## 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 1769

Spoiler for New Scientist Enigma 1769: “Crossing lines” (Follow the link to see the puzzle.)

Easier than it looks at first.

Consider at first the three lines that cross at a point. On their own they create 6 edge regions and 0 non-edge regions. Call these lines 1 through 3.

Now add new lines one at a time, each crossing all the other lines but not at a point already containing a crossing. (Can this be done? I think so; you can always make each line nearly parallel to a previous one, close enough in spacing and angle to cross the remaining lines within some epsilon of the previous line’s crossings, yet not at the same angle, so it can cross the previous line at some point. These distances and angles might need to get very small, though.) Call these lines 4 through n. The kth line, at the time it’s added, crosses (k–1) previous lines and two edges. Since the created regions can never be nonconvex, each region that is crossed is crossed only once. Therefore the kth line crosses k regions, cutting each into two. Two of the crossed regions are edge regions, and each is cut into two edge regions, so the number of edge regions increases by 2. The remaining (k–2) regions are non-edge regions, each of which is cut into two non-edge regions, so the number of non-edge regions increases by (k–2). That should remind you of triangular numbers. The number of edge regions after the nth line is 2n and the number of non-edge regions is (n–2)(n–1)/2–1. Then the solution is obtained by solving

3(2n) = (n–2)(n–1)/2–1

with the result n = 15.

Do not expect a diagram of this.

## Enigma 1766

Spoiler for New Scientist Enigma 1766: “Triangular sums” (Follow the link to see the puzzle.)

There are several possibilities to consider. The single digit number could be the first in the sequence. Then it must be triangular: 1, 3, or 6. The second number must be two digits, with no zero, which when added to the first must be triangular, and the between the two numbers there must be no repeated digits.

The single digit number could be the second in the sequence. Then the first must be triangular, with a single digit difference between it and another triangular number; it can only be 10, 15, 21, 28, or 36. But 10 is ruled out because it has a zero, and 28 is ruled out because the corresponding single digit number, 8, shares a digit with it.

Could the single digit number be the third in the sequence? Then the first two numbers must add to 10, 15, 21, 28, or 36. They must have two digits each, which is impossible for 10 and 15, and they must not have the same first digit, which is impossible for 21 and 28. The first must be triangular, so all we’re left with is 21+15=36 (in either order), but those share a digit. So no, the single digit number cannot be third. Nor can it be fourth or fifth.

Then we’re left with these possibilities:

Sequence Sum Possible next numbers
1, 27, … 28 38
1, 35, … 36 69 84
1, 54, … 55 23
1, 65, … 66 87
3, 12, … 15 none
3, 18, … 21 none
3, 25, … 28 17
3, 42, … 45 91
3, 52, … 55 81
3, 75, … 78 none
6, 15, … 21 none
6, 39, … 45 none
6, 49, … 55 23 81
6, 72, … 78 13 58
6, 85, … 91 14 29
15, 6, … 21 34
21, 7, … 28 38
36, 9, … 45 none

For each first two numbers we consider which 2-digit numbers can be added to bring the sum up to another triangular number, and we rule out ones that repeat digits with themselves or the first two numbers, or which share a common factor with the first 2-digit number. The possibilities are shown in the fourth column.

So we have 16 sequences to consider, and in each case there are four unused digits, so we can quickly scan differences of triangular numbers (I used a spreadsheet) to see which ones we can use. There are only three cases where we can get a fourth number satisfying the puzzle requirements:

3, 25, 17, 46 but this leaves only 89 or 98, neither of which works

6, 49, 23, 75 and then 18 will give a triangular number — but it shares a common factor with 75

6, 49, 23, 58 and then 17 will give a triangular number. No common factors.

The only solution is 6, 49, 23, 58, 17 with sums 6, 55, 78, 136, 153.

## 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).