# Thread: Biggest natural number whrer each of her figures is smaller ...

1. ## Biggest natural number whrer each of her figures is smaller ...

Hello dear Sirs!

We try to solve this Problem, may be, you could help us to find a beginning of the solution?

Which is the biggest natural number with the quality that each of her figures is smaller except the first one and the last than the arithmetic means of her both neighbouring figures? Remark: The correctness of the result is to be proved.

Sincerly, Enoimreh

2. I would start with 3 digit numbers, then 4, etc.

3. Originally Posted by enoimreh7
Hello dear Sirs!

We try to solve this Problem, may be, you could help us to find a beginning of the solution?

Which is the biggest natural number with the quality that each of her figures is smaller except the first one and the last than the arithmetic means of her both neighbouring figures? Remark: The correctness of the result is to be proved.

Sincerly, Enoimreh
The first step in solving a problem is to determine what it means. We recently discussed what sounds like the same problem, but stated in a way that made it impossible to be sure what it meant! This version seems much clearer, but I would clarify it slightly (as I read it) like this:

What is the greatest natural number such that each of its figures [digits], other than the first and the last, is smaller than the arithmetic mean of its neighbouring figures?

So, first, choose a random number and show how you would decide whether it has the required property. Then use what you learn from doing that, to write a small number that has this property.

Once you've done that, we can start discussing how to find larger numbers that satisfy the requirement. Only then will we be able to propose and then prove the largest.

4. ## Thank you

Dear Mr. Peterson!

Thank you, for your answer. Yes, the way you described the problem, is the way I understood it.
So far I went on trying. Like
97679
976 works, as 7 < 7,5
767 works, as 6 < 7
679 works, as 7 < 7,5

9 in the middle never works - finished - as soon, as you need 10 or 11 or 12 it does not continue.

But I guess this is not the highest number and I will continue.
Sincerly, Enoimreh

5. Originally Posted by enoimreh7
Dear Mr. Peterson!

Thank you, for your answer. Yes, the way you described the problem, is the way I understood it.
So far I went on trying. Like
97679
976 works, as 7 < 7,5
767 works, as 6 < 7
679 works, as 7 < 7,5

9 in the middle never works - finished - as soon, as you need 10 or 11 or 12 it does not continue.

But I guess this is not the highest number and I will continue.
Sincerly, Enoimreh
That's a very nice beginning; you've done some good thinking. I haven't tried fully solving the problem, but your example provides some useful ideas.

Here are a few observations on this very interesting little problem:

• If you have a number that satisfies the requirements (e.g. 976), the reverse number (679) will also work. (Your example is a palindrome, so that the reverse number is the same; the final answer may be one, too.)
• Therefore, if the last number is larger than the first number, you should reverse it to get a larger number.
• One obvious goal is to start with the largest possible digit, as you did; but it is more important to have as many digits as possible. The latter may even override the former!
• I find it interesting to graph the digits on a grid, in your case (1,9), (2,7), (3,6), (4,7), (5,9). Can you see what the requirement looks like on this graph?

Having pursued these ideas, and some others that I'll let you discover (if you haven't already!), I think I may have the answer.

Note that what we are doing is how interesting (non-routine) problems are typically solved: play with the ideas in order to get a feel for how it works, then experiment with possibilities, keeping your mind open, and finding alternative perspectives or models (like my graph and another I haven't mentioned ...); then take specific ideas as far as they will go.

6. Dear Sirs, dear Dr. Peterson!
Thank you very much for your idea. Now I know something about palindromes and for the solution and the proof I found the following: The number must be as long as possible, should start with a figure 9, the figures in a graph have the form of a parabolic curve (?). My next suggestion is: 96433469. I think, this is the solution, but do not know why.
Sincerly, Enoimreh

7. Originally Posted by enoimreh7
Dear Sirs, dear Dr. Peterson!
Thank you very much for your idea. Now I know something about palindromes and for the solution and the proof I found the following: The number must be as long as possible, should start with a figure 9, the figures in a graph have the form of a parabolic curve (?). My next suggestion is: 96433469. I think, this is the solution, but do not know why.
Sincerly, Enoimreh
Yes, that is the answer I got. The differences between successive digits must be increasing by 1 each time, which in fact makes it a parabola. If we made it longer, the end digits would have to be greater than 9.

8. Originally Posted by Dr.Peterson
Yes, that is the answer I got. The differences between successive digits must be increasing by 1 each time, which in fact makes it a parabola. If we made it longer, the end digits would have to be greater than 9.
The most difficult part is the proof and I have no idea, how to grab this.

9. Originally Posted by enoimreh7
The most difficult part is the proof and I have no idea, how to grab this.
Well, I told you the main ideas in what you just quoted. See if you can put them together! Show why what I said is true, and why it implies the solution.

10. If a number has to be big, there have to be as many digits as possible, and, not so important, it should start with the biggest digit 9.
I made now the following row. Each number is smaller than the arithmetic mean of the number before and after it:
1, 2, 4, 7, 11, 16, 22, 29, 36, ... and so on.
I can see, that the differences between the numbers increases by 1:
1, 1+1=2, 2+2=4, 4+3=7, 7+4=11, 11+5=16, ...
But the numbers of this row have to become figures/digits, so there can only be used the numbers with only one digit: 1, 2, 4, 7
So now I've got four digits for a number: 1247. Each digit of it is smaller than the arithmetic mean of the digit before and after it. But there's another one: 7421. The condition is also fulfilled by this one. You can see, these are the first numbers of my row at the beginning.
Because the searched number has to be as big as possible, I put them together: 74211247
But the row could also start with an other number:
2, 3, 5, 8, ...
3, 4, 6, 9, ...
4, 5, 7, 10, ... and so on.
So I could also create the numbers 85322358 or 96433469, but not 107545710, because than the 1 and 0 aren't bigger than the arithmetic mean of their neigbours.
96433469 is bigger than 85322358 and 74211247, so it's the solution.

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