Question 327163
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There is more than one answer to this problem.  In fact, there are an infinite number of answers.


Start with 5 times 8 equals 40.  All of the divisors of 40 are:


1, 2, 4, 5, 8, 10, 20, and 40, for a total of 8 divisors.  But the number we seek has exactly 10 total divisors.  Hence you can select any prime number other than 2 and 5 which we will call *[tex \Large p] and multiply this number times 40.  Now we will have a number we can call *[tex \Large x] which has exactly 10 divisors, to wit: 1, 2, 4, 5, 8, 10, 20, 40, *[tex \Large p], and *[tex \Large x\ =\ 40p].


For example:  Let *[tex \Large p\ =\ 7].  Then *[tex \Large x\ =\ 280].  The prime factorization of *[tex \Large 280] is *[tex \Large 2^3,\ 5,\ 7].


Or let *[tex \Large p\ =\ 11].  Then *[tex \Large x\ =\ 440], the prime factorization of which is *[tex \Large 2^3,\ 5,\ 11]


And so on.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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