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Puzzle NS1481

 

This puzzle was published in New Scientist, issue of 16 February 2008, as Enigma 1481 by Adrian Somerfield:

There are 10 cards, each with a single digit from 0 to 9.  6 is distinghuishable from 9.  Six cards with consecutive digits are selected.  From them three numbers of two digits are formed.  These three numbers have a factor in common that is not 1.  The numbers are then multiplied and the product is an odd number that has as a factor the cube of a number that is not 1 and not the common factor observed earlier.  What are the numbers?

Solution:

Let's make the grid of all two-digit numbers:

00 10 20 30 40 50 60 70 80 90
01 11 21 31 41 51 61 71 81 91
02 12 22 32 42 52 62 72 82 92
03 13 23 33 43 53 63 73 83 93
04 14 24 34 44 54 64 74 84 94
05 15 25 35 45 55 65 75 85 95
06 16 26 36 46 56 66 76 86 96
07 17 27 37 47 57 67 77 87 97
08 18 28 38 48 58 68 78 88 98
09 19 29 39 49 59 69 79 89 99

Since we selected six consecutive digits, there can be no number that has two digits that are further than 5 apart, e.g. 06 cannot be formed from any selection of six consecutive cards.  This eliminates two triangles of numbers:

00 10 20 30 40 50 60 70 80 90
01 11 21 31 41 51 61 71 81 91
02 12 22 32 42 52 62 72 82 92
03 13 23 33 43 53 63 73 83 93
04 14 24 34 44 54 64 74 84 94
05 15 25 35 45 55 65 75 85 95
06 16 26 36 46 56 66 76 86 96
07 17 27 37 47 57 67 77 87 97
08 18 28 38 48 58 68 78 88 98
09 19 29 39 49 59 69 79 89 99

Since all cards are different, there can be no number with two digits the same, eliminating all multiples of 11:

00 10 20 30 40 50 60 70 80 90
01 11 21 31 41 51 61 71 81 91
02 12 22 32 42 52 62 72 82 92
03 13 23 33 43 53 63 73 83 93
04 14 24 34 44 54 64 74 84 94
05 15 25 35 45 55 65 75 85 95
06 16 26 36 46 56 66 76 86 96
07 17 27 37 47 57 67 77 87 97
08 18 28 38 48 58 68 78 88 98
09 19 29 39 49 59 69 79 89 99

The numbers have a common factor that is not 1, therefore they must all be greater than 1 which eliminates 01.  Then, since the product is odd, the three numbers must be odd, which eliminates all even numbers:

00 10 20 30 40 50 60 70 80 90
01 11 21 31 41 51 61 71 81 91
02 12 22 32 42 52 62 72 82 92
03 13 23 33 43 53 63 73 83 93
04 14 24 34 44 54 64 74 84 94
05 15 25 35 45 55 65 75 85 95
06 16 26 36 46 56 66 76 86 96
07 17 27 37 47 57 67 77 87 97
08 18 28 38 48 58 68 78 88 98
09 19 29 39 49 59 69 79 89 99

The numbers have a common factor p, therefore they look like a.p, b.p, c.p.  The product is then a.b.c.p3 but the product also has a factor that is a cube of some number q whereby q is different from p.  Hence the product looks like p3.q3.r.  Note that q is not necessarily also a factor of each of the numbers, q3 could be a factor of only one of the numbers.

What candidates do we have for p?  Since all numbers must be odd, p must be odd.  All numbers must be odd multiples of p.  It is now useful to mark the prime numbers in our grid, though they are not eliminated:

00 10 20 30 40 50 60 70 80 90
01 11 21 31 41 51 61 71 81 91
02 12 22 32 42 52 62 72 82 92
03 13 23 33 43 53 63 73 83 93
04 14 24 34 44 54 64 74 84 94
05 15 25 35 45 55 65 75 85 95
06 16 26 36 46 56 66 76 86 96
07 17 27 37 47 57 67 77 87 97
08 18 28 38 48 58 68 78 88 98
09 19 29 39 49 59 69 79 89 99

The smallest odd number that is not 1 is 3, so if 3 is the value of p then our numbers must come from the multiples of 3 that remain in our grid:

03 15 21 27 45 51 57 63 69 75 87

If we write these as products of their prime factors we get:

3   3.5   3.7   3.3.3   3.3.5   3.17   3.19   3.3.7   3.5.5   3.29

If we take away the common factor p=3 we are left with:

1   5   7   3.3   3.5   17   19   3.7   5.5   29

We need to look for the cube of q, meaning that in any selection of three numbers there must be three times the factor q, but these three q's could be spread over two or three of the numbers or could occur all three in a single number.

q cannot be equal to p, but it could be a multiple of p, say p.k. 

Even if q is a multiple k of p, then we must still find three occurrences of k in this sequence.  There are two candidates for k or q:  3 and 5.

If we consider 3 then the candidates are 27, 45 and 63.  They do satisfy the condition that they are composed of six consecutive digits (2 to 7) and are therefore a solution.  The common factor is then p=3 and the cube factor of the product is then q=9.

Are there other solutions?

We look at values of p:

p=3

With this value of p the other possibility for q is 5 and the numbers would have to be
15, 45 and 75
or
45, 75 and one other number.

Both these sequences contain more than one digit 5.

p=5

p cannot be 5 because the three numbers would then end in 5 and there can be only one digit 5.

p=7

This leaves us with 21 35 49 63.  Because the largest difference between digits is 5 we cannot choose both 21 and 49, but because 35 and 63 both contain the digit 3 we cannot choose both of those either.

p=9

This leaves 27 45 63 and that is the solution we already have.

p=11

This is obviously already eliminated.

p=13

This leaves only two numbers: 13 and 65.

p=15

This leads to the problem with multiple digits 5.

All higher values of p leave only one or two numbers.

Q.E.D.

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