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SandhyAvinash
Joined: 05 Oct 2016
Last visit: 19 Sep 2019
Posts: 68
Given Kudos: 135
Location: United States (OH)
Schools: CBS '20 Fisher '20 Kelley '20
GPA: 3.58
12 Dec 2017, 09:51
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
55% (02:31) correct
45%
(02:13)
wrong
based on 438
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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
A)1
B)2
C)3
D)4
E)6
12 Dec 2017, 10:15
Kudos
Bookmarks
SandhyAvinash
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
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.
General Discussion
SandhyAvinash
Joined: 05 Oct 2016
Last visit: 19 Sep 2019
Posts: 68
Given Kudos: 135
Location: United States (OH)
Schools: CBS '20 Fisher '20 Kelley '20
GPA: 3.58
12 Dec 2017, 16:36
Kudos
Bookmarks
Bunuel
SandhyAvinash
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
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.
