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Re: How many integers less than 100 have exactly 4 odd factors b [#permalink]
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: How many integers less than 100 have exactly 4 odd factors b [#permalink]
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I had this question in 800score's first test. QA is 15. Does this mean that they are wrong?
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Re: How many integers less than 100 have exactly 4 odd factors b [#permalink]
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efet
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: How many integers less than 100 have exactly 4 odd factors b [#permalink]
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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: How many integers less than 100 have exactly 4 odd factors b [#permalink]
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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
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: How many integers less than 100 have exactly 4 odd factors b [#permalink]
Bunuel
efet
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
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: How many integers less than 100 have exactly 4 odd factors b [#permalink]
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pawankumargadiya
Bunuel
efet
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
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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How many integers less than 100 have exactly 4 odd factors b [#permalink]
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Re: How many integers less than 100 have exactly 4 odd factors b [#permalink]
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In order to achieve requested 4 factors the number should be in the form p^3 or p*q (giving 3+1 and 2+2 total number of factors), where p, and q are odd prime factors.
In first case we have only one match - 3^3 = 27
Second case we’ll calculate in ascending order to evade double counting:
3*5=15
3*7=21
3*11=33
….
3*31=93
____________
5*7=35
5*11=55

5*19=95
____________
7*11=77
7*13=91
Total 16

Total number of integers = 16+1=17

Answer D.
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Re: How many integers less than 100 have exactly 4 odd factors b [#permalink]
vitaliyGMAT
In order to achieve requested 4 factors the number should be in the form p^3 or p*q (giving 3+1 and 2+2 total number of factors), where p, and q are odd prime factors.
In first case we have only one match - 3^3 = 27
Second case we’ll calculate in ascending order to evade double counting:
3*5=15
3*7=21
3*11=33
….
3*31=93
____________
5*7=35
5*11=55

5*19=95
____________
7*11=77
7*13=91
Total 16

Total number of integers = 16+1=17

Answer D.


Why didn't you include 5, 9 = 45 and 7, 9 = 63? That makes sense right?
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Re: How many integers less than 100 have exactly 4 odd factors b [#permalink]
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