#math trivia #120 solution

One way to calculate the number of positive divisors of 120 is to consider its factors:

120 = 2^3 * 3 * 5

There can be 0, 1, 2, or 3 2’s, 0 or 1 3’s, and 0 or 1 5’s.  The product 4*2*2 gives the number of positive divisors.

In order to get more than 16 positive divisors, the number must have at least three prime factors, because the most divisors possible with just two prime factors occurs at 2^3 * 3^3 = 216, which is again 16.  (Increasing either exponent produces a number greater than 366.  Increasing only the first exponent requires too much of a decrease to the second exponent.)

The numbers with four prime factors can only have exponents of 1, which again reaches just 16.

There are no day-numbers with five prime factors; the smallest number with five prime factors is 2310.

So, we’re back to three prime factors.  The total of the exponents can’t be more than 6, because the smallest value possible with three prime factors and an exponent total of 7 is 2^5 * 3 * 5 = 480, which is too large.  On the other hand, the total must be at least 5, because the most divisors possible with a total of 4 is 3*2*2 = 12.

The potential exponent combinations are thus:

(2,2,1) in any order — 18 divisors
(2,2,2) — 27 divisors
(3,2,1) in any order — 24 divisors
(4,1,1) in any order — 20 divisors

Do any of these exponent combinations produce day-numbers?

(2,2,1):  2^2 * 3^2 * 5 = 180; 2^2 * 3^2 * 7 = 252; 2^2 * 3 * 5^2 = 300
(2,2,2):  no possibilities in range
(3,2,1):  2^3 * 3^2 * 5 = 360
(4,1,1):  2^4 * 3 * 5 = 240; 2^4 * 3 * 7 = 336

The record number of positive divisors among day-numbers is thus held by 360, with 24, and five other day-numbers have more positive divisors than 120 does.

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