Enigma 1773

Spoiler for New Scientist Enigma 1773 “Cutting Corners”  (Follow the link to see the puzzle.)

We need a triangle with one side 65 mm (we’ll work in millimeters to make everything an integer), another side a+b, and another side c+d such that the following are Pythagorean triples:

• 56, ac+d
• 56, b, 65
• 60, ca+b
• 60, d, 65

So immediately we have b = 33 and d = 25. Now we just scan a table of Pythagorean triples for ones with 56 as one leg and a as the other, where a+33 is the hypotenuse of a Pythagorean triple which has 60 as a leg. We find 56, 42, 70; 42+33 = 75 is the hypotenuse of the triple 45, 60, 75. So the triangle’s other two sides (in cm) are 7.0 and 7.5.

Is that all there is to it? Seems too easy.

Enigma 1772

Spoiler for New Scientist Enigma 1772 (Follow the link to see the puzzle.)

Surprisingly easy for a Susan Denham puzzle. 5 across must be divisible by 5 so must end with either 0 or 5. If our even digit is 0 then one of 4 across and 5 across must be 00, which is presumably disallowed implicitly — otherwise there would be multiple trivial solutions of the puzzle, e.g. 50 at 5 across and zeros everywhere else.

Assuming all the numbers have to be nonzero, 5 across must end in 5 and the odd digit is 5. For 6 down, the only 2-digit multiple of 6 containing a 5 is 54, so our even digit is 4 and 5 across is 45. Excluding any more 5s from the third and fifth columns and third and fourth rows, 2 down is 44, as is 4 across, and 3 down is 44544.

7 across can only be 54444 or 45444 or 44454, but only the second of these is a multiple of 7. Likewise 1 down can only be 45444. The second digit of 1 across must be 4, and for divisibility by 3 the fourth digit must be 5.

So the (no-zeros) solution is unique:

 4 4 4 5 4 5 – 4 – 4 4 4 – 4 5 4 – 5 – 4 4 5 4 4 4

Enigma 1771

Spoiler for New Scientist Enigma 1771: “Squares either way” (Follow the link to see the puzzle.)

For the first part the smallest and largest numbers that represent dates in either format are 010100 and 121299, whose square roots are 100+ and 348+. We could just examine the squares of all the numbers from 100 to 348, but let’s make it easier.

We want squares of the form 0x0xyz or 1x1xyz. For instance, are there any squares of the form 0101yz? Well, √010100 > 100 and √010199 < 101, so no. Likewise for all the other possibilities except: √040400 > 200 and √040499 < 202, so 2012 = 0404yz, specifically 040401, as stated in the enigma; and √070700 > 265 and √ 070799 < 267, so 2662 = 070756. So answer to part 1): 7 July 2056.

For the other two parts, we’re looking for the next (chronologically) square date in a particular format where the last two digits are the year. Here we can consider numbers up to 311299, whose square root is 557+, so we could examine the squares of all numbers from 100 to 557. Or we could be smart about it.

Is the next square date this year? No, because no square ends in 3.

Next year? Squares can end in 4. But you’d need to square a number of the form xy2 or uv8. In the first case the second to last digit of the square is 4y mod 10; in the second case it’s 16v+6 mod 10. Either way it’s even, so can’t be 1 to give us a date in 2014.

2015? Here the square root must be of the form xy5, and the second to last digit of the square of xy5 is 10y+2 mod 10 = 2. Not 1.

2016? Now we have xy4 or uv6. In the first case the second to last digit of the square is 8y+1 mod 10, and in the second case it’s 12v+3 mod 10. Either way, it’s odd. For it to be 1 we need y=0 or 5, or v=4 or 9. So we have the following candidate numbers: 104, 146, 154, 196, 204, 246, 254, 296, 304, 346, 354, 396, 404, 446, 454, 496, 504, 546, 554.

The squares of these numbers that constitute legal dates in one form or the other are: 010816, 021316, 041616, 060516, 092416. It’s coincidental, I think, that the squares of all the candidate numbers from 346 on up have large values for the middle two digits (32 or larger), whereas for all but 346 and 354 we would need the middle two digits to be 12 or less. For part 2), 060516 = 6 May 2016 is the next day:month:year square date (and the only other such square date in 2016 is 1 August). For part 3), 010816 = 8 Jan 2016 is the next month:day:year square date (and the other square dates are 13 Feb, 16 Apr, 5 Jun, and 24 Sep).

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 1760

Spoiler for New Scientist Enigma 1759: “Squares and cubes” (Follow the link to see the puzzle.)

The number which is both a square and a cube must be a sixth power: 26 = 82 = 43 = 64, or 36 = 272 = 93 = 729.

Note that the only cubes with 2 or 3 digits are 27, 64, 125, 216, 343, 512, and 729.

Try 64 at 6 down. Then 6 across must be a square: 625 or 676. If 625 then 5 down has middle digit 5 which means it’s 256. Then 7 across can only be 64, which duplicates. If 6 across is 676 then there is no possibility for 2 down. So 6 down is not 64.

Try 64 at 3 down. Then 4 across is 144, 324, 484, or 784. If 144 then 2 down must be 216 or 512. The former leaves nothing for 1 across; the latter forces 1 across to be 25. 5 down must be 441 or 484 (400 is ruled out because we don’t allow a leading zero for 7 across), but no 3-digit cube ends with 8 so it’d have to be 441. Then 7 across must be 16, 6 across is 324, and 6 down is 36. If 4 across is 324 the middle digit of 2 down is 3, but there are no 3-digit squares or cubes with 3 as the middle digit. If 4 across is 484 or 784 then 5 down is 841 and 7 across is 16. In the 484 case 2 down must be 144 or 441 or 841, none of which leaves anything for 6 across. In the 784 case 2 down must be 576 or 676 either of which leaves nothing for 6 across.

The other possibilities are 729 at 2 down or 5 down. 5 down doesn’t work because it leaves nothing for 7 across.

With 729 at 2 down, 6 across can only be 196, with 6 down being 16. Then 5 down must be 169, 361, or 961. The first leaves nothing for 7 across while the latter two leave only 16, which duplicates.

So there is indeed only one solution:

 2 5 6 1 4 4 3 2 4 6 1 6

Enigma 1759

Spoiler for New Scientist Enigma 1759: “Cell count” (Follow the link to see the puzzle.)

I can’t see an easy way to do this by hand, but it can be done in a spreadsheet; the key is to figure out the needed formula.

Consider just the upper right quadrant; of course the number of cells crossed for the whole circle will be 4 times the number for the quadrant. Now consider each increment along the x axis from 0 to 1 to 2 to … to r, the radius. When x goes to (x+1), y goes from √(r2x2) to √(r2–(x+1)2). The number of cells passed through in this column is 1 greater than the number of times the circle crosses a horizontal line (not counting lines started or ended on). Think about it hard enough and you realize the number of cells crossed in the file starting at x is:

√(r2x2)√(r2–(x+1)2)

where xand xare respectively the ceiling and floor functions. Multiply that by 4 and sum over all x from 0 to r–1 and you have the number of cells crossed.

So put that into a spreadsheet and graph the result: For a radius of less than 50, the only place the number of cells decreases when the radius increases is in going from radius 24 (188 cells) to 25 (180 cells).