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

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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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New post 12 Dec 2017, 09:51
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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

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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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New post 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.

Answer: E.
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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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New post 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.
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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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New post 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.
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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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New post 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.

Answer: E.

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]

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New post 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.

Answer: E.

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

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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:
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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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New post 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?
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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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New post 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
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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