Spoiler for New Scientist Enigma 1775 “Third Symphony” (Follow the link to see the puzzle.)
First consider 3-digit cubic numbers, and cubic numbers plus 3. Filtering out all but the ones with one 3, no repeated digits, no zeros, and non multiples of 3, all that’s left is 732. But 732 is not triangular, is not a single-digit prime followed by a 2-digit prime (not (p)(pp)), does not start or end with 3 (not 3xx or xx3), and is not (pp)(p). So none of these work.
Now consider 3-digit triangular numbers. After filtering we have 153, 231, 351, 378, 435. Of these 231 and 435 are products of three primes; neither is 3xx or xx3; one is (p)(pp) and the other is (pp)(p). So these are two of the numbers.
The remaining number must be non triangular, non cubic or cubic plus 3, and have 3 of the 4 remaining properties. Enumerating (p)(pp) and (pp)(p) numbers that pass our filter, there are two numbers with both properties, 237 and 537. But both are not 3xx or xx3, and both are products of only two primes. So our fourth number must be (p)(pp) or (pp)(p) but not both, and must be 3xx or xx3, and must be a product of 3 primes.
Of the (p)(pp) and (pp)(p) numbers, only one is a product of 3 primes, and it’s xx3: 273.
So the three numbers are 231, 273, and 435.