A number problem.

bobbym · 2013-03-02 17:02:23

Hi;

This one was posed in another thread and solved there by anonimnystefy.

Find the smallest natural number for which, if you move its last digit to the beginning, you get a number that is 6 times the original. Or prove that no such number can exist.

We will follow his fine idea and see how we can utilize a CAS to help.

Suppose there were a 3 digit number that was the answer then:

or

No insight yet on the problem so we try a 4 digit number.

One more a 5 digit number:

We see a pattern developing and this is an important part of experimental mathematics.

The pattern is there is always a multiple of 59 on the left and a predictable form of 10^k - 6 on the right.

We know that

We also know that a0 which is the front digit of the new number must be one of these {1,2,3,4,5,6,7,8,9}.

The whole thing now presents itself as an algorithm that leads to this pseudocode:

1) Begin k loop

2) h = 10^k - 6 / 59

3) Times h by 1,2,3,4,5,6,7,8,9 and see if the products are integers.

4) Yes they are, print k and stop

5) They are not go back to 1.

Last edited by bobbym (2013-03-02 17:32:08)

Agnishom · 2013-03-02 19:44:59

Is k the number of digits?

bobbym · 2013-03-02 19:49:01

Hi;

The answers are 58 digits long, but k+1 is the number of digits.

Agnishom · 2013-03-02 20:37:51

Is k coming as 16?

bobbym · 2013-03-02 20:42:11

Hi;

You probably got that answer because of precision problems.
Your programming language is just not carrying enough digits. You need to have a language that has higher precision than 17 significant digits.

When I invented this problem I had no idea that such an answer would be necessary.

Agnishom · 2013-03-02 20:47:57

I never knew python is that bad

I will import decimal module and do again

bobbym · 2013-03-02 20:48:56

Does python support big integer arithmetic?

Agnishom · 2013-03-02 20:52:38

Ya good, Thank You I am getting 57

P.S: Defining numbers in Python isn't a problem at all. How big do you want your integer to be?

Last edited by Agnishom (2013-03-02 20:54:36)

bobbym · 2013-03-02 20:53:36

That is correct, for the first part!

Agnishom · 2013-03-02 20:55:29

Hmm... Let me see if I can find the number

bobbym · 2013-03-02 20:56:22

There is a little trick, you will have to refer to the math in post #1.

Agnishom · 2013-03-02 21:05:32

1694915254237288135593220338983050847457627118644067796610169491525423728813559322033898305084745762711864406779661

169491525423728813559322033898305084745762711864406779661

Is that it?

Agnishom · 2013-03-02 21:08:13

I am close but something is wrong

bobbym · 2013-03-02 21:09:55

Hi;

The great part of doing this type of work is that you never need to ask that question. You always have an
infallible way to know. Did you check that number?

Agnishom · 2013-03-02 21:23:38

Answer:

A. 1694915254237288135593220338983050847457627118644067796601

B. Something is still wrong

But B is 116 digits long

Last edited by Agnishom (2013-03-02 21:28:05)

bobbym · 2013-03-02 21:30:26

Hi;

Do not worry, we will sort it all out.

You remember from the math:

The a0 can take any value from 1 - 9. You only picked 1. The left has every digit except for a0. When you choose
a digit for a0 it goes on the back of the number.

Last edited by bobbym (2013-03-02 21:30:57)

Agnishom · 2013-03-02 21:36:02

you wrote:

3) Times h by 1,2,3,4,5,6,7,8,9 and see if the products are integers.

Isn't this number of times the value of a_0?

bobbym · 2013-03-02 21:37:22

Hi;

It is a0!

Agnishom · 2013-03-02 22:34:44

(10^57 - 6)*1 / 59 is an Integer.

So why won't 1 serve as a0?

Last edited by Agnishom (2013-03-02 22:37:14)

bobbym · 2013-03-02 22:40:13

It sure could but that does not mean it solves the problem. As a matter of fact that are 4 solutions in this area.

anonimnystefy · 2013-03-02 22:43:54

My solution was with a0=6 and a0=9.

bobbym · 2013-03-02 22:44:21

There are 2 more solutions.

anonimnystefy · 2013-03-02 23:06:22

I figured that out. And I think that I will be able to find them as soon as I look at my notebook.

bobbym · 2013-03-02 23:14:13

Once Agnishom figures it out then I can post another one.

anonimnystefy · 2013-03-02 23:26:23

I have a feeling it will be monstrous.

Math Is Fun Forum

#1 2013-03-02 17:02:23

A number problem.

#2 2013-03-02 19:44:59

Re: A number problem.

#3 2013-03-02 19:49:01

Re: A number problem.

#4 2013-03-02 20:37:51

Re: A number problem.

#5 2013-03-02 20:42:11

Re: A number problem.

#6 2013-03-02 20:47:57

Re: A number problem.

#7 2013-03-02 20:48:56

Re: A number problem.

#8 2013-03-02 20:52:38

Re: A number problem.

#9 2013-03-02 20:53:36

Re: A number problem.

#10 2013-03-02 20:55:29

Re: A number problem.

#11 2013-03-02 20:56:22

Re: A number problem.

#12 2013-03-02 21:05:32

Re: A number problem.

#13 2013-03-02 21:08:13

Re: A number problem.

#14 2013-03-02 21:09:55

Re: A number problem.

#15 2013-03-02 21:23:38

Re: A number problem.

#16 2013-03-02 21:30:26

Re: A number problem.

#17 2013-03-02 21:36:02

Re: A number problem.

#18 2013-03-02 21:37:22

Re: A number problem.

#19 2013-03-02 22:34:44

Re: A number problem.

#20 2013-03-02 22:40:13

Re: A number problem.

#21 2013-03-02 22:43:54

Re: A number problem.

#22 2013-03-02 22:44:21

Re: A number problem.

#23 2013-03-02 23:06:22

Re: A number problem.

#24 2013-03-02 23:14:13

Re: A number problem.

#25 2013-03-02 23:26:23

Re: A number problem.

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