## Enigma 1775

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.

## Enigma 1705

Spoiler for New Scientist Enigma 1705:

Harry and Tom have each found a 4-digit perfect square, a 3-digit perfect cube and a 3-digit triangular number that use 10 different digits, but with no leading zero. A triangular number is one that fits the formula n(n+1)/2, such as 1, 3, 6, 10, 15.

Even if I told you Harry’s triangular number you would not be able to deduce with certainty which perfect square he has found. Even if I told you Tom’s perfect square you would not be able to deduce with certainty which triangular number he has found.

What are (a) Harry’s triangular number, and (b) Tom’s perfect square?

Break out the tables of squares, cubes, and triangular numbers…

There are only four 3-digit perfect cubes with no repeated digits: 125, 216, 512, and 729. Notice that each of these uses 2.

The 4-digit perfect squares with no repeated digits and no 2 are: 1089, 1369, 1764, 1849, 1936, 3481, 4096, 4356, 4761, 5041, 5184, 5476, 6084, 7056, 7396, 7569, 8649, 9604, 9801. Notice that each of these uses either 1 or 6 (or both), so the cube 216 can’t be used.

If the cube is 125 or 512, then the squares that could be used are 4096, 6084, 7396, 8649, and 9604. For 6084 and 7396 there are no 3-digit triangular numbers that use the remaining three digits. For 8649 there is 703, giving two sets of numbers (square, cube, triangular):

• 8649, 125, 703
• 8649, 512, 703

For 4096 and 9604 there is 378, giving four sets:

• 4096, 125, 378
• 4096, 512, 378
• 9604, 125, 378
• 9604, 512, 378

If the cube is 729, then the squares that could be used are 3481, 4356, 5041, 5184, and 6084. For 3481, 4356, and 5041 there are no 3-digit triangular numbers that use the remaining three digits. For 5184 there is 630, giving one set of numbers:

• 5184, 729, 630

For 6084 there are 153 and 351, giving two sets of numbers:

• 6084, 729, 153
• 6084, 729, 351

If I’ve done this right these are all the sets of such numbers that use all ten digits with no leading zero.

So Harry’s triangular number is 378 (his square could be 4096 or 9604), and Tom’s square is 6084 (his triangular number could be 153 or 351).