## Magic Cuboid by Brimstone

Posted by shirleycurran on 13 March 2015

Yes, one of the Numpties, even with a reasonable pass in O level maths, takes a week off when there is a numerical, and the other Numpty puts in a few grumbling hours with calculator, pencil, paper and a glass of the good stuff (well, Brimstone obviously didn’t qualify for the Listener Setters’ Tipsy Crew with his completely alcohol-free clues so someone has to!) – see fig. 1, sheaves of scribbled notes finally leading to the desired solution.

Fortunately there is an able friend who made less of a pig’s ear of it and kindly sent me his two and a half pages of neatly rationalised solution (see fig. 2).

Many thanks to Brimstone. The consensus is that this was nowhere near so tough as the preamble initially suggested it would be and that the rotating of the square slices was very satisfying.

Here’s the polished solution path:

Magic Cuboid by Brimstone

Entries in the original grid will be shown with spaces to indicate the cell boundaries. For example if a 3-cell entry were 12345, I would show it as one of 1 23 45, 12 3 45 or 12 34 5.

(a,b) denotes the numbers between a and b. E.g. (1,9) mean 1 thru 9.

[r,c] means the entry in row r, column c.

16a > 2*(F^7)^2 = 2*F^14. So F^14 < 50000 => F<3. F=2

6d = 16 + (V-2)^2 + 4V^6 < 10000 => V^6<50000 => V<7. So V = 3 or 5. If V=5, 18a=48V^2=1200 but 00 is not a valid cell entry. V=3. 18a=4 32 or 43 2. 16d = 3 42 or 34 2. These two entries must match so 18a= 43 2 and 16d=34 2. 6d=29 33.

16a = 324U^2 + 32768 and starts with 34 so is either in (34000,34999) or (340000,349999). If 16a <= 34999, then 324U<=(34999-32768)=2231 => U^2<6 => U=2 which is F. So 16a is in (340000,349999) =>

324U^2 is in (307232, 317231) => U^2 is in (948,979) => U is in (30,32) => U=31, 16a=34 41 32.

5a =4Q^2 + 9Q^2 = 13Q^2. 3d = 36 + Q^4. Since 11^4>10000, Q<11 => Q=5 or 7. If Q=5, 5a=325 and 3d=661 which do not match. So Q=7, 3d=24 37, 5a=6 37.

1d=1 48 or 14 8. So 1a=5J^2 is either in (100,199) or (1400,1499) => J^2 is in (20,39) or (2800,2999) =>

J is in (4,6) or (52,54). J=5 or 53. J=53 would make 8a much too big. J=5, 1a=1 25, 1d=1 48, 4a=28 12, 8a=14 10, 10a= 8 30 (not 83 0 because 0 is not a valid cell value), 12d=2 27 or 22 7, 15d=4 36 or 43 6.

Note that we now have the first three cells in column 1: 1, 48 and 29. So the magic square total in each case is 78. So 15d=4 36 (not 43 6).

10d=4K^2 + 162 is in (800,899) => 4K^2 is in (638,737) =>K^2 is in (160,184) => K=13, 7a=13 19,

10d=8 38, 17a=18 40 36.

6a = 9 + 25E^2 + 250000 is in (290000,299999) =>25E^2 is (39991,49990) => E^2 is in (1600,1999) =>

E is in (40,44). E=41 or 43. If E=43, 2d=22 91, but no cell entry is more than 51. E=41, 6a=29 20 34,

2d=21 23.

In column 1 we have [4,1]=33 and [5,1]=14. So the [6,1] = 31 (to give the magic total). Hence 9a starts with 31 => 9a=4 + 9R^2 + 2025 < 3200 => 9R^2 < 1171 => R^2 < 131 => R<12 => R=11. 9a=31 18, 11a=27 11 22, 12d=22 7 (not 2 27).

[8:5]=34 and [9:5]=2, so cell [7:5]=78-34-2=42. So 14a = 169 + 3969 + (2M+7)^2 + 8100 is in (42007,42997) (the last cell is 7 from 12d)=> (2M+7)^2 is in (29762, 30759) => 2M+7 is in (172,175) => 2M is in (165,168) => M=83, 13a=4 43 38 or 44 3 38, 14a=42 16 7.

If 13a=4 43 38, row 8 column 1 would have to be 56 which is too big. 13a=44 3 38.

The rest of the cells can be easily deduced from the magic square constraint.

1 25 21 45 17 26 24

48 33 23 28 12 6 37

29 20 34 5 49 46 17

33 50 19 13 19 36 47

14 10 51 35 32 31 9

31 18 8 30 27 11 22

44 3 38 20 42 16 7

16 35 4 15 34 41 32

18 40 36 43 2 21 39

If a column is abcdefghi, the three possible rotations of the magic square are gdahebifc, ihgfedcba and cfibehadg. Of course the middle cell (e) is constant – row 5 is fixed. Since 1 must remain in the top left cell, column 1 is not changed. The 9 in column 7 is unchanged. So all of the other single digit cells must change. These seven cells add to 190 (200 – 1 – 9); there should be two primes; two squares and three non-square multiples of 4. We get these possibilities – the ones marked in red are not valid because they are not prime, square, multiples of 4 or they are squares that are multiple of 4.

2 -> 49,17,42

3 -> 40,50,25

4 -> 8,23,19

5 -> 45,20,43

6-> 36,41,11

7 -> 39,17,24

8 -> 23,19,4

The two squares are 49 and 24. One of the primes goes where the 6 is. So 4->23, 8->19 will not work. Thus 4->8 and 8->23. 5 and 7 get multiples of 4: 5->20; 7->24

So there are two possibilities for the values in the original cells with single digit

1 49 25 8 20 41 24 23 9-> 200 Winner!

1 49 25 8 20 11 24 23 9-> 170

1 20 38 43 42 21 17

48 18 19 15 19 41 22

29 40 21 20 17 16 39

33 33 4 30 34 11 37

14 10 51 35 32 31 9

31 35 23 13 12 36 32

44 25 36 5 2 46 24

16 50 8 28 27 6 47

18 3 34 45 49 26 7

Magic Cuboid by Brimstone

Entries in the original grid will be shown with spaces to indicate the cell boundaries. For example if a 3-cell entry were 12345, I would show it as one of 1 23 45, 12 3 45 or 12 34 5.

(a,b) denotes the numbers between a and b. E.g. (1,9) mean 1 thru 9.

[r,c] means the entry in row r, column c.

16a > 2*(F^7)^2 = 2*F^14. So F^14 < 50000 => F<3. F=2

6d = 16 + (V-2)^2 + 4V^6 < 10000 => V^6<50000 => V<7. So V = 3 or 5. If V=5, 18a=48V^2=1200 but 00 is not a valid cell entry. V=3. 18a=4 32 or 43 2. 16d = 3 42 or 34 2. These two entries must match so 18a= 43 2 and 16d=34 2. 6d=29 33.

16a = 324U^2 + 32768 and starts with 34 so is either in (34000,34999) or (340000,349999). If 16a <= 34999, then 324U<=(34999-32768)=2231 => U^2<6 => U=2 which is F. So 16a is in (340000,349999) =>

324U^2 is in (307232, 317231) => U^2 is in (948,979) => U is in (30,32) => U=31, 16a=34 41 32.

5a =4Q^2 + 9Q^2 = 13Q^2. 3d = 36 + Q^4. Since 11^4>10000, Q<11 => Q=5 or 7. If Q=5, 5a=325 and 3d=661 which do not match. So Q=7, 3d=24 37, 5a=6 37.

1d=1 48 or 14 8. So 1a=5J^2 is either in (100,199) or (1400,1499) => J^2 is in (20,39) or (2800,2999) =>

J is in (4,6) or (52,54). J=5 or 53. J=53 would make 8a much too big. J=5, 1a=1 25, 1d=1 48, 4a=28 12, 8a=14 10, 10a= 8 30 (not 83 0 because 0 is not a valid cell value), 12d=2 27 or 22 7, 15d=4 36 or 43 6.

Note that we now have the first three cells in column 1: 1, 48 and 29. So the magic square total in each case is 78. So 15d=4 36 (not 43 6).

10d=4K^2 + 162 is in (800,899) => 4K^2 is in (638,737) =>K^2 is in (160,184) => K=13, 7a=13 19,

10d=8 38, 17a=18 40 36.

6a = 9 + 25E^2 + 250000 is in (290000,299999) =>25E^2 is (39991,49990) => E^2 is in (1600,1999) =>

E is in (40,44). E=41 or 43. If E=43, 2d=22 91, but no cell entry is more than 51. E=41, 6a=29 20 34,

2d=21 23.

In column 1 we have [4,1]=33 and [5,1]=14. So the [6,1] = 31 (to give the magic total). Hence 9a starts with 31 => 9a=4 + 9R^2 + 2025 < 3200 => 9R^2 < 1171 => R^2 < 131 => R<12 => R=11. 9a=31 18, 11a=27 11 22, 12d=22 7 (not 2 27).

[8:5]=34 and [9:5]=2, so cell [7:5]=78-34-2=42. So 14a = 169 + 3969 + (2M+7)^2 + 8100 is in (42007,42997) (the last cell is 7 from 12d)=> (2M+7)^2 is in (29762, 30759) => 2M+7 is in (172,175) => 2M is in (165,168) => M=83, 13a=4 43 38 or 44 3 38, 14a=42 16 7.

If 13a=4 43 38, row 8 column 1 would have to be 56 which is too big. 13a=44 3 38.

The rest of the cells can be easily deduced from the magic square constraint.

1 25 21 45 17 26 24

48 33 23 28 12 6 37

29 20 34 5 49 46 17

33 50 19 13 19 36 47

14 10 51 35 32 31 9

31 18 8 30 27 11 22

44 3 38 20 42 16 7

16 35 4 15 34 41 32

18 40 36 43 2 21 39

If a column is abcdefghi, the three possible rotations of the magic square are gdahebifc, ihgfedcba and cfibehadg. Of course the middle cell (e) is constant – row 5 is fixed. Since 1 must remain in the top left cell, column 1 is not changed. The 9 in column 7 is unchanged. So all of the other single digit cells must change. These seven cells add to 190 (200 – 1 – 9); there should be two primes; two squares and three non-square multiples of 4. We get these possibilities – the ones marked in red are not valid because they are not prime, square, multiples of 4 or they are squares that are multiple of 4.

2 -> 49,17,42

3 -> 40,50,25

4 -> 8,23,19

5 -> 45,20,43

6-> 36,41,11

7 -> 39,17,24

8 -> 23,19,4

The two squares are 49 and 24. One of the primes goes where the 6 is. So 4->23, 8->19 will not work. Thus 4->8 and 8->23. 5 and 7 get multiples of 4: 5->20; 7->24

So there are two possibilities for the values in the original cells with single digit

1 49 25 8 20 41 24 23 9-> 200 Winner!

1 49 25 8 20 11 24 23 9-> 170

1 20 38 43 42 21 17

48 18 19 15 19 41 22

29 40 21 20 17 16 39

33 33 4 30 34 11 37

14 10 51 35 32 31 9

31 35 23 13 12 36 32

44 25 36 5 2 46 24

16 50 8 28 27 6 47

18 3 34 45 49 26 7

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