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Re: integer with exactly 4 odd factors
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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.
Re: integer with exactly 4 odd factors
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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.... _________________
Re: integer with exactly 4 odd factors
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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.
Re: integer with exactly 4 odd factors
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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.
_________________
Re: integer with exactly 4 odd factors
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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.
Re: integer with exactly 4 odd factors
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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
Re: integer with exactly 4 odd factors
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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.
Re: integer with exactly 4 odd factors
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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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