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Re: Find the number of factors of 180 that are in the form (4*k + 2), wher
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12 Dec 2017, 09:15
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4
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?
A)1 B)2 C)3 D)4 E)6
4k + 2 = 2(2k + 1) = 2*odd. So, we are looking for even factors which are not multiples of 4.
180 = 2^2*3^2*5. Consider the part without 2^2. Now, 3^2*5 has (2 + 1)(1 + 1) = 6 factors: 1, 3, 5, 9, 15, 45. Any of them paired with 2 will be even factor of 180 which is not a multiple of 4, so 2, 6, 10, 18, 30 and 90.
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher
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12 Dec 2017, 15:36
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Bunuel wrote:
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?
A)1 B)2 C)3 D)4 E)6
4k + 2 = 2(2k + 1) = 2*odd. So, we are looking for even factors which are not multiples of 4.
180 = 2^2*3^2*5. Consider the part without 2^2. Now, 3^2*5 has (2 + 1)(1 + 1) = 6 factors: 1, 3, 5, 9, 15, 45. Any of them paired with 2 will be even factor of 180 which is not a multiple of 4, so 2, 6, 10, 18, 30 and 90.
Answer: E.
Bunuel how did you find 6 factors as 1,3,5,9,15,45. please help me in this.
_________________
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher
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12 Dec 2017, 19:30
2
SandhyAvinash wrote:
Bunuel wrote:
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?
A)1 B)2 C)3 D)4 E)6
4k + 2 = 2(2k + 1) = 2*odd. So, we are looking for even factors which are not multiples of 4.
180 = 2^2*3^2*5. Consider the part without 2^2. Now, 3^2*5 has (2 + 1)(1 + 1) = 6 factors: 1, 3, 5, 9, 15, 45. Any of them paired with 2 will be even factor of 180 which is not a multiple of 4, so 2, 6, 10, 18, 30 and 90.
Answer: E.
Bunuel how did you find 6 factors as 1,3,5,9,15,45. please help me in this.
I just manually listed the factors of 90 = 3^2*5. You don't have to find the factors for this question though, it's done here just to illustrate. The fact that 90 has 6 factors (does not matter what they really are) is enough to get that 180 will have twice as many even factors which are not multiples of 4.
_________________
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher
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12 Dec 2017, 20:09
1
Bunuel wrote:
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?
A)1 B)2 C)3 D)4 E)6
4k + 2 = 2(2k + 1) = 2*odd. So, we are looking for even factors which are not multiples of 4.
180 = 2^2*3^2*5. Consider the part without 2^2. Now, 3^2*5 has (2 + 1)(1 + 1) = 6 factors: 1, 3, 5, 9, 15, 45. Any of them paired with 2 will be even factor of 180 which is not a multiple of 4, so 2, 6, 10, 18, 30 and 90.
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher
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12 Dec 2017, 20:16
1
2
Spongebob02 wrote:
Bunuel wrote:
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?
A)1 B)2 C)3 D)4 E)6
4k + 2 = 2(2k + 1) = 2*odd. So, we are looking for even factors which are not multiples of 4.
180 = 2^2*3^2*5. Consider the part without 2^2. Now, 3^2*5 has (2 + 1)(1 + 1) = 6 factors: 1, 3, 5, 9, 15, 45. Any of them paired with 2 will be even factor of 180 which is not a multiple of 4, so 2, 6, 10, 18, 30 and 90.
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: Find the number of factors of 180 that are in the form (4*k + 2), wher
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13 Dec 2017, 08:40
2
Bunuel wrote:
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?
A)1 B)2 C)3 D)4 E)6
4k + 2 = 2(2k + 1) = 2*odd. So, we are looking for even factors which are not multiples of 4.
180 = 2^2*3^2*5. Consider the part without 2^2. Now, 3^2*5 has (2 + 1)(1 + 1) = 6 factors: 1, 3, 5, 9, 15, 45. Any of them paired with 2 will be even factor of 180 which is not a multiple of 4, so 2, 6, 10, 18, 30 and 90.
Answer: E.
how would you solve this question? Q: Find the number of factors of 180 that can be expressed in the form of 3*even number?
_________________
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher
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04 Aug 2018, 21:20
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?
A)1 B)2 C)3 D)4 E)6
I did this question with simple substitution like below -
we know K >=0 and is an integer.
therefore for K = 0 ; (4*k + 2) = 2 - A factor of 180 K = 1 ; (4*k + 2) = 6 - A factor of 180 K = 2 ; (4*k + 2) = 10 - A factor of 180 K = 4 ; (4*k + 2) = 18 - A factor of 180
No we know answer can be D or E.
If I am able to find one more my answer will be E.
Hence I factorized 180 = 2x3x3x2x5
Next number after 18 I can form is 30 that will be a factor of 180 and I get that with k = 7.
Hence E.
FYI - I did this question under two minutes. So IMO not a lengthy method.
_________________
Regards AD --------------------------------- A Kudos is one more question and its answer understood by somebody !!!
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher
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06 Aug 2018, 06:51
Damn, time consuming one. Got the right answer but @ 03:50..... I lost 1 minute trying to figure out all the factors of 180, then its crucial to find a pattern quickly when you plug in numbers for 4*k+2.... Lesson learned
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher
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14 Aug 2018, 00:20
i did this manually by listing all the factor pairs of 180, then choosing factors from the list which are 2 more than multiples of 4 but it took me 3:29 to solve _________________
Hasnain Afzal
"When you wanna succeed as bad as you wanna breathe, then you will be successful." -Eric Thomas
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher
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10 Oct 2018, 08:27
I can find the answer as 6.
But many people have provided solution as below:
4k + 2 = 2(2k + 1) = 2*odd. So, we are looking for even factors which are not multiples of 4. 180 = 2^2*3^2*5. Consider the part without 2^2. Now, 3^2*5 has (2 + 1)(1 + 1) = 6 factors
So, my question is why "not multiples of 4"??
gmatclubot
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher &nbs
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10 Oct 2018, 08:27
close
50% (01:42) correct 
but it took me 3:29 to solve 
