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How many factors of 80 are greater than square_root 80?
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How many factors of 80 are greater than \(\sqrt{80}\)? A. Ten B. Eight C. Six D. Five E. Four
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Originally posted by vanidhar on 16 Sep 2010, 04:04.
Last edited by Bunuel on 11 Nov 2014, 03:33, edited 1 time in total.
Edited the question.




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Re: How many factors of 80 are greater than square_root 80?
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16 Sep 2010, 07:46
vanidhar wrote: how many factors of 80 are greater than square_root 80?
a)5 No need to find all factors of 80. \(\sqrt{80}\) is more than 8 and less than 9. So we are asked to find # of factors of 80 which are more than 8. Now, \(80=16*5=2^4*5\) > # of factors of 80 is \((4+1)(1+1)=10\) (see below how to find the # of factors of an integer). Out of these 10, following 5 factors are less or equal to 8: 1, 2, 4, 5, and 8. So other 5 factors are more than 8. Answer: 5. Finding the Number of Factors of an IntegerFirst 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: How many factors of 80 are greater than square_root 80?
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16 Sep 2010, 04:28
vanidhar wrote: how many factors of 80 are greater than square_root 80?
a)5 The square root condition means you have to find divisors >= 9 So start with 80 and keep dividing till you hit the condition 80 > can be divided in 2 ways by 2 or by 5 to get (40, 16) 16 > cant be divided into anything greater than or equal to 9 40 > can be divided by 2 or by 5 to get (20,8). The 8 doesnt count 20 > can be divided by 2 or by 5 to get (10,4). The 4 doesnt count 10 > cant be divided into anything greater than or equal to 9 So we get : 10,20,40,16,80 Hence 5 is answer It is easier to do this if you make a tree structure on paper ... comes much more naturally
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Re: How many factors of 80 are greater than square_root 80?
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16 Oct 2010, 01:18
8. How many different positive integers are factors of 342? A. 9 B. 11 C. 12 D. 20 E. 22
Bunuel logic gave me only 7 but the andser says 12 .. here the explaination given :
C. From the answers we can see that the list of factors will be relatively small, so it’s easiest just to list them out. The pairs of factors are 1 and 342, 2 and 171, 3 and 114, 6 and 57, 9 and 38, and 18 and 19. That makes 12 factors.



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Re: How many factors of 80 are greater than square_root 80?
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16 Oct 2010, 01:21
342 = 2 x 3^2 x 19 Number of factors = (1+1)(2+1)(1+1) = 12 Posted from my mobile device
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Re: How many factors of 80 are greater than square_root 80?
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17 Oct 2010, 05:55
vanidhar wrote: 8. How many different positive integers are factors of 342? A. 9 B. 11 C. 12 D. 20 E. 22
Bunuel logic gave me only 7 but the andser says 12 .. here the explaination given :
C. From the answers we can see that the list of factors will be relatively small, so it’s easiest just to list them out. The pairs of factors are 1 and 342, 2 and 171, 3 and 114, 6 and 57, 9 and 38, and 18 and 19. That makes 12 factors. It's not MY logic, it's MATH. According to the formula in my previous post as \(342=2*3^2*19\) then # of factors of 342 equals to \((1+1)(2+1)(1+1)=12\). Answer: C.
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Re: How many factors of 80 are greater than square_root 80?
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18 Oct 2010, 08:12
apologies for the wording .. My fault .. I did a calculation mistake..



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Re: How many factors of 80 are greater than square_root 80?
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07 Jan 2015, 09:38
Bunuel wrote: vanidhar wrote: how many factors of 80 are greater than square_root 80?
a)5 No need to find all factors of 80. \(\sqrt{80}\) is more than 8 and less than 9. So we are asked to find # of factors of 80 which are more than 8. Now, \(80=16*5=2^4*5\) > # of factors of 80 is \((4+1)(1+1)=10\) (see below how to find the # of factors of an integer). Out of these 10, following 5 factors are less or equal to 8: 1, 2, 4, 5, and 8. So other 5 factors are more than 8. Answer: 5. Finding the Number of Factors of an IntegerFirst 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. Bunuel, Did not get the part where you said root 80 is between 8 and 9..the value of root 80 is 4 root 5..Am i missing something here? Thanks in Advance.



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Re: How many factors of 80 are greater than square_root 80?
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07 Jan 2015, 09:42
Ralphcuisak wrote: Bunuel wrote: vanidhar wrote: how many factors of 80 are greater than square_root 80?
a)5 No need to find all factors of 80. \(\sqrt{80}\) is more than 8 and less than 9. So we are asked to find # of factors of 80 which are more than 8. Now, \(80=16*5=2^4*5\) > # of factors of 80 is \((4+1)(1+1)=10\) (see below how to find the # of factors of an integer). Out of these 10, following 5 factors are less or equal to 8: 1, 2, 4, 5, and 8. So other 5 factors are more than 8. Answer: 5. Finding the Number of Factors of an IntegerFirst 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. Bunuel, Did not get the part where you said root 80 is between 8 and 9..the value of root 80 is 4 root 5..Am i missing something here? Thanks in Advance. \(4\sqrt{5}\approx{8.94}\). \(\sqrt{81}=9\), thus \(\sqrt{80}\) is a bit less than 9.
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Re: How many factors of 80 are greater than square_root 80?
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01 Aug 2015, 00:16
Honestly it's not really necessary to look at actual factors or even check the square root of \(80\) once you count factors (as Bunuel explained, \(80=2^4*5^1\) and thus has \((4+1)*(1+1)=10\) factors).
Since factors pair off, every integer will always have exactly half of its factors less than its square root. (Round down in the case of an odd number of factors  i.e. a perfect square.)



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Re: How many factors of 80 are greater than square_root 80?
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03 Aug 2015, 03:16
IMHO, what you are writing is true only if the number is a non perfect square. If I change the question like  How many factors of 100 will be less than \sqrt{100, answer would be equal to ((total number of factors of 100)1/2). Total number of factors of 100 is 9. Hence total number of factors which are less than or greater than square root of 100 will be (91)/2which is equal to 4. Bunuel wrote: vanidhar wrote: how many factors of 80 are greater than square_root 80?
a)5 No need to find all factors of 80. \([square_root]80}\) is more than 8 and less than 9. So we are asked to find # of factors of 80 which are more than 8. Now, \(80=16*5=2^4*5\) > # of factors of 80 is \((4+1)(1+1)=10\) (see below how to find the # of factors of an integer). Out of these 10, following 5 factors are less or equal to 8: 1, 2, 4, 5, and 8. So other 5 factors are more than 8. Answer: 5. Finding the Number of Factors of an IntegerFirst 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: How many factors of 80 are greater than square_root 80?
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03 Aug 2015, 06:09
gmatcrackerindia wrote: IMHO, what you are writing is true only if the number is a non perfect square. If I change the question like  How many factors of 100 will be less than \sqrt{100, answer would be equal to ((total number of factors of 100)1/2). Total number of factors of 100 is 9. Hence total number of factors which are less than or greater than square root of 100 will be (91)/2which is equal to 4. Bunuel wrote: vanidhar wrote: how many factors of 80 are greater than square_root 80?
a)5 No need to find all factors of 80. \([square_root]80}\) is more than 8 and less than 9. So we are asked to find # of factors of 80 which are more than 8. Now, \(80=16*5=2^4*5\) > # of factors of 80 is \((4+1)(1+1)=10\) (see below how to find the # of factors of an integer). Out of these 10, following 5 factors are less or equal to 8: 1, 2, 4, 5, and 8. So other 5 factors are more than 8. Answer: 5. Finding the Number of Factors of an IntegerFirst 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. You can't generalise it.... Bunnuel gave the easiest solution possible. Lets take an example of 144. As per the factors  2^4*3^2 = Total factors including 1 & 144 = (4+1)(2+1) = 15. Now go to the factors less than root 144 i.e. 12 = 2^2*3^1 = 6.. Where is your (n1)/2 logic in this???



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Re: How many factors of 80 are greater than square_root 80?
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03 Aug 2015, 09:33
Quote: You can't generalise it....
Bunnuel gave the easiest solution possible.
Lets take an example of 144.
As per the factors  2^4*3^2 = Total factors including 1 & 144 = (4+1)(2+1) = 15.
Now go to the factors less than root 144 i.e. 12 = 2^2*3^1 = 6..
Where is your (n1)/2 logic in this??? This is a perfect example of exactly why taking half of the total factors (as noted, round down in the case of a perfect square) is better than trying to list them all by hand. Actually, the seven factors of \(144\) that are less than \(\sqrt{144}\) are: \(1\), \(2\), \(3\), \(4\), \(6\), \(8\), and \(9\). Not all factors of \(144\) that are less than \(12\) are necessarily factors of \(12\).



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Re: How many factors of 80 are greater than square_root 80?
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04 Aug 2015, 01:03
Cadaver wrote: gmatcrackerindia wrote: IMHO, what you are writing is true only if the number is a non perfect square. If I change the question like  How many factors of 100 will be less than [square_root]100, answer would be equal to ((total number of factors of 100)1/2). Total number of factors of 100 is 9. Hence total number of factors which are less than or greater than square root of 100 will be (91)/2which is equal to 4.
You can't generalise it....
Bunnuel gave the easiest solution possible.
Lets take an example of 144.
As per the factors  2^4*3^2 = Total factors including 1 & 144 = (4+1)(2+1) = 15.
Now go to the factors less than root 144 i.e. 12 = 2^2*3^1 = 6..
Where is your (n1)/2 logic in this??? What I am trying to say is: (a)If the number is not a perfect square, then exactly half of the factors will be smaller than the square root of the number, and other half will be greater than the square root of the number. (b) If the number is a Perfect Square, then half of (total factors  1) will be smaller than the square root of the number, and half of (total factors  1) will be greater than the square root of the number. One of the factors will be equal to the square root of the number.



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