Name: james
Who is asking: Student
Level: Secondary
Question:
If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have?
(a) 32 (b) 34 (c) 35 (d) 36 (e) 38
Hi James,
We have two solutions for you. One by Penny and a second by Haragauri.
Penny's Solution:
If you write n as a product of powers of primes, i.e.,
then a divisor m of n must be of the form
where
Thus, if you write d(n) for the number of divisors of n then
Writing a^b for and * for multiplication
we see for example that if n = 2^4*5^7*11^2 then d(n) = 5*8*3 = 120.
Since d(2n) = 28 and 28 can be written as a product of integers larger than 1 in four ways, (2x2x7, 2x14, 4x7, and 28), we get the following table of possibilities. (Notice that since 2 divides 2n, 2 is one of the prime factors of 2n.)

In this table p and q are distinct primes neither of which is 2. 

In this table s and t are distinct primes neither of which is 3. 
Haragauri's Solution:
Suppose n=(2^s)*(3^t)*Q where Q is an integer not divisible by 2 or 3 then, using the the function d defined in Penny's solution, d(n) = d(2^s)*d(3^t)*d(Q). Hence d(2*n)=d(2^(s+1))*d(3^t)*d(q) = 28 or
(s+2)(t+1)d(Q)=28, and
d(3*n)=d(2^s)*d(3^(t+1))*d(q) = 30 or
(s+1)(t+2)d(Q)=30.
Thus d(Q) divides both 28 and 30 and hence d(Q) divides 2=gcd(28,30). Hence d(Q) is either 1 or 2.
Suppose d(Q)=2 then
(s+2)(t+1)=14 and
(s+1)(t+2)=15. Subtracting these equation yields
st=1 or s=t+1. Thus
14=(s+2)(t+1)=(s+2)s which is impossible since s is an integer. Thus d(Q)=1 and hence Q=1.
Since d(Q)=1,
(s+2)(t+1)=28 and
(s+1)(t+2)=30. Again subrtacting gives
st=2 or s=t+2. Thus
30=(s+1)(t+2)=(s+1)s and therefore s=5 and t=3. hence, as in Penny's solution
n = (2^5)(3^3)
Cheers.
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