stonecold wrote:
A number is said to be prime saturated if the product of all the different positive prime factors of n is less than the square root of n.
If p is a prime saturated,what is the value of p?
A)p is a perfect cube.
B)2≤p≤15
Source-> Self-Made
A) \(p\) is a perfect cube. Then \(p\) could be \(3^3=27\) or \(2^3=8\). Insufficient.
B) \(2 \leq p \leq 15\)
First, \(p\) can't be a prime. Out 2, 3, 5, 7, 11, 13, left 4, 6, 8, 9, 10, 12, 14, 15
Second, \(p\) can't be in prime factorization of \(a \times b\) where \(a\) and \(b\) are prime. Out 6, 10, 14, 15. Left 4, 8, 9, 12.
Third, \(p\) can't be in form of \(a^2\) where \(a\) is prime. Out 4, 9. Left 8, 12.
8 is a cube, so 8 is prime saturated.
\(12=2^2 \times 3\). We have \(2 \times 3 = 6 > \sqrt{12}=2\sqrt{3}\) so 12 is not prime saturated.
Hence, \(p=8\). Sufficient.
The answer is B.
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