If n has 15 factors (1 and n inclusive ) and 2n has 20 factors. What is the number of factors of 4n?
Also, what can be the number of factors of 5n ?
If n has 15 factors, it is either (p1)^2 * (p2)^4, where p1 and p2 are primes, or (p1)^14, as each prime can be used either not at all in a proposed factor, up to the full power in which it is found in the number n itself.
If 2 were not a factor of the original number, multiplying it by 2 would double the number of factors, but that's not the case here, so either p1 or p2 is 2. If n were 2^14, the number of factors would merely increase from 15 to 16, so that's not the case either.
If p2 in the first choice given above were the 2, then the factors would go from 3*5=15 to 3*6=18, which is again not the case.
That leaves n as being 2^2 * (p2)^4. When that becomes 2^3 * (p2)^4, the number of factors becomes 4*5=20 as stated in the problem.
So, to get to the first question, when 2*n is then doubled to 4*n, or 2^4 * (p2)^4, the number of factors becomes 5*5=25.
In the second part, I assume that the conditions and results of the first part still hold: that is, n = 2^2 * (p2)^4, where p2 is any prime other than 2.
If p2 is 5, then 5*n has 3*6 = 18 factors. If, however, p2 is some other prime than 5, say 3, then 5*n has 3*5*2 = 30 factors.
Examples:
n 2n 4n 5n
number 324 648 1296 1620
factors 15 20 25 30
number 2500 5000 10000 12500
factors 15 20 25 18
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Posted by Charlie
on 2010-10-05 17:22:19 |
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