Burt Kaliski Jr. (@modulomathy)

4/23/12 3:25 AM

#math trivia for #April23: #114 is a product of three primes. How many other numbers 1-366 have this form? What if prime powers are allowed?

The first form to consider is p*q*r where p, q, and r are distinct primes. The numbers with this form in the range 1-366 are

— 30 = 2*3*5

— 42 = 2*3*7

— 66 = 2*3*11

— 78 = 2*3*13

— 102 = 2*3*17

— 105 = 3*5*7

— 114 = 2*3*19

— 138 = 2*3*23

— 154 = 2*7*11

— 165 = 3*5*11

— 182 = 2*7*13

— 184 = 2*3*29

— 195 = 3*5*13

— 196 = 2*3*31

— 222 = 2*3*37

— 238 = 2*7*17

— 246 = 2*3*41

— 255 = 3*5*17

— 258 = 2*3*43

— 266 = 2*7*19

— 285 = 3*5*19

— 286 = 2*11*13

— 292 = 2*3*47

— 318 = 2*3*53

— 322 = 2*7*23

— 345 = 3*5*23

— 354 = 2*3*59

— 366 = 2*3*61

Note that one of the primes must be 2 or 3, because the smallest product not involving these primes is 5*7*11, which is larger than 366.

The second form also includes powers of p, q, and r (which still must be distinct). The numbers to be added depend on the factors already present in the list above:

— 30 = 2*3*5: 60, 120, 240, 90, 270, 150, 180, 360, 300

— 42 = 2*3*7: 84, 168, 336, 126, 294, 252

— 66 = 2*3*11: 132, 264, 198

— 78 = 2*3*13: 156, 312, 234

— 102 = 2*3*17: 204, 306

— 105 = 3*5*7: 315

— 114 = 2*3*19: 228, 342

— 138 = 2*3*23: 276

— 154 = 2*7*11: 308

— 182 = 2*7*13: 364

Neither 165 nor any number in the list that is greater than 182 can admit a prime power, because the product would be larger than 366.

The most distinct prime factors a day-number can have is four, as seen in 210 and 330. Many have three as noted in the lists above. How many have two or one?