## digitalmars.D.learn - puzzles (8-15-08)

• Wyverex (9/9) Aug 15 2008 1) Consider the number 142857. We can right-rotate this number by
• Leandro Lucarella (14/16) Aug 15 2008 Hi! No offense, but I find this constant Puzzle posts a little annoying.
• BCS (4/15) Aug 15 2008 A NG is really nice for this type of thing (not all posts are shown unti...
• Christopher Wright (8/21) Aug 15 2008 10^100 is greater than ulong.max.
• BCS (4/28) Aug 16 2008 The only practical way to do this would be an intelligent search.
Wyverex <wyverex.cypher gmail.com> writes:
```1)  Consider the number 142857. We can right-rotate this number by
moving the last digit (7) to the front of it, giving us 714285.
It can be verified that 714285=5*142857.
This demonstrates an unusual property of 142857: it is a divisor of its
right-rotation.

Find the digits of the set of all integers n, 11 < n < 10^100, that have
this property.

2 and 3 from the other day are time consuming but are good practice for
new programmers!
```
Aug 15 2008
Leandro Lucarella <llucax gmail.com> writes:
```Wyverex, el 15 de agosto a las 11:30 me escribiste:
2 and 3 from the other day are time consuming but are good practice for new
programmers!

Hi! No offense, but I find this constant Puzzle posts a little annoying.
Is there any chance you can move it to another list where just interested
people can subscribe?

Thank you!

--
Leandro Lucarella (luca) | Blog colectivo: http://www.mazziblog.com.ar/blog/
----------------------------------------------------------------------------
GPG Key: 5F5A8D05 (F8CD F9A7 BF00 5431 4145  104C 949E BFB6 5F5A 8D05)
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Hello?
Is there anybody in there?
Just nod if you can hear me.
Is there anyone at home?
```
Aug 15 2008
```Reply to Leandro,

Wyverex, el 15 de agosto a las 11:30 me escribiste:

2 and 3 from the other day are time consuming but are good practice
for new programmers!

Hi! No offense, but I find this constant Puzzle posts a little
annoying. Is there any chance you can move it to another list where
just interested people can subscribe?

Thank you!

A NG is really nice for this type of thing (not all posts are shown until
you open them and it's threaded)

Hay Walter! Any Way to talk you into adding an OT newsgroup?
```
Aug 15 2008
Christopher Wright <dhasenan gmail.com> writes:
```Wyverex wrote:
1)  Consider the number 142857. We can right-rotate this number by
moving the last digit (7) to the front of it, giving us 714285.
It can be verified that 714285=5*142857.
This demonstrates an unusual property of 142857: it is a divisor of its
right-rotation.

Find the digits of the set of all integers n, 11 < n < 10^100, that have
this property.

2 and 3 from the other day are time consuming but are good practice for
new programmers!

10^100 is greater than ulong.max.

I tried a version that went up to ulong.max. It logged every billion
checks. That took a few minutes (for the first billion, I mean). Getting
all the way to ulong.max would take overnight; to 10^100, ages, even if

Granted, I was using ulongs on my 32-bit machine, so it wasn't optimal.
But bignums are a lot slower.
```
Aug 15 2008
```Reply to Christopher,

Wyverex wrote:

1)  Consider the number 142857. We can right-rotate this number by
moving the last digit (7) to the front of it, giving us 714285.
It can be verified that 714285=5*142857.
This demonstrates an unusual property of 142857: it is a divisor of
its
right-rotation.
Find the digits of the set of all integers n, 11 < n < 10^100, that
have this property.

2 and 3 from the other day are time consuming but are good practice
for new programmers!

10^100 is greater than ulong.max.

I tried a version that went up to ulong.max. It logged every billion
checks. That took a few minutes (for the first billion, I mean).
Getting all the way to ulong.max would take overnight; to 10^100,
ages, even if you had a ucent primitive.

Granted, I was using ulongs on my 32-bit machine, so it wasn't
optimal. But bignums are a lot slower.

The only practical way to do this would be an intelligent search.

get out the discrete book and start manipulating this:
(D*10^n + M) % (10*M + D) && M < 10^n
```
Aug 16 2008