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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

Find a natural number for which, if you move its first digit at the end, you get a number that is half the original one

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

*Last edited by anna_gg (2012-12-02 03:16:06)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

Hi anna_gg;

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

bobbym wrote:

Hi anna_gg;

Bobbym,

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

Hi;

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

bobbym wrote:

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,262

Hi;

There are more solutions.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**scientia****Member**- Registered: 2009-11-13
- Posts: 222

Suppose the number is

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

*Last edited by scientia (2012-12-02 16:54:02)*

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

Here is another variation: Find the smallest natural number for which, if you move its last digit at the beginning, you get a number that is 5 times the original.

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**scientia****Member**- Registered: 2009-11-13
- Posts: 222

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

scientia wrote:

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

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 3,847

Hi anna_gg,

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
```

*Last edited by phrontister (2013-03-04 00:40:44)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

Excellent! Thanks!

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So, we generalize the problem as follows:

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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