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Manager  G
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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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Question Stats: 49% (02:37) correct 51% (02:05) wrong based on 331 sessions

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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

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Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher  [#permalink]

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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.

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Manager  G
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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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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.

Bunuel how did you find 6 factors as 1,3,5,9,15,45. please help me in this.
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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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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.

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.
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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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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.

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

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Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher  [#permalink]

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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.

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 ! ! !
Ultimate GMAT Quantitative Megathread
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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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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.

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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GMAT 1: 200 Q1 V1 Find the number of factors of 180 that are in the form (4*k + 2), wher  [#permalink]

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1
1
Here's what I did:

$$180 = (2^2)(3^2)(5)$$

$$4k + 2 = 2(2k + 1)$$ --> We have 2 * odd. How many ways can we make an odd number using the prime factors of 180?

possible powers of 3: {0,1,2} --> $$3^0, 3^1, 3^2$$
possible powers of 5: {0,1} --> $$5^0, 5^1$$

3 x 2 = 6

Originally posted by aserghe1 on 07 Mar 2018, 17:40.
Last edited by aserghe1 on 04 Aug 2018, 12:05, edited 1 time in total.
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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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Bunuel

Number of factors of 180 = 2^2*3^2*5 = (2+1)(2+1)(1+1) = 18

Can't we write this factors in terms of question (4*k + 2) as 18 = (4* 4 + 2) and arrives at option D ?

What am I missing here?

Thanks a lot
Intern  B
Joined: 04 Jul 2018
Posts: 1
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher  [#permalink]

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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
Manager  G
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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:
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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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 Manager  S
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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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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

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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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aserghe1 wrote:
Here's what I did:

$$180 = (2^2)(3^2)(5)$$

$$4k + 2 = 2(2k + 1)$$ --> We have 2 * odd. How many ways can we make an odd number using the prime factors of 180?

possible powers of 3: {0,1,2} --> $$3^0, 3^1, 3^2$$
possible powers of 5: {0,1} --> $$5^0, 5^1$$

3 x 2 = 6

Hi aserghe1,

How did you know that you need to plug in the odd factors of 180 and not any odd numbers to get the number of factors of 180 fitting in 2(2k+1)?

Intern  B
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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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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"?? Re: Find the number of factors of 180 that are in the form (4*k + 2), wher   [#permalink] 10 Oct 2018, 09:27
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