## Enigma 1780

Spoiler for New Scientist Enigma 1780 “Pure hedronism”  (Follow the link to see the puzzle.)

Not even with a bang…

Here’s the diagram with blanks labelled a through n. I also copied the 3 and 2 from the right to the left, and the 2 and 5 from the left to the right, to make the connections clearer.

Now it’s pretty much just reading it off. a must be 4 because we already have 1, 2, 3, and 5 along with a around a vertex. Then b and c must be 1 and 3, and b can’t be 1, so it’s 3 and is 1.

Likewise n is 1, and l and m must be 4 and 5, but m is not 4 so is 5 and l is 4.

d and e must be 2 and 5, but d is not 2, so is 5 and e is 2. That means h is 3.

Likewise j is 2 and k is 3. Then g is 5.

Finally f has to be 4 and i must be 1. ENIGMA is 324125.

Each of the four rows of triangles pointing in the same direction (there are two such rows interleaved in the middle) contains a cyclic permutation of 1-5-3-4-2.

And that’s it for Enigma.

## Enigma 1779

Spoiler for New Scientist Enigma 1779 “Four triangles”  (Follow the link to see the puzzle.)

Edited to correct two omissions, one pointed out by Jim Olson.

Is there an easier way to do this than to find or build a list of all Pythagorean triples with hypotenuse < 100 and start looking through it?

I find four chains of four if you allow the hypotenuse of each smaller triangle to be either leg of the next:

• 9-12-15 / 15-20-25 / 25-60-65 / 65-72-97
• 9-12-15 / 15-36-39 / 39-52-65 / 65-72-97
• 12-16-20 / 15-20-25 / 25-60-65 / 65-72-97
• 12-16-20 / 20-48-52 / 39-52-65 / 65-72-97

and the first two satisfy the requirement that the hypotenuse of the smaller triangle be the smaller leg of the next. In either case, the smallest and largest sides are 9 and 97.

Here’s a diagram of the second of these:

## Enigma 1778

Spoiler for New Scientist Enigma 1777 “Around the houses”  (Follow the link to see the puzzle.)

Kind of an odd puzzle and I wasn’t entirely sure how to interpret some of the wording.

But I think it’s pretty easy. First, if there are any houses 90 degrees apart then the number of houses must be a multiple of 4. Number of houses N=4n.

Second, if there are any digit reversal pairs 90 degrees apart then the house numbers differ by n: (10a+b)–(10b+a)=n. Then n=9a–9b. So n is a multiple of 9, and N=4n=36k.

Then there are only two possibilities, N=36 or 72. Examining these two — and I assume we don’t use leading zeros so 10 is not a reversal of 01, for instance, though in fact allowing that doesn’t permit any solution to the puzzle — we find:

For N=36, there are 2 90-degree pairs ((12, 21) and (23, 32)) and 1 180-degree pairs ((13, 31)).

For N=72, there are 5 90-degree pairs ((13, 31), (17, 71), (24, 42), (35, 53), and (46, 64)) and 2 180-degree pairs ((15, 51) and (26, 62)).

Now, does “the higher digit of the lowest-numbered house involved” refer to the larger digit or the digit in the higher place? If the latter then in either case it’s 1 and that’s the difference between the two counts for the N=36 case. If the former, then it’s 2 in the N=36 case, which isn’t the count difference, or 3 in the N=72 case, which is.

think they mean the larger digit, in which case the answer is 72.

I suppose one should check the cases of N=18, 54, and 90, for which the counts would be 0 for 90 degrees apart (there are no houses 90 degrees apart) and nonzero for 180 degrees, but even if those cases turn up more solutions, I doubt if they’re what was intended… and besides, I’m not that interested!

## Enigma 1777

Spoiler for New Scientist Enigma 1777 “Powerful tombola”  (Follow the link to see the puzzle.)

Let’s first ask, how many perfect squares are less than or equal to j? More generally, what is Nk(j), the number of kth powers less than or equal to j? Well, clearly, for example, if j = 50, there are 7 perfect squares less than or equal to j because √50 = 7.071…;  72 < 50 but 82 > 50. So Nk(j) = j1/k.

Now one’s first thought might be to add up Nk(j) for all k to get N(j), the number of powers less than or equal to j. But this gets into trouble.

First, if you ask how many fourth powers there are less than 50, all of the ones you find also are squares, so they don’t add to the total number of powers under 50. Likewise any other even power. So to avoid double counting, we consider only prime powers.

Here’s another double counting problem to handle: 1 is a power of every number.

And here’s another: consider (prime) powers of (prime) powers. For instance, 43 = (22)3 = (23)2 = 82. So counting up powers of 2 and 3 will include 64 twice.

The (incorrect) formula N(j) = 1 + Σ(j1/k–1), where the sum is over k = primes, takes care of the first and second types of double counting. Then we can correct for the third type. The smallest prime powers of prime powers are:

• (22)= (23)= 64
• (22)= (25)= 1024
• (32)= (33)= 729
• (42)= (43)= 4096

Those are all of them less than 15000; should be enough? Hope so. Now let’s do a binary search.

For j = 100, we have 100(1/2) = 10, 100(1/3) = 4, 100(1/5) = 2; then N(100) = 1+(9+3+1)-1 (the -1 corrects for double counting 64) = 13. That’s 13% winning numbers. We need j larger.

For j = 1000, we have 100(1/2) = 31, 100(1/3) = 10, 100(1/5) = 3; 100(1/7) = 2; then N(100) = 1+(30+9+2+1)-2 (the -2 corrects for double counting 64 and 729) = 41. That’s 4.1% winning numbers. We need j larger.

Similarly for 10000 there are 125 winners, getting close to our 1%. We need j larger.

Since we only have the corrections up to 15000 let’s try 15000. Turns out there are … 150 winners. Bang on 1%! (The sum is: 1+(121+23+5+2+1+1)-4).

There were 15000 tickets.

## Enigma 1776

Spoiler for New Scientist Enigma 1776 “Elevenses”  (Follow the link to see the puzzle.)

For a warmup puzzle see the comments below!

(At least) one number must be a multiple of 55. But the even multiples of 55 (through 990) have repeated digits, so it must be an odd multiple. So one number must be 165, 275, 385, 495, 605, 715, 825, or 935. The digit 5 must occur only in the units place of one number.

One number must be a multiple of 99, but not 594 (5 not in units place) or 990 (repeated digits). So one number must be 198, 297, 396, 495, 693, 792, or 891. The digit 9 must occur only in the tens place of one number.

There must be a multiple of 88, but not 440, 616, or 880 (repeated digits) and not 352 or 528 (5 not in units place) and not 968 (9 not in tens place). So one number must be 176, 264, 704, or 792.

297 shares one or more digits with each of these multiples of 88, so is ruled out.

If the multiples of 88 and 99 are distinct numbers, the only options are:

• 176 495; the latter is a multiple of 55. For the third number we need a multiple of 6, 7, and 11: 6 x 7 x 11 = 462 and 2 x 6 x 7 x 11 = 924, each of which uses two digits already used, so this is ruled out.
• 264 198 or 264 891; ruled out because no multiple of 55 can be used as the third number.
• 704 198 or 704 891; ruled out because no multiple of 55 can be used as the third number.
• 704 396 or 704 693; multiple of 55 must be 825. But 704 396 825 ruled out because no multiple of 77 and 704 693 825 ruled out because no multiple of 66.

So we have to use 792 as our multiple of both 88 and 99. Compatible multiples of 77 are 308, 385, and 693. Compatible multiples of 55 are 165, 385 and 605. But 605 is incompatible with all three multiples of 77 and 693 is incompatible with all three multiples of 55, so they are eliminated. So we’re down to these possibilities:

• 792 385 xxx
• 792 308 165

In the first case 792 is a multiple of 2, 3, 4, 6, 8, 9, and 11, and 385 is a multiple of 5, 7, and 11, so our third number only has to be a multiple of 11 using three of the digits 0, 1, 4, and 6. But there are none.

In the second case 792 is a multiple of 2, 3, 4, 6, 8, 9, and 11;  308 is a multiple of 2, 4, 7, and 11; and 165 is a multiple of 3, 5, and 11. So these three numbers are the only solution.

## 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 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).