Solution of the Week #341 - Mastermind

From guess 3, one is in the correct place. It cannot be the initial Red as that would have shown as in the correct place in guess 1. Similarly it cannot be the final purple, as guess 2 would have shown it. So it must be the red in position 2.

Since the first two guesses cover all of the colours, but only show two correct pegs, there must be a colour repeated. It cannot be the red, or else that would have shown in guess 3. It must be one of the colours from guess 2, but since they must occupy positions 1 and 3 in the answer, they cannot be Orange or Purple, as they would have shown as in the correct place in guess 2.

The final answer is therefore Blue Red Blue

Solution of the Week #340 - Infinite Coin Push

All three coins will be at (13,7).

The key to solving this is to compute the x and y coordinates separately. We are given the first three x coordinates, 0, 3, 9 and each subsequent x coordinate will be 3/4 of the previous two less 1/2 of the one before that. Eventually this settles on an x coordinate of 13. Doing the same for the y coordinates will eventually settle on 7.

 

In general, given starting positions of A, B, C, the coins will converge on: (-2A+B+4C)/3.

 

Solution of the Week #339 - Envelope Centroid

The centroid of a shape is the weighted average of the centroids of its component parts.

 

A_1.C_1 + A_2.C_2 = A.C

 

Where A_1 is the area of the unshaded region, C_1 is the centroid height of the unshaded region etc.

 

We require C_2, so making that the subject:

 

C_2 = (A.C - A_1.C_1)/A_2

 

We can calculate the area and centroid height (A and C) of the overall figure from first principles, as it is simply an equilateral triangle, and we can find A_2 by simply taking A_1 from A.

 

C_2 = (rt(3)/4 x rt(3)/6 - rt(3)/6 x rt(3)/10) / (rt(3)/12)

 

C_2 = 3rt(3)/10 =~ 0.5196…

 

 

Solution of the Week #334 - Skeleton Cube

When viewed such that the cubes where the spheres are centred are at the top and bottom of the figure, those points are square root of 3 apart. The unit lengths lines from the top and bottom vertices describe the extent of the overlap of the two spheres (they are actually cone shaped but from this angle appear to be equilateral triangles). The circle in the middle of the cube is viewed here edge-on, and so its diameter is 1.

cube without spheres sol.jpg

Solution of the Week #333 - Non-seqitur

Since 5 is a prime number, the only way a number can have exactly five factors is if it is the fourth power of a prime number, in which case the factors will be 1, p, p^2, p^3 and p^4.

Therefore, to answer the question we need to find out how many prime numbers lie between the fourth root of 10^4 and the fourth root of 10^6. The fourth root of 10^4 is clearly 10. The fourth root of 10^6 is 10^(6/4) which is 10^(3/2) or 10 x 10^(1/2), in other words 10 x the square root of 10. The square root of 10 is given in the question as about 3.16, so therefore the fourth root of 10^6 is about 31.6.

So the answer we seek is the number of prime numbers between 10 and 31.6, which is straightforward enough to simply count: 11, 13, 17, 19, 23, 29, 31 which gives the answer of 7.

Solution of the Week #332 - Even Binary Puzzle

The highest number that will terminate at 25 is 123:

123  1111011    111 x 1011 7 x 11

77   1001101    1001 x 101 9 x 5

45   101101     101 x 101  5 x 5

25   11001      stop

 

The highest number that will stop at 32 is 965:

965  1111000101      111 x 1000101   7 x 69

483  111100011       111 x 100011    7 x 35

245  11110101        111 x 10101     7 x 21

147  10010011        100100 x 11     36 x 3

108  1101100         110 x 1100      6 x 12

72   1001000         100 x 1000      4 x 8

32   100000          stop

 

Solution of the Week #331 - Sixth Powers

All sixth powers are either a multiple of 13 or one away from a multiple of 13:

 

1 = 13x0 +1

64 = 13x5 -1

729 = 13x56 +1

4096 = 13x315 +1

5^6 = 13x1202 -1

6^6 = 13x3589 -1

7^6 = 13x9050 -1

8^6 = 13x20265 -1

etc.

 

To check that it works for ALL sixth powers, you only need to check the first 13, since:

(n+m)^a = (n)^a (mod m)

 

Proof: if you multiply out the left-hand side, one term will be (n)^a, plus several other terms which will all contain m, and therefore be a multiple of m, and not affect its value, modulo m.

In actual fact you only have to look at the first 6, since

(m-n)^a = (-n)^a (mod m), for precisely the same reason.

To be sure that there are no numbers higher than 13 that work, you only need to look at the first couple of sixth powers (and the numbers either side of them) and see what their factors are.

The only numbers that divide into (63, 64, OR 65) AND (728, 729 OR 730) are 1,2,3,4,5,7,8,9,13.

13 is the highest of this list, so as long as it proves to be a valid solution, it will be the highest such solution.

Solution of the Week #329 - p+2=sp

Since we are told that a pair of 3s are amongst the prime factors of n, the semiprime 9 is a factor, and since 9 is 2 greater than a prime, 7 is also a factor. So another semiprime factor is 21, therefore 19 is a prime factor. Of the semiprime factors involving 19, 133 is 2 greater than a prime number, so 131 is also a prime factor of n. Of the semiprimes involving 131, there are none that are 2 greater than a prime, so we can stop there.

n is therefore 3x3x7x19x131 = 156807.

There could be other factors, for instance if 41 were also a prime factor, the rule would still be satisfied, since 41 in not 2 less than a semiprime, and each of the semiprimes involving 41 would not be 2 greater than a prime, so 156807 is merely the smallest.

 

For the second challenge, we need to introduce another prime factor, as small as possible.

It cannot be 2, as that would imply 4 divides too, but we know we can’t have any more repeat prime factors.

If it was 5, that implies 13 is also a factor, so we’ll put that to one side.

If it was 11, 31 would also be a factor.

13 implies 5 is a factor, as we’ve already seen.

If 17 is a factor, none of the semiprimes formed with it are 2 greater than a prime, and 17 itself is not 2 less than a semiprime, so we can stop there.

Therefore multiplying the previous answer by 17 gives the next smallest: 2665719.

 

Solution of the Week #327 - False Conjecture

The 58th number: 64,320,987,654,320,987,654,320,987,654,320,987,654,320,987,654,320,987,654,321

is the first in the sequence to be prime.

 

The 40th number: 44,320,987,654,320,987,654,320,987,654,320,987,654,321

is tempting as it doesn’t have any prime factors smaller than 78 million, but it is in fact composite.

Several others only have two prime factors.

Solution of the Week #326 - Inch Centimetre Prime

To be a whole number of centimetres, the number of inches must be a multiple of 50. For the number of centimetres to be even, the number of inches can be further narrowed down to multiples of 100. For both the number of inches and the number of centimetres to avoid being one more than a multiple of 3, the number of inches must be a multiple of 300. This straight away narrows down the task to the point where we can check individual cases.

If the number of inches were 2100, that would convert to 5334cm: 2099, 5333 and 7433 are all prime.

So the distance is 2100 inches or 5334 cm.

This is the only possible distance under 200m as the next set of numbers that fit the criteria are: 9000 inches (22860cm).

Solution of the Week #325 - Uprooted

Let √x=c

Let √y=d

The two given equations become:

c^2d+cd^2=14

c^3+d^3=22

 

If you cube (c+d), you will get an expression purely based on the two above equations:

(c+d)^3 = c^3+d^3+3c^2d+3cd^2

(c+d)^3 = 22 + 14*3

(c+d)^3 = 64

(c+d) = 4

 

Since the first equation can also be written as cd(c+d) = 14,

cd = 14/4 = 3.5

 

If you square (c+d) you get:

(c+d)^2 = c^2 + d^2 + 2cd

But c^2 = x and d^2 = y, and we know (c+d) and cd, so therefore we can rearrange to:

(x+y) = (c+d)^2 - 2cd = 16 - 7

(x+y) = 9

Solution of the Week #324 - Number Hunt

5623109 = 23 x 41 x 67 x 89

There are in total 12 ways of combining four two-digit primes with no digits in common, but only one of the twelve products contains no repeating digits.

This is how you can reduce the number of possible arrangements to just twelve (which can then be easily checked to see if the product contains no repeated digits), just with a little logic:

Once we are past single digit primes, all primes must end in 1, 3, 7, or 9. Since we are after four primes with no repeated digits, they must end in these four digits, and therefore they cannot start with those four digits. They also cannot start with a 0 as they wouldn’t be two-digit numbers, so the start numbers must be four of the following five: 2,4,5,6,8. Since the primes starting with 2,5,8 all end in 3 or 9, only two of those three start numbers will be used, and the 3 and 9 will also come from those. Therefore 4 and 6 must be in any selection, and must be followed by 1 and 7.

In summary, two of the numbers must start with 2,5,8 and end with 3,9 (6 possibilities), and two must start with 4,6 and end with 1,7 (2 possibilities). Overall there are therefore 6x2= 12 possibilities.