Number of divisors?

Spoiler:

I'm getting 15n has either 45 or 48 positive divisors and that n = (3^3)(5^7) or is of the form n = (3^2)(5^4)(p) where p is a prime not equal to 3 or 5. I'll see if I can write up a solution later. The solution I have involves some casework so I suspect there may be a slicker way to solve it.

Addtion: Let's do some casework!

Case 1: 5 does not divide n

Let's first write n in the form n = (p_1)^(e_1) (p_2)^(e_2) ...(p_n)^(e_n) where the p_i's are primes and the e_i's are positive integers. Define d(n) to be the number of positive integer divisors of n. Then d(n) = (e_1 + 1)(e_2 + 1)...(e_n + 1). We can write 5n = 5(p_1)^(e_1)*...*(p_n)^(e_n), so

d(5n) = (1 + 1)(e_1 + 1)(e_2 + 1)...(e_n + 1) = 2d(n) and

d(5n) = 36, so

d(n) = 18.

If 3 does not divide n, then we we could do a similar analysis to find that d(n) = 20, which contradicts the above statement that d(n) = 18.

If 3 does divide n, then we can write 3n = (p_1)^(e_1)*...*(p_i)(e^i + 1)*...*(p_n)^(e_n) where p_i = 3. In this case d(3n) = (e_1 + 1)...(e_i + 2)...(e_n + 1) = 40. Let's compare this with the similar eqn for d(n) above by dividing them:

d(3n)/d(n) = (e_i + 2) / (e_i + 1) <= 3/2 < 40/18 where we determine (e_i +2) / (e_i + 1) <= 3/2 by knowing the rational expression is monotonically decreasing over the positive integers. Here again we reached a contradiction.

We conclude that if there are any successful values of n, then 5|n for those values of n.

Case 2: 5 does divide n.

If 3 does not divide n, then we know from the analysis above that d(n) = d(3n) / 2 = 40/2 = 20. Assuming 5 does divide n we can write 5n = (p_1)^(e_1)*...*(p_j)^(e_j + 1)*...*(p_n)^(e_n) where p_j = 5. It follows that d(5n) = (e_1 + 1)...(e_j + 2)...(e_n + 1) = 36. Let's divide the d(5n) eqn by the d(n) eqn:

d(5n) / d(n) = (e_j + 2) / (e_j + 1) <= 3/2 < 36/20.

Here also we reach an impossibility, so we don't have any solutions when 3 does not divide n. Now we must consider the condition where 3 does divide n (the last possibility).

If there are any n with d(5n) = 36, d(3n) = 40 we must have 3|n and 5|n. Thus, we can write

n = (3^a) (5^b) (p_1)^(e_1)... (p_n)^(e_n) for postive integers a and b. We will work more closely with the following eqns:

d(3n) = (a + 2)(b + 1)(e_1 + 1)...(e_n + 1) = 40.....(1) and

d(5n) = (a + 1)(b + 2)(e_1 + 1)...(e_n + 1) = 36.....(2).

Observe that when we divide these expressions we get (a+2)(b+1) / (a+1)(b+2) = (ab + 2b + a + 2) / (ab + a + 2b + 2) = (s + b) / (s + a) = 40 / 36 > 1 so b > a (where s = ab + a + b + 2).

Let's analyze (1). Observe that 40 = (2^3)(5). We wish to split 40 into pairs, triples, etc. of factors such that each factor is at least 2. We can do this in six ways:

40 = (2)(20) = (4)(10) = (8)(5) = (5)(2)(4) = (2)(2)(10) = (2)(2)(2)(5). Now we must consider the six subcases where d(3n) is split under these conditions.

(i) (a+2)(b+1) = (2)(20). If a+2 = 2, then a is 0 which we can't have. If a+2 = 20, then a > b which we can't have either. I will skip this analysis in the subsequent cases since it's pretty quick. The restrictions are that a, b, and the p_i are positive integers and that a < b.

(ii) (a+2)(b+1) = (4)(10). This gives (a, b) = (2, 9).

(iii) (a+2)(b+1) = (8)(5). This gives (a, b) = (3, 7).

(iv) (a+2)(b+1)(e_1 + 1) = (5)(2)(4). This implies (a, b, e_1) = (2, 4, 1).

(v) (a+2)(b+1)(e_1 + 1) = (2)(2)(10). We don't generate any successful triples here.

(vi) (a+2)(b+1)(e_1 + 1)(e_2 + 1) = (2)(2)(2)(5). Here again we don't generate any successful quadruples.

Our list of possibilities is reduced to (a, b) = (2, 9) or (3, 7) or (a, b, e_1) = (2, 4, 1). Let's check which of these satisfy (2). I'll associate with these pairs (triple) their corresponding subcase above:

(ii) (a, b) = (2, 9). Subsitute into (a+1)(b+2) to get (3)(11) = 33 which does not equal 36.

(iii) (a, b) = (3, 7). Substitute into (a+1)(b+2) to get (4)(9) = 36, so this works.

(iv) (a, b, e_1) = (2, 4, 1). Substitute into (a+1)(b+2)(e_1 + 1) to get (3)(6)(2) = 36 which also works.

We have the following two forms for n:

n = (3^3)(5^7) or n = (3^2)(5^4)(p) where p is a prime not equal to 3 or 5. This answers the second part of the question. Now for the first part:

If n = (3^3)(5^7), then 15n = (3^4)(5^8) which has (4+1)(8+1) = 45 positive integer divisors. If n = (3^2)(5^4)(p), then 15n = (3^3)(5^5)(p) which has (3+1)(5+1)(1+1) = 48 positive integer divisors.

for 5n,

36 = 2*2*3*3 divisors

for 3n,

40 = 2*2*2*5 divisors

let n = a^x * b^y * c^z, then n will have :

(x+1)(y+1)(z+1) divisors, where a, b, and c are all primes. and x, y, z ≥ 0

i just take 3 variables, because both 36 and 40 have 4 prime factors. assuming the difference in the multiplicators (3 and 5) gave the fourth factors, all the other 3 came from n.

if there aren't any 3(s) or 5(s) in n, then

number of divisors of 3n and 5n = 2(x+1)(y+1)(z+1). but we know this is not the case, so in n, there must exist at least 1 "3" or/and 1 "5".

if either one is a factor but not the other, the number of divisors would be

2(x+1)(y+1)(z+1) against (x+2)(y+1)(z+1), assigning x to be the exponent of repeated prime factor. we have 5 in 40, that is an abnormality(?). so if this is really the case, we'll have to "eliminate" 5, by being one of the retorted (x+1) or (x+2).

x+2 could be 5, 6 or 10, and x+1 be 4, 5 or 9 (respectively).

then (y+1)(z+1) must be 8, 6, or 4 (in that order). we clearly see that this is definitely NOT the case.

this only mean 1 thing; 2 of a, b, c must be 3 and 5. suppose a = 3 and b = 5. now x, y ≥ 1.

for 5n,

36 = 2*2*3*3 divisors

= (x+1)(y+2)(z+1)

= (z+1)[(xy+x+y+2) + x]

for 3n,

40 = 2*2*2*5 divisors

= (x+2)(y+1)(z+1)

= (z+1)[(xy+x+y+2) + y]

z+1 is either 1, 2, or 4. but 5 is certainly (x+2) OR (y+1). must be or. that will lead us to x = 3 OR y = 4.

and also, from

36 = (z+1)[(xy+x+y+2) + x] and

40 = (z+1)[(xy+x+y+2) + y],

we get y = x + [4/(z+1)].

the OR rule nullified (z+1) = 4. so z+1 is either 1 or 2.

if x = 3, y = 5 OR 7. <-- (edited : wrote 5 or 9 before)

if y = 4, x = 2 OR 0.

x = 0 is contradictory (we already established of 3 AND 5 being repeated primes). for (x,y,z), we have 3 bawai;

(2,4,1), (3,5,1) or (3,7,0).

from all 3 sets, (x,y,z) = (2,4,1) and (3,7,0) gave the right answers.

n = 3^2 * 5^4 * c^1

= 75^2 * c

= 5625c

or

n = 3^3 * 5^7 * c^0

= 2109375

answer = many (with 1 fixed value), as long as c is prime other than 3 or 5.

number of divisors of 15n

= (3+1)(5+1)(2)

= 4 * 6 * 2

= 48

or (for 2109375)

= (4+1)(8+1)(1)

= 45