Integer problem 2

What is the smallest positive integer that has just 15 factors?

(Note: 1 and the integer itself are included in the factors. For example, the number 6 has 4 factors, that are 1, 2, 3, and 6.)

Solution 1

Noticing what integer has odd number of factors

Let a(n) be the number of factors a positive integer 'n' has,
a(1)=1, a(2)=2, a(3)=2, a(4)=3, a(5)=2, a(6)=4, a(7)=2, a(8)=4, a(9)=3, a(10)=4, ...
If 'p' is one of the factors of 'n', then n/p is also the factor of 'n', which means a(n) is an even number.
However, if 'n' is a square number, p=n/p when 'p' is the square root of 'n', which means a(n) is an odd number.
Therefore, the given condition a(n)=15 requires 'n' to be square numbers.
For square numbers 'n' beyond 10, a(n)s are as follows:
a(16)=5, a(25)=3, a(36)=9, a(49)=3, a(64)=7, a(81)=5, a(100)=9, a(121)=3, a(144)=15, ...
Hence, the solution is 144.


Solution 2

Utilizing the relation between a factorized integer and the number of factors the integer has

Suppose 'p', 'q', 'r', ... are distinct primes.
The factors of n=paqbrc... are given as the expanded terms of
(pa+pa-1+...+p+1)(qb+...+1)(rc+...+1)...
And the number of the factors is (a+1)(b+1)(c+1)...
If this equals to 15, then a=14 or (a, b)=(4, 2).
In the case of a=14, 'n' minimizes when p=2, or the smallest prime:
n=214=16384.
In the case of (a, b)=(4, 2), 'n' minimizes when (p, q)=(2, 3):
n=24*32=144.
Hence, the solution is 144.


Mathematical Problems

© 2006 Takashi Shimazaki
Updated: April 14, 2013