#permalink]

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2
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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Re: integer with exactly 4 odd factors  [#permalink]

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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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Manager  Joined: 14 Jul 2010
Posts: 67
Re: integer with exactly 4 odd factors  [#permalink]

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I had this question in 800score's first test. QA is 15. Does this mean that they are wrong?
Math Expert V
Joined: 02 Sep 2009
Posts: 60647
Re: integer with exactly 4 odd factors  [#permalink]

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efet wrote:
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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Director  Joined: 17 Feb 2010
Posts: 753
Re: integer with exactly 4 odd factors  [#permalink]

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1
Hey Bunuel,

Quote:
xy = odd<100, where x and y are odd prime numbers, then # of factors will be ;

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.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 60647
Re: integer with exactly 4 odd factors  [#permalink]

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seekmba wrote:
Hey Bunuel,

Quote:
xy = odd<100, where x and y are odd prime numbers, then # of factors will be ;

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.

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.

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.

seekmba wrote:
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.

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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Director  Joined: 17 Feb 2010
Posts: 753
Re: integer with exactly 4 odd factors  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 60647
Re: integer with exactly 4 odd factors  [#permalink]

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Tags list by forum: viewforumtags.php

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

PS number properties questions: search.php?search_id=tag&tag_id=59
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Intern  Joined: 29 Mar 2011
Posts: 15
Re: integer with exactly 4 odd factors  [#permalink]

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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
Intern  Joined: 26 May 2011
Posts: 2
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Re: integer with exactly 4 odd factors  [#permalink]

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1
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
Intern  Joined: 29 Mar 2011
Posts: 15
Re: integer with exactly 4 odd factors  [#permalink]

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assiduo wrote:
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
Retired Moderator Joined: 20 Dec 2010
Posts: 1533
Re: integer with exactly 4 odd factors  [#permalink]

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assiduo wrote:
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

Actually 99 has 6 factors
1
99
9
11
3
33

Manager  Status: ==GMAT Ninja==
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Re: integer with exactly 4 odd factors  [#permalink]

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Bunuel wrote:
efet wrote:
how many integers less than 100 have exactly 4 odd factors but no even factors?

a. 11
b. 13
c. 15
d. 17
e. 19

--

Is 9 with exactly 4 odd factors

gurpreetsingh wrote:
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

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 $$9+5+2+1=17$$.

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.

Kind of perfect explanation Kudos to your approach bunuel    Manager  Status: ==GMAT Ninja==
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Posts: 171
Schools: ISB, IIMA ,SP Jain , XLRI
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Re: integer with exactly 4 odd factors  [#permalink]

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fluke wrote:
assiduo wrote:
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

Actually 99 has 6 factors
1
99
9
11
3
33

Dear Fluke

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

please clarify  Manager  B
Joined: 10 Mar 2014
Posts: 177
Re: integer with exactly 4 odd factors  [#permalink]

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Bunuel wrote:
efet wrote:
how many integers less than 100 have exactly 4 odd factors but no even factors?

a. 11
b. 13
c. 15
d. 17
e. 19

--

Is 9 with exactly 4 odd factors

gurpreetsingh wrote:
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

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 $$9+5+2+1=17$$.

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.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 60647
Re: integer with exactly 4 odd factors  [#permalink]

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Bunuel wrote:
efet wrote:
how many integers less than 100 have exactly 4 odd factors but no even factors?

a. 11
b. 13
c. 15
d. 17
e. 19

--

Is 9 with exactly 4 odd factors

gurpreetsingh wrote:
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

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 $$9+5+2+1=17$$.

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.

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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Posts: 56
Re: How many integers less than 100 have exactly 4 odd factors b  [#permalink]

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1
efet wrote:
How many integers less than 100 have exactly 4 odd factors but no even factors?

A. 11
B. 13
C. 15
D. 17
E. 19

OA is wrong. Answer should be D.
Please do not post questions with wrong answers to avoid any confusion.
Math Expert V
Joined: 02 Sep 2009
Posts: 60647
Re: How many integers less than 100 have exactly 4 odd factors b  [#permalink]

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chanakya84 wrote:
efet wrote:
How many integers less than 100 have exactly 4 odd factors but no even factors?

A. 11
B. 13
C. 15
D. 17
E. 19

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
_________________ Re: How many integers less than 100 have exactly 4 odd factors b   [#permalink] 26 Apr 2014, 09:37

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