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Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
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12 Dec 2017, 10:15
2
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 [#permalink]
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12 Dec 2017, 16:36
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.
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 [#permalink]
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12 Dec 2017, 20:30
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 [#permalink]
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12 Dec 2017, 21:09
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 [#permalink]
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12 Dec 2017, 21:16
1
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 [#permalink]
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13 Dec 2017, 09:40
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 [#permalink]
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07 Mar 2018, 17:40
Here's what I did:
\(180 = (2^2)(3^2)(5)\)
\(4k + 2 = 2(2k + 1)\) --> This means that \(2^1\) must be a factor and any other power of \(2\) must not be a factor. Also, any odd number must be a factor. So,
possible powers of 2: {1} possible powers of 3: {0,1,2} possible powers of 5: {0,1}
1 x 3 x 2 = 6
Answer: E
gmatclubot
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher
[#permalink]
07 Mar 2018, 17:40
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42% (01:35) correct 
