Help with word Problem

[sallyk57]

The cube of a positive integer has 5 times as many divisors as the interger does itself. How many divisors does the square of the original number have?

I don't know how i should start the problem. I started out with listing a bunch of intergers from 1- 30 and listing their divisors but it wouldnt work.

 

[stapel]

I would start by picking some number of divisors, and working from there.

If the original integer had only one divisor, then it must be "1". But 1³ = 1, which still has only one divisor.

If the original integer had two divisors, then it must be a prime, p. But the cube p³ has divisors 1, p, p², and p³, four divisors. This is only twice as many as you'd started with.

So the integer can't be a prime number; it must be composite.

Suppose the composite number has two prime factors, so the number is mn, with divisors 1, m, n, and mn. Then (mn)³ = m³n³, with divisors, 1, m, m², m³, n, n², n³, mn, mn², etc. How many divisors do you get? Is it enough, or do you need to consider some other case?

There may be a better way to approach this, but that's how I'd start.

Eliz.

 

[soroban]

Hello, Sally!

Where did this problem come from?
It requires some knowledge of "number theory".

The cube of a positive integer has 5 times as many divisors as the interger does itself.
How many divisors does the square of the original number have?

I started out with listing a bunch of intergers from 1- 30 and listing their divisors but it wouldnt work.

But it should have worked! . . . \(\displaystyle \;N\,=\,24\) is the first number to be found.

The divisors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24 . . . \(\displaystyle 8\) divisors.

The divisors of \(\displaystyle 24^3\,=\,13,824\) are: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48,
\(\displaystyle \;\;\)54, 64, 72, 96, 108, 128, 144, 192, 216, 256, 288, 384, 432, 512, 576. 768. 864,
\(\displaystyle \;\;\)1152, 1536, 1728, 2304, 3456, 4608, 6912, 13824 . . . \(\displaystyle 40\) divisor.

Then the divisors of \(\displaystyle 24^2\,=\,576\) are: 1, 2, 3, 4, 6, 8, 9, 12, 16, 28, 24, 32, 36,
\(\displaystyle \;\;\)48, 64, 72, 96, 144, 192, 288, 576 . . . \(\displaystyle 21\) divisors.


Now, I'll show how I found it . . . certainly not by plug-and-chug.

Here is some of the necessary Number Theory:

If the prime factorization of a whole number is: \(\displaystyle \,N\:=\^a\cdot q^b\cdot r^c\,\cdots\)

\(\displaystyle \;\;\)where \(\displaystyle p,\,q,\,r,\,\cdots\) are primes and \(\displaystyle a,\,b,\,c,\,\cdots\) are positive integers,

\(\displaystyle \;\;\)then the number of divisors of \(\displaystyle N\) is:\(\displaystyle \,d(N)\:=\a+1)(b+1)(c+1)\.\cdots\)

(A rather neat formula . . Add one to each exponent and multiply them together.)


I tried it with \(\displaystyle N\) consisting of one prime factor, but there no solutions.

I conjectured that: \(\displaystyle N\;=\;p^a\cdot q^b\) . . . with two prime factors.
\(\displaystyle \;\;\)Then: \(\displaystyle \,d(N)\;=\;(a\,+\,1)(b\,+\,1)\)

Then \(\displaystyle \,N^3\:=\^{3a}\cdot q^{3b}\) . . . and: \(\displaystyle \,d(N^3)\:=\3a\,+\,1)(3b\,+\,1)\)

Since this second quantity is to be five times the first, we have:
\(\displaystyle \;\;\;(3a\,+\,1)(3b\,+\,1)\;=\;5(a\,+\,1)(b\,+\,1)\)

Solve for \(\displaystyle b\), and we get: \(\displaystyle \,b\:=\:\frac{a\,+\,2}{2a\,-\,1}\)

The only integer solutions are: \(\displaystyle \,(a,b)\:=\1,3),\;(3,1)\) ... which turn out to be "equivalent".


Hence, \(\displaystyle N\) is of the form: \(\displaystyle \,p^1\cdot q^3\) . . . which has: \(\displaystyle (1\,+\,1)(3\,+\,1)\:=\:8\) divisors.

Then \(\displaystyle N^3\:=\^3\cdot q^9\) . . . which has: \(\displaystyle (3\,+\,1)(9\,+\,1)\:=\;40\) divisors . . . check!


Therefore: \(\displaystyle N^2\:=\^2\cdot q^6\) will have: \(\displaystyle (2\,+\,1)(6\,+\,1)\:=\:\)21 divisors . . . . ta-DAA!
.

 

[Denis]

Lowest case for 10 times and 20 times:
120 has 16, 120^3 has 160 (10 times)
840 has 32, 840^3 has 640 (20 times)

Interesting(?):
28^3 has 28 divisors
40^3 has 40 divisors
I think these are the only 2 cases where n^3 has n divisors.