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How many integers less than 100 have exactly 4 odd factors b

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

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

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New post 17 Jul 2010, 12:51
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 \(9+5+2+1=17\).

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

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New post 17 Jul 2010, 12:08
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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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Re: integer with exactly 4 odd factors  [#permalink]

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New post 17 Jul 2010, 12:55
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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Re: integer with exactly 4 odd factors  [#permalink]

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New post 17 Jul 2010, 20:24
I had this question in 800score's first test. QA is 15. Does this mean that they are wrong?
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Re: integer with exactly 4 odd factors  [#permalink]

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New post 20 Jul 2010, 02:30
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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Re: integer with exactly 4 odd factors  [#permalink]

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New post 20 Jul 2010, 10:58
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.
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Re: integer with exactly 4 odd factors  [#permalink]

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New post 20 Jul 2010, 11:58
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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Re: integer with exactly 4 odd factors  [#permalink]

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New post 20 Jul 2010, 12:29
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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Re: integer with exactly 4 odd factors  [#permalink]

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New post 20 Jul 2010, 12:35
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Re: integer with exactly 4 odd factors  [#permalink]

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New post 29 Jun 2011, 00:46
Hello,
the solution is the extension of the property that every number is a factor of 1 and itself right ?

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

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New post 30 Jun 2011, 04:39
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
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Re: integer with exactly 4 odd factors  [#permalink]

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New post 30 Jun 2011, 09:02
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.

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

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New post 30 Jun 2011, 09:09
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

Please see Bunuel's solution above.
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Re: integer with exactly 4 odd factors  [#permalink]

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New post 05 Jul 2011, 06:28
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 \(9+5+2+1=17\).

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.



Kind of perfect explanation Kudos to your approach bunuel :beer :beer :beer :beer
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Re: integer with exactly 4 odd factors  [#permalink]

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New post 05 Jul 2011, 06:38
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

Please see Bunuel's solution above.



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

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New post 07 Apr 2014, 04:48
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 \(9+5+2+1=17\).

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.


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

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New post 07 Apr 2014, 05:35
pawankumargadiya wrote:
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 \(9+5+2+1=17\).

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.


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

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New post 26 Apr 2014, 06:32
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.
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Re: How many integers less than 100 have exactly 4 odd factors b  [#permalink]

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

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