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19 Jan 2020, 06:51
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (02:00) correct
63%
(01:50)
wrong
based on 112
sessions
History
Date
Time
Result
Not Attempted Yet
How many factors of 4500 have odd number of factors?
A) 4
B) 6
C) 8
D) 12
E) 16
Source: https://www.GMATinsight.com
A) 4
B) 6
C) 8
D) 12
E) 16
Source: https://www.GMATinsight.com
19 Jan 2020, 07:28
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4500= (2^2)*(3^2)*(5^3)
For factors having odd factors, factor must have even powers.
So such factors are 1, 2^2, 3^2, 5^2, (2^2)*(3^2), (3^2)*(5^2), (2^2)*(5^2), (2^2)*(3^2)*(5^2) - Total 8 factors
1 is considered because 1 has only one factor that is 1.
C is answer.
For factors having odd factors, factor must have even powers.
So such factors are 1, 2^2, 3^2, 5^2, (2^2)*(3^2), (3^2)*(5^2), (2^2)*(5^2), (2^2)*(3^2)*(5^2) - Total 8 factors
1 is considered because 1 has only one factor that is 1.
C is answer.
20 Jan 2020, 04:17
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GMATinsight
Property: Perfect squares (i.e. squares of Integers) always have odd number of Factors. See ref: https://gmatclub.com/forum/how-many-int ... l#p1549663
i.e. All factors of 4500 which are squares of some integers need to be counted to answer this question
\(4500 = 2^2*3^2^5^3\)
So the factors made up by \(2^0\), \(2^2\), \(3^0\), \(3^2\), \(5^0\), \(5^2\),
We can take one number from each of the three colour codings and make combinations such as
\(2^0*3^0*5^0\)
\(2^0*3^2*5^0\)
\(2^2*3^0*5^0\)
\(2^2*3^2*5^0\)
\(2^0*3^0*5^1\)
\(2^0*3^2*5^1\)
\(2^2*3^0*5^1\)
\(2^2*3^2*5^1\)
i.e. Total combinations = 2*2*2 = 8 (as mentioned above
Answer: Option C
