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# Find the number of factors of 180 that are in the form (4*k + 2), wher

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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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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.

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.
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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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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.

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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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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.

How did you got (2+1)(1+1) ?? Pls explain

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Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
There are 6 possibilities, where f(x)=4x+2 is a factor of 180.

f(0)=2
F(1)=6
f(2)=10
f(4)=18
f(7)=30
f(22)=80
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Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
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.
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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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Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
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