The Palindromic Diamond Cube

Question

A wealthy, famous eccentric had a huge cube of equal sized diamonds.* Due to her eccentricities the number of diamonds was known to be both palindromic in base ten, and a multiple of $7$ - her lucky number.

* That is a cube of $n^3 | n \in \Bbb{N}$ diamonds.

One day $7$ thieves stole the cube and got away in a spaceship. They planned to break the cube up to share the diamonds out equally, since it would be too difficult to sell such a unique piece anywhere in the system.

During the flight $2$ of the thieves decided to do away with the other $5$ thieves and share the diamonds equally, however they found that $1$ diamond would be left over.

As they discovered this a third thief discovered them, and suggested they do away with the other $4$ thieves instead and again share the diamonds equally. Again there would be $1$ diamond left over, and again as they discovered this a fourth thief discovered them.

The process repeated until the seventh thief found the others plotting against him, and still with a single remaining diamond after splitting them into $6$ equal shares.

What was the minimum number of diamonds in the cube?

The Seventh thief found them and found there'd be 1 left after spitting it in SIX? equal shares? What about thief #7 doesn't he get diamonds? — dcfyj, Jun 10 '16 at 12:07
@dcfyj When the $6$ were sharing equally there would be $1$ remaining diamond. They already know the number of diamonds is divisible by $7$ due to the well publicised peculiarities of the famous eccentric. — Jonathan Allan, Jun 10 '16 at 12:12

Gareth McCaughan · Accepted Answer · 2016-06-10T12:43:57.507

7

I believe there were

1033394994933301 diamonds, in a cube 101101 on a side.

Solution by semi-brainless computer search. Here's the code (though the code I actually used -- and briefly had here -- had a stupid bug which by mere good luck happened not to make it find the wrong answer).

for n in range(1,100000000):
  m = 420*n+301
  m = m**3 # number is a cube
  s = str(m)
  if s != s[::-1]: continue
  print(n,m)
  break

The only non-obvious things here are:

m = 420*n+301: for k=2,3,4,5,6 it happens that $n^3=1$ mod k iff $n=1$ mod k, so the side length of our cube must be 1 mod 60. It also happens to be a multiple of 7, which means it needs to be 301 mod 420 as per the earlier puzzle linked from the question here.
s[::-1]: terse but unobvious Pythonese for reversing a sequence.

edited Jun 10 '16 at 12:43

answered Jun 10 '16 at 12:32

Gareth McCaughan

119,120
7
313
454

$1685^3=4,784,094,125$, not 1033394994933301. – Trenin Jun 10 '16 at 12:34
Er, no, sorry, 1685 is the value of n, not of 60n+1. Will fix. – Gareth McCaughan Jun 10 '16 at 12:35
1685 per side there are 6 sides to a cube – dcfyj Jun 10 '16 at 12:35
@dcfyj Um... what? – Trenin Jun 10 '16 at 12:36
Fixed now. dcfyj, by "side" I meant the length of the cube edge. – Gareth McCaughan Jun 10 '16 at 12:36
Yep per edge, a palindromic number and per side a palindromic number. I did say she was eccentric. – Jonathan Allan Jun 10 '16 at 12:37
1

And 7 divides 101101, too. – Rosie F Jun 10 '16 at 12:38
Note that you do not need to check anywhere near that many n. – Jonathan Allan Jun 10 '16 at 12:38
@JonathanAllan What's sad is i had found the cube root of that number. Trenin, Gareth had written per side, he had never said that 1685 was the cube root. – dcfyj Jun 10 '16 at 12:39
you could have used m = 301 + n*420 since it's directly a multiple of seven among the 60*n+1 serie, and that would have gave you n=240. – Red-Jeweled Jun 10 '16 at 12:39
Jonathan: oops, forgot to exploit the divisibility by 7. – Gareth McCaughan Jun 10 '16 at 12:39
yes - you can start at 301 and add 420 each step :) – Jonathan Allan Jun 10 '16 at 12:41
What actually happened was that at first I did, then I put a bug in my program that made it actually force n to be 1 mod 7, then I made it a bit less stupid and had by then managed to forget the "multiple of 7" condition from the side length, noticed that the thieves never tried to split the diamonds 7 ways, and therefore removed the mod-7 condition altogether. My stupidity is a source of constant amazement to me. – Gareth McCaughan Jun 10 '16 at 12:41
By the way, what an absurd amount of diamonds that is ... At the point of having 2 thieves sharing, they shouldn't had to argue over a single piece, and flee with 516697497466650 pieces in hand. – Red-Jeweled Jun 10 '16 at 12:44
I've edited the code to be correct, with a note explaining that it used to be wrong so that the comments above aren't made too confusing for future readers. – Gareth McCaughan Jun 10 '16 at 12:44
Anton, perhaps diamonds are much less valuable in this hypothetical future world. – Gareth McCaughan Jun 10 '16 at 12:44
@Anton they were not necessarily arguing over the remaining diamond, but as they discovered the fact the next thief showed up anyway. – Jonathan Allan Jun 10 '16 at 12:46

score 2 · Answer 2 · answered Jun 10 '16 at 12:31

2

You don't actually say that when the 5th thief discovered the other 4 and demanded that the diamonds be divided into 5, that there would be 1 left over. So I suggest that there are 343 diamonds.

answered Jun 10 '16 at 12:31

Rosie F

8,621
2
18
50

"The process repeated" – Jonathan Allan Jun 10 '16 at 12:32
upv for creativity though :) – Jonathan Allan Jun 10 '16 at 12:39

score 1 · Answer 3 · answered Jun 10 '16 at 12:19

1

I believe the answer is:

10801 diamonds

because

if divided by 7 you get 1543 and if divided by any of the other numbers you get a remainder of 1. And also because it's the lowest palindromic number that fits the bill

I found the answer by using this:

vb.net code
For i As Integer = 0 To 50000
If i Mod 7 = 0 AndAlso i Mod 6 = 1 AndAlso i Mod 5 = 1 AndAlso i Mod 4 = 1 AndAlso i Mod 3 = 1 AndAlso i Mod 2 = 1 Then
txtPass.Text &= i.ToString() & "| "
End If
Next

answered Jun 10 '16 at 12:19

dcfyj

7,756
23
64

Unfortunately the answer you have given is not a cube of a natural number. – Jonathan Allan Jun 10 '16 at 12:21
1

That wasn't stated in the question, but ok. – dcfyj Jun 10 '16 at 12:22
1

I will add it explicitly to the text, I'd thought "cube of equal sized diamonds" was enough to specify this fact. – Jonathan Allan Jun 10 '16 at 12:23
1

To me that just sounded like a cube, as in a boxed shaped object – dcfyj Jun 10 '16 at 12:24

The Palindromic Diamond Cube

3 Answers3