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26 May 2017, 04:48
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When the product of different prime factors of n is smaller than \(\sqrt{n}\), \(n\) is called "a prime-saturated number". What is the sum of all factors of the biggest 2-digit prime-saturated number?
A. 96
B. 97
C. 132
D. 252
E. 326
A. 96
B. 97
C. 132
D. 252
E. 326
26 May 2017, 04:49
Kudos
Bookmarks
Official Solution:
When the product of different prime factors of n is smaller than \(\sqrt{n}\), \(n\) is called "a prime-saturated number". What is the sum of all factors of the biggest 2-digit prime-saturated number?
A. 96
B. 97
C. 132
D. 252
E. 326
The biggest 2-digit integer of the prime-saturated number is 96. This is because \(96 = (2^{5})(3)\), and each different prime factor of 96 are 2 and 3, and \((2)(3) = 6 < \sqrt{96}\). If so, the sum of all the factors of 96 from \(96 = (2^{5})(3)\) is \((1+2+2^{2}+2^{3}+2^{4}+2^{5})(1+3) = (63)(4) = 252\). The answer is D.
Answer: D
When the product of different prime factors of n is smaller than \(\sqrt{n}\), \(n\) is called "a prime-saturated number". What is the sum of all factors of the biggest 2-digit prime-saturated number?
A. 96
B. 97
C. 132
D. 252
E. 326
The biggest 2-digit integer of the prime-saturated number is 96. This is because \(96 = (2^{5})(3)\), and each different prime factor of 96 are 2 and 3, and \((2)(3) = 6 < \sqrt{96}\). If so, the sum of all the factors of 96 from \(96 = (2^{5})(3)\) is \((1+2+2^{2}+2^{3}+2^{4}+2^{5})(1+3) = (63)(4) = 252\). The answer is D.
Answer: D
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