Enigma 1758

Spoiler for New Scientist Enigma 1758: “Path-o-logical” (Follow the link to see the puzzle.)

If a is a north-south side of the park and b is the length of a path then a and b are a side and hypotenuse of a right triangle. Call its other side x. Likewise b and c, where c is an east-west side of the park, must be a side and hypotenuse of another right triangle. Now, if you draw a perpendicular to the south side of the park through the path intersection, then the second of these triangles is split into two. All these triangles are similar and in fact the one in the southeast corner is congruent to the one in the northwest, so that perpendicular line segment must also be and c, not a, is the long side of the park: ca + 25. Then the southwest triangle has sides x and 25. Call its hypotenuse (the distance from the southwest corner to the path intersection) y.

Summarizing, the sides of these three similar triangles are:

  • a : x : b
  • x : 25 : y
  • b : y : a+25

a and b must be integers (but nothing says x or y must be).

From similarity of the first two triangles we get x2 = 25a. From the Pythagorean Theorem applied to the first triangle, b2 – a2 = x2 = 25a. If we write ba + δ then 2δ+δ2/a = 25. δ then must be an integer in the range [1, 12] and of these only 12 and 10 give integer values for a in that equation, namely 144 and 20, with corresponding values for b of 156 and 30.

Then x = 60 or 10√5 respectively. Using the Pythagorean Theorem for the second triangle, y = 65 or 15√5 ~ 33.54 respectively. But for the path intersection to be inside the park we need yb. So the only solution that works is: a = 144, b = 156:

park

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Enigma 1757

Spoiler for New Scientist Enigma 1757: “Power point” (Follow the link to see the puzzle.)

The only 3-digit powers of 1-digit numbers are: 128, 256, and 512 (2^7, 2^8=4^4, and 2^9 = 8^3), 243 and 729 (3^5 and 3^=9^3), 125 and 625 (5^3 and 5^4), 216 (6^3), and 343 (7^3).

The four numbers with 2 as the middle digit (125, 128, 625, and 729) can connect to each other but none can connect to anything else, so they’re eliminated, leaving us with the remaining five numbers. Now all we need is the order.

512 can connect only to 216, so if we don’t start with it we have to end with it. But we start with an odd number, so 512 must be the last number and 216 must be second to last.

343 can connect only to 243, so if we don’t end with it we have to start with it. But we end with 512, so 343 must be the first number and 243 must be second.

Then 256 must be in the middle: 343–243–256–216–512.

 

Enigma 1755

Spoiler for New Scientist Enigma 1755: “Sudoprime II” (Follow the link to see the puzzle.)

Label the columns a through e (left to right) and the rows 1 through 5 (top to bottom). We’re given a1 = 7 and b3=3.

We’ll make lots of use of the fact that the last digit of a prime (other than 2) must be 1, 3, 7, or 9.

To make a1 down prime, a2 must be 1, 3, or 9. If 1: a2 across is 11, 13, 17, or 19; only 19 works. If 3: a2 across is 31 or 37; only 31 works. If 9: a2 across is 97, which doesn’t work. So a1 down and a2 across are 71 and 19, or 73 and 31.

Try 71 and 19:

b2 down must be 937. a5 can be 3 or 9, but if 9, there is no value for a4 that makes a4 across and a4 down both prime. If a5 is 3 then a4 can only be 4.

Now e5 can only be 1, and then e4 must be 3. And now d4 can’t be anything!

Try 73 and 31:

b2 down can only be 137 or 139. If it’s 139 then a5 is 1, but now there are no valid values for a4. So b4 is 7. a4 across and a4 down can be 17 and 19, 47 and 41, or 67 and 61.

d4 and e5 each can only be 3 or 9. Then e4 must be 1. That means d4 is 3 and e5 is 9.

a4 cannot be 1, so a5 is 1.

e2 can only be 7. d2 then is 4, 6, or 9. d3 can be 1, 7, or 9. Checking the nine possibilities for d2 down, the only ones that work are 613 and 673. So d2 is 6. d3 can be 1 or 7. a4 can’t be 6 so must be 4, and then e1 can only be 3.

b3 across is 3_1 or 3_7. Checking the possible values for c3 the only one that works is 0, with d3 equal to 7.

That’s everything. The solution is

a b c d e
1 7 3
2 3 1 6 7
3 3 0 7
4 4 7 3 1
5 1 9

and the shaded regions required are 307 and 41.

 

Enigma 1752

Spoiler for New Scientist Enigma 1751: “Pentagon of squares” (Follow the link to see the puzzle.)

I’m not happy about my solution to this puzzle. I think I have the right answer, but I had to resort to a computer program to get it. I’m probably thinking about it stupidly.

Here’s a picture of what’s going on:

e1752The red line segments are the pentagon; as required, it doesn’t contain the center of the circle it’s inscribed in . The interior angles (angles between the red line segments) are marked a2, b2, c2, d2, and x. Since the non-square angle x is the smallest, I’ve made it one of the angles on the long side. We know all of the interior angles are less than 180°, and all the ones that are squares must be larger than x and therefore larger than 1, so abc, and d must be in the range [2..13]. Also marked are the radii through each vertex, and the radial angles (between the radii) α, β, ɣ, and δ.

We know the interior angles of a pentagon must sum to 540°, but that’s not enough of a constraint; we also have to make sure the pentagon can be inscribed in a circle. For that we use a theorem that says interior angles subtending the same chord are equal to one another, and to half the central angle subtending that chord. So for instance interior angle aand central angle α + β + ɣ subtend the same chord, so a= (α + β + ɣ)/2. Likewise x = (β + ɣ + δ)/2, and similarly (pun intended) b= 180 – (ɣ + δ)/2, c= 180 – (β + ɣ)/2, and d= 180 – (α + β)/2. From these relations we can get expressions for the radial angles:

α = 2(a2+c2)–360

β = 720–2(a2+c2+d2)

ɣ = 2(a2+d2)–360

δ = 720–2(a2+b2+d2)

So given choices for abc, and d, we can calculate α, β, ɣ, δ, and x. We require all of these to be positive; xa2b2c2, and d2; and α+β+ɣ+δ < 180 to make the center lie outside the pentagon.

There may be a clever way from here to find the one and only solution, but I don’t see it. Instead I wrote a Perl script that came up with:

a= 64, b2 = 169, c2 = 144, and d= 121; then α = 56, β = 62, ɣ = 10, and δ = 12 (summing to 140), and = 42. That in fact is exactly the pentagon I’ve illustrated above. In case you’re suspicious, the interior angles do indeed sum to 540.The answer they’re looking for is: 42, 64, and 121.

 

Enigma 1751

Spoiler for New Scientist Enigma 1751: “Rainbow square” (Follow the link to see the puzzle.)

We have R-O-Y-G-B-V across the top row. Where can the consecutive R-O-Y-B be on one of the diagonals? Not starting at the upper left corner; that would put two Os, Ys, and Bs in the second, third, and fourth columns. Not starting at the lower left corner for similar reasons (plus two Rs in the first column.) It can’t start anywhere else on the upper left/lower right diagonal, because that would put two Rs on that diagonal. It can’t start in the second row, fifth column; that puts two Ys in column 3. It can’t start in the third row, fourth column; then the B on the upper right/lower left diagonal would have to be in column 5, but there’s already a B in that column. The only place it can start, then, is in the fifth row, second column. Then the lower left square must be B, and the V-B-G-Y-O-R row must be the third row. So we have:

R O Y G B V
? ? ? ? G ?
V B G Y O R
? ? O ? ? ?
? R ? ? ? ?
B ? ? ? ? ?

Where can the B on the upper left/lower right diagonal go? Only in the fourth row, fourth column. Likewise, the O in that diagonal must go in the sixth row, sixth column.

R O Y G B V
? ? ? ? G ?
V B G Y O R
? ? O B ? ?
? R ? ? ? ?
B ? ? ? ? O

Now the G in the bottom row can only go in the second column, and then the Y in the bottom row can only go in the fifth column.

R O Y G B V
? ? ? ? G ?
V B G Y O R
? ? O B ? ?
? R ? ? ? ?
B G ? ? Y O

But that means the fifth row, fifth column can’t be Y, so must be V, and the second row, second column must be Y. The diagonal not containing R-O-Y-G, then, is R-Y-G-B-V-O.

There isn’t a unique solution for the remaining squares. Here’s one possibility:

R O Y G B V
O Y R V G B
V B G Y O R
G V O B R Y
Y R B O V G
B G V R Y O

or in color:

(Another solution is the same but with the G and Y squares in the first and last columns interchanged. There are others.)

Enigma 1750

Spoiler for New Scientist Enigma 1750: “Navigating the grid” (Follow the link to see the puzzle.)

The digits 1 through 9 sum to 45, so all the numbers in the list are divisible by 9, 3, and of course 1. If the final digit were even then each number in the list would be divisible by 2, but then it would also be divisible by 6, and therefore would have five factors in the 1 to 9 range. Since some of the numbers have only four such factors, the final digit must be odd, and the five factors of the five-factor numbers must be 1, 3, 5, 7, and 9. Those numbers must end with a 5; therefore so do the four-factor numbers, whose factors then must be 1, 3, 5, and 9.

How many such numbers are there? We can start from 5 and traverse the paths backwards. From 5 (in the center square) we can go to a corner square or to an edge square. First consider corner squares; there are four options, equivalent under rotational symmetry. For each there are two possibilities for the next square, namely the two adjacent edge squares, which are equivalent under reflection symmetry. So let’s count the paths that start 5–1–2 and multiply that by 8. From 2 we could go to 6 but then could not complete a path to all 9 squares. Or we go to 3, then 6, then (similarly) 9, then 8. From there we can go to 7 and then 4, or to 4 and then 7. Or from 2 we go to 4, then 7–8–9–6–3 is forced. So there are three paths starting with 5–1–2, or 24 starting with a corner.

If from 5 we go to an edge square there again are eight symmetry equivalent ways to do the first two moves, and from there it’s forced; e.g. 5–2–3–6–9–8–7–4–1. So there are just eight paths starting with a step to an edge square.

The total number of paths then is 32, eight of each of the types shown here:

Four path types

The list of numbers is:

236987415
741236895
748963215
896321475
963214875
968741235
123698475
123698745
124789635
142369875
147896235
147896325
214789635
321478695
321478965
326987415
362147895
369874125
369874215
412369875
478963215
632147895
698741235
741236985
784123695
789632145
789632415
874123695
963214785
986321475
987412365
987412635

of which the first six are divisible by 1, 3, 5, 7, and 9, and the rest by 1, 3, 5, and 9.

Enigma 1749

Spoiler for New Scientist Enigma 1749: “Genotype” (Follow the link to see the puzzle.)

This is pretty easy. Writing it as a multiplication and filling in the numbers we’re already given we have

9E0 x E9IGMA = 5CIE9TI5T

The left side is a multiple of 10, so the right side is too: T = 0:

9E0 x E9IGMA = 5CIE90I50

or, dividing by 10,

9E x E9IGMA = 5CIE90I5

 Now either E or A (or both) must be 5, and the other must be odd; but to get 5 as the first digit on the right E must be 5:

95 x 59IGMA = 5CI590I5

 and whatever IGMA is, 95 x 59IGMA is between 5605000 and 56999905 so C is 6:

95 x 59IGMA = 56I590I5

 Of the 10 possibilities for I, only one makes 56I590I5 a multiple of 95: I = 3, with 56359035/95 = 593253.

So G = 2 and GENETICS = 25950365.