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

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

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12 Dec 2017, 08:51
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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
[Reveal] Spoiler: OA

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Kudos [?]: 40 [1], given: 121

Math Expert
Joined: 02 Sep 2009
Posts: 43348

Kudos [?]: 139663 [2], given: 12794

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, 09:15
2
KUDOS
Expert's post
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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Kudos [?]: 139663 [2], given: 12794

Manager
Joined: 05 Oct 2016
Posts: 76

Kudos [?]: 40 [1], given: 121

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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, 15:36
1
KUDOS
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.

_________________

Kudos APPRECIATED!

Kudos [?]: 40 [1], given: 121

Math Expert
Joined: 02 Sep 2009
Posts: 43348

Kudos [?]: 139663 [2], given: 12794

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, 19:30
2
KUDOS
Expert's post
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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Kudos [?]: 139663 [2], given: 12794

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Joined: 09 Aug 2017
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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, 20:09
1
KUDOS
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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Kudos [?]: 1 [1], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 43348

Kudos [?]: 139663 [1], given: 12794

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, 20:16
1
KUDOS
Expert's post
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.

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

Sent from my Redmi 3S using GMAT Club Forum mobile app

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.

2. Properties of Integers

5. Divisibility/Multiples/Factors

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
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Kudos [?]: 139663 [1], given: 12794

Manager
Joined: 05 Oct 2016
Posts: 76

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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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13 Dec 2017, 08:40
1
KUDOS
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 would you solve this question?
Q: Find the number of factors of 180 that can be expressed in the form of 3*even number?
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Kudos [?]: 40 [1], given: 121

Re: Find the number of factors of 180 that are in the form (4*k + 2), wher   [#permalink] 13 Dec 2017, 08:40
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