#math trivia #114 solution

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?

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