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.

Solution of the Week #322 - Number Hunt

Since p-1 will be even for any odd prime, 2 must be a prime factor of n, and as it is also one less than a prime, so is 3.

Since 2x3 is one less than a prime, 7 is also a prime factor of n. Of the products of subsets of 2,3,7, only 2x3x7 is one less than a prime, so 43 is also a prime factor of n.

Of the products of subsets of 2,3,7,43, none are one less than a prime, so we can stop there.

n = 2x3x7x43 = 1806.

Solution of the Week #315 - Shaded Area

If you draw a vertical line between the apexes of the two triangles you will exactly divide the shaded area in two. Each half will have a side of 1 and another of x, and also a fixed angle (opposite the side 1) of 150 degrees. In order to maximise the area of this triangle, we must maximise the distance between the vertex with angle 150, and the side of length 1. This will be achieved when the triangle is isosceles, such that the distance between the two apexes is also x.

Since cosine of 150 degrees is -sqrt(3)/2, it’s possible to calculate x to be sqrt(2-sqrt(3)), which is approximately 0.5176.

 

Solution of the Week #312 - Wolf Tooth

Below is a diagram showing the necessary arrangement of the colours. The cube colours are in the squares, and the octahedron colours are in the hexagons.

The colours on the octahedron vertex in the middle of the green cube face are: Red, Orange, Silver, Blue.

WOLF TOOTH SOLUTION.JPG

And here is an image of the final cube, courtesy of Philip Morris Jones

PMJ solution.JPG