Find the smallest natural number of two or more digits such that if you move the last digit to the front, the resulting number will be *n* times the original number.

Here are my calculations:

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Try the following in Excel (I use v2007):-

And here's a little program in BASIC, but testing more numbers, with the same single result (142857):

```
FOR n = 10 TO 5000000000 STEP 5
n$ = STR$(n)
a$ = RIGHT$(n$,LEN(n$)-1) + LEFT$(n$,1)
IF VAL(a$)*5 - n = 0 THEN PRINT a$
NEXT n
END
```

CORRECT! I found it by using Excel but now am trying to formulate it.

]]>NB:

(i) Since the number must be even, I write

(ii)

and cannot be zero but the other can be 0.So we want

i.e.

As the LHS is divisible by 19, so must the RHS, and as

we must have divisible by 19. The smallest suchPS: Yes, there are infinitely many solutions, because there are infinitely many *n* such that

There are more solutions.

]]>Hi;

Great! It is said that there are infinitely many solutions. I have used Excel for the calculations but have not been able to find any at the range 1-5,000,000. Then I gave up

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Hi anna_gg;

Bobbym,

Sorry, I had made a mistake - please read the new description!

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(e.g. 81345--->13458 but the resulting number must be the half of the original).]]>