There is a number which has 8 divisors including, 8, itself, and 1.
There is also a number which has 18 divisors, including 18, itself and 1.
The difference between these numbers is 28.
What are the two numbers?
Since AB=28 and A is divisible by 4, then B is also divisible by 4. Therefore: A=2^{3}p, B=2^{2}3^{2}q. Let's [A] is the number divisors of [A], so [A]=8, [B]=18. The number of divisors for any number is:
[P_{1}^{k1}P_{2}^{k2}....P_{n}^{kn}]=(k_{1}+1)*(k_{2}+1)...(k_{n}+1)
where P_{n} are prime factors.
Thus, [A]=4[p] and [B]=9[q] therefore p and q are primes (p>2 and q>3). So, if two primes, p>2 and q>3, satisfy 8p36q=28 or
2p9q=7 (1)
then 8p and 36q satisfy the condition of the problem. The first ten such prime pairs are:
19 5
53 11
73 17
89 19
107 23
127 29
163 37
181 41
197 43
269 59
So the first ten solutions are:
152 180
424 396
584 612
712 684
856 828
1016 1044
1304 1332
1448 1476
1576 1548
2152 2124
One can suggest that there are infinite number of solutions but I'm not going to prove it.<o:p></o:p>
Edited on September 18, 2006, 3:26 pm
Edited on September 18, 2006, 3:30 pm

Posted by Art M
on 20060918 15:21:19 