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05 Oct 2019, 15:03
Kudos
Bookmarks
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:
38% (03:01) correct
62%
(02:35)
wrong
based on 21
sessions
History
Date
Time
Result
Not Attempted Yet
A certain number has a total of 16 factors. The square of this number cannot have which one of the following total factors?
(A) 31
(B) 45
(C) 49
(D) 63
(E) 87
(A) 31
(B) 45
(C) 49
(D) 63
(E) 87
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05 Oct 2019, 19:19
Kudos
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Hovkial
Formula: Number of factors of N = 2^a*3^b*5^c*7^d*. . . = (a+1)(b+1)(c+1)(d+1)......
Note: The formula is independent of any prime number as BASE. It only depends on the power(s) of prime number(s).
Possible solutions for 16 factors are
Case1: If N= 2^1*3^1*5^1*7^1 = (1+1)(1+1)(1+1)(1+1) = 16
Square of N = 2^2*3^2*5^2*7^2
Number of factors = 3*3*3*3 = 81
Case2: If N= 2^1*3^1*5^3 = (1+1)(1+1)(3+1) = 16
Square of N = 2^2*3^2*5^6
Number of factors = 3*3*7 = 63
Case3: If N= 2^1*3^7 = (1+1)(7+1) = 16
Square of N = 2^2*3^14
Number of factors = 3*15 = 45
Case4: If N= 2^3*3^3 = (3+1)(3+1) = 16
Square of N = 2^6*3^6
Number of factors = 7*7 = 49
Case4: If N= 2^15 = (15+1) = 16
Square of N = 2^30
Number of factors = 30+1 = 31
IMO Option E
Posted from my mobile device
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