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

enoimreh7

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Nov 14, 2017
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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? :shock:Remark: The correctness of the result is to be proved.

Sincerly, Enoimreh
 

lev888

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

Dr.Peterson

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Nov 12, 2017
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2,754
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? :shock: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.
 

enoimreh7

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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
 

Dr.Peterson

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Nov 12, 2017
Messages
2,754
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.
 

enoimreh7

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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
 

Dr.Peterson

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Nov 12, 2017
Messages
2,754
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.
 

enoimreh7

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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. :(
 

Dr.Peterson

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Nov 12, 2017
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2,754
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.
 

enoimreh7

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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.
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
2,754
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.
Yes, that is essentially my reasoning as well. I thought of it largely in terms of a graph, making this curve to be as long as possible (by curving as little as possible), and then pushing it up as high as possible.
 

enoimreh7

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Hello! I tried to find an equation for the graph, but did not succeed. It is 3 + ????
Sincerly, Enoimreh
 

Dr.Peterson

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Hello! I tried to find an equation for the graph, but did not succeed. It is 3 + ????
Sincerly, Enoimreh
Do you mean you want to find the digit as a function of the digit position, so that f(1) = 9, f(2) = 6, and so on? I had considered doing that, but decided it wasn't necessary. But it is interesting, so I'm glad you saw it that way, too!

As I have indicated, it is in fact a parabola (do you know how I knew that?). So you can take any three points and solve for the coefficients in f(n) = an^2 + bn + c. Or, you can recognize where the vertex is, and use the form f(n) = a(n - h)[SUP]2[/SUP] + k with two known points.
 

enoimreh7

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Thank you very much. So far, I do not know, wherefrom you know that it is a parabola except for drawing it. May be? Excuse me, but what stands h and what stands k for? Enoimreh
 

Dr.Peterson

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Thank you very much. So far, I do not know, wherefrom you know that it is a parabola except for drawing it. May be? Excuse me, but what stands h and what stands k for? Enoimreh
When the "second differences" of a sequence are constant, it is quadratic; there are standard methods for finding the formula for such a sequence. This method is called by various names, such as "finite differences" or "common differences" (though those terms are also used for other concepts, so they are dangerous to search for).

Once you know that it is quadratic, as these and other sites show, you can suppose that the formula is f(x) = ax^2 + bx + c, put three pairs into this formula to obtain three equations in the unknown a, b, and c, and then solve for a, b, and c.

As for f(x) = a(x - h)[SUP]2[/SUP] + k, the pair (h,k) is the vertex of the parabola. You don't need to use that form.

Give it a try!
 
Last edited:

enoimreh7

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Dear Mr. Dr. Peterson,


thank you for the Websites.

I now tried to do it:

If you draw a graph with the figures of this number (96433469) a parable arises. (Not only this number, it is the whole figure row (3, 4, 6, 9, 13, 18, ...) of the number which forms a parable. )
Because there is a formula to every parable, there must be a formula for the figure row. Or, actually, vice versa: If I have found a formula for the row, I know with certainity whether this is a parable or something else.
With the finite differences method I try to get the equation:
I make a table with the figures:

3 // 4 // 6 // 9
// 1 // 2 // 3
/////1 // 1

For the second line I subtract the first number of the second, the second of the third etc.
I repeat this for the third line.
Now the values, so the second differences, are always the same, so I am ready. Because the second differences are always the same, I know that the Polynom is a square for this sequence of values. Therefore, it must have this form:

an^2 + bn + C for a, B and C

F(n) = an^2 + bn + C
F(1) = a+b+c
F(2) = 4a+2b+c
F(3) = 9a+3b+c

a+b+c // 4a+2b+c // 9a+3b+c
//////3a+b /////// 5a+b

//////////////2a

Therefore, is a = 0. 5 because 2a=1.
Therefore, B = is-0,5 because 3a+b=1 and a 0. 5 is
Therefore, C = is 3 because a+b+c=3 and a 0. 5 is and B-0,5 is.
I use these values now in the origin equation:
F(n) = an^2 + bn + C = 0,5n^2-0,5+3 = 1 / 2n^2+2,5
n=1 F(1) =0,5+2,5=3
n=2 F(2) =0,5x4+2,5=4,5
n=3 F(3) =0,5x9+2,5=7

But this is not correct, because if n=3 then F(3) has to be 6 and not 7, the same with n=2.
I'm stuck, because I don't now where my mistake is. Could you please help me once another time?

Thank you very much!

Sincerely enoimreh
 
Last edited:

Dr.Peterson

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2,754
If you draw a graph with the figures of this number (96433469) a parable arises. (Not only this number, it is the whole figure row (3, 4, 6, 9, 13, 18, ...) of the number which forms a parable. )
Because there is a formula to every parable, there must be a formula for the figure row. Or, actually, vice versa: If I have found a formula for the row, I know with certainity whether this is a parable or something else.
With the finite differences method I try to get the equation:
I make a table with the figures:

3 // 4 // 6 // 9
// 1 // 2 // 3
/////1 // 1

For the second line I subtract the first number of the second, the second of the third etc.
I repeat this for the third line.
Now the values, so the second differences, are always the same, so I am ready. Because the second differences are always the same, I know that the Polynom is a square for this sequence of values. Therefore, it must have this form:

an^2 + bn + C for a, B and C

F(n) = an^2 + bn + C
F(1) = a+b+c
F(2) = 4a+2b+c
F(3) = 9a+3b+c

a+b+c // 4a+2b+c // 9a+3b+c
//////3a+b /////// 5a+b

//////////////2a

Therefore, is a = 0. 5 because 2a=1.
Therefore, B = is-0,5 because 3a+b=1 and a 0. 5 is
Therefore, C = is 3 because a+b+c=3 and a 0. 5 is and B-0,5 is.
I use these values now in the origin equation:
F(n) = an^2 + bn + C = 0,5n^2-0,5+3 = 1 / 2n^2+2,5
n=1 F(1) =0,5+2,5=3
n=2 F(2) =0,5x4+2,5=4,5
n=3 F(3) =0,5x9+2,5=7

But this is not correct, because if n=3 then F(3) has to be 6 and not 7, the same with n=2.
First, when I worked out the equation, I did it for the digits of the number you found, 96433469. That is, f(1) = 9, f(2) = 6, and so on. You will, of course, get a different result.

Second, you made a very small error: you omitted n in the second term. You should have, not 0,5n^2-0,5+3, but 0,5n^2-0,5n+3. There should not be any like terms to combine!

Using this,
f(1) = 0,5(1)^2 - 0,5(1) + 3 = 3
f(2) = 0,5(2)^2 - 0,5(2) + 3 = 4
f(3) = 0,5(3)^2 - 0,5(3) + 3 = 6
and so on.

So you were very close to the answer.

My formula for the nth digit is (1/2)n^2 - (3/2)n + 13.
 

enoimreh7

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Dear Mr. Dr. Peterson!

Thank you!
Now, if I don't forget the n, then it's correct.

Well, I tried to do it like you did, so with all the digits of 96433469 and not only 3469.
But then, my function is different. Here is my work:

9///6///4///3///3///4///6///9
/-3//-2//-1///0///1///2///3
////1///1///1///1///1///1

a+b+c///4a+2b+c///9a+3b+c///16a+4b+c///25a+5b+c///36a+6b+c///49a+7b+c///64a+8b+c
/////3a+b////////5a+b///////7a+b/////////9a+b////////11a+b///////13a+b////////15a+b
/////////////2a///////////2a///////////2a///////////2a/////////////2a/////////////2a

F(n) = an^2 + bn + c

3.row:
a=0,5 because of 2a=1
a=0,5 because of 2a=1

2.row:
b=-4,5 because of 3a+b=-3 and a=0,5
b=-4,5 because of 5a+b=-2 and a=0,5
b=-4,5 because of 7a+b=-1 and a=0,5
b=-4,5 because of 9a+b=0 and a=0,5
b=-4,5 because of 11a+b=1 and a=0,5
b=-4,5 because of 13a+b=2 and a=0,5
b=-4,5 because of 15a+b=3 and a=0,5

1.row:
c=13 because of a+b+c=9 and a=0,5 and b=-4,5
c=13 because of 4a+2b+c=6 and a=0,5 and b=-4,5
c=13 because of 9a+3b+c=4 and a=0,5 and b=-4,5
c=13 because of 16a+4b+c=3 and a=0,5 and b=-4,5
c=13 because of 25a+5b+c=3 and a=0,5 and b=-4,5
c=13 because of 36a+6b+c=4 and a=0,5 and b=-4,5
c=13 because of 49a+7b+c=6 and a=0,5 and b=-4,5
c=13 because of 64a+8b+c=9 and a=0,5 and b=-4,5

F(n) = an^2 + bn + c
F(n) = 0,5n^2 - 4,5n + 13

F(1)=0,5-4,5+13=9
F(2)=0,5x4-9+13=6
F(3)=0,5x9-13,5+13=4
F(4)=0,5x16-18+13=3
F(5)=0,5x25-22,5+13=3
F(6)=0,5x36-27+13=4
F(7)=0,5x49-31,5+13=6
F(8)=0,5x64-36+13=9

My function is F(n)= 0,5n^2 - 4,5n +13

Could you please explain me why I've got a different function for n?
Where's my mistake?

I've got another question. I still didn't really understand what the function prooves more than the explaination with the digits?

Thank you very much,

Sincerly
Enoimreh
 

Dr.Peterson

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Nov 12, 2017
Messages
2,754
Dear Mr. Dr. Peterson!

Thank you!
Now, if I don't forget the n, then it's correct.

Well, I tried to do it like you did, so with all the digits of 96433469 and not only 3469.
But then, my function is different. Here is my work:

9///6///4///3///3///4///6///9
/-3//-2//-1///0///1///2///3
////1///1///1///1///1///1

a+b+c///4a+2b+c///9a+3b+c///16a+4b+c///25a+5b+c///36a+6b+c///49a+7b+c///64a+8b+c
/////3a+b////////5a+b///////7a+b/////////9a+b////////11a+b///////13a+b////////15a+b
/////////////2a///////////2a///////////2a///////////2a/////////////2a/////////////2a

F(n) = an^2 + bn + c

3.row:
a=0,5 because of 2a=1
a=0,5 because of 2a=1

2.row:
b=-4,5 because of 3a+b=-3 and a=0,5
b=-4,5 because of 5a+b=-2 and a=0,5
b=-4,5 because of 7a+b=-1 and a=0,5
b=-4,5 because of 9a+b=0 and a=0,5
b=-4,5 because of 11a+b=1 and a=0,5
b=-4,5 because of 13a+b=2 and a=0,5
b=-4,5 because of 15a+b=3 and a=0,5

1.row:
c=13 because of a+b+c=9 and a=0,5 and b=-4,5
c=13 because of 4a+2b+c=6 and a=0,5 and b=-4,5
c=13 because of 9a+3b+c=4 and a=0,5 and b=-4,5
c=13 because of 16a+4b+c=3 and a=0,5 and b=-4,5
c=13 because of 25a+5b+c=3 and a=0,5 and b=-4,5
c=13 because of 36a+6b+c=4 and a=0,5 and b=-4,5
c=13 because of 49a+7b+c=6 and a=0,5 and b=-4,5
c=13 because of 64a+8b+c=9 and a=0,5 and b=-4,5

F(n) = an^2 + bn + c
F(n) = 0,5n^2 - 4,5n + 13

F(1)=0,5-4,5+13=9
F(2)=0,5x4-9+13=6
F(3)=0,5x9-13,5+13=4
F(4)=0,5x16-18+13=3
F(5)=0,5x25-22,5+13=3
F(6)=0,5x36-27+13=4
F(7)=0,5x49-31,5+13=6
F(8)=0,5x64-36+13=9

My function is F(n)= 0,5n^2 - 4,5n +13

Could you please explain me why I've got a different function for n?
Where's my mistake?

I've got another question. I still didn't really understand what the function prooves more than the explaination with the digits?
Your result is different from mine only because I copied mine wrong. (I wrote it too small on my scrap paper.) I meant (1/2)n^2 - (9/2)n + 13.

The function proves nothing at all, and was not needed. You chose to do it; I think I just mentioned it in passing, because I thought it was interesting.

But since we are talking about it, I'll show you my approach, which took less work. I just took three points, (1,9), (2,6), (3,4), and put them into the formula y = ax^2 + bx + c:
a(1)^2 + b(1) + c = 9 --> a + b + c = 9
a(2)^2 + b(2) + c = 6 --> 4a + 2b + c = 6
a(3)^2 + b(3) + c = 4 --> 9a + 3b + c = 4

Then I subtracted each equation from the one below it:
3a + b = -3
5a + b = -2

Then I subtracted the first from the second:
2a = 1 --> a = 1/2

Then I put that into 3a + b = -3 and solved:
3/2 + b = -3 --> b = -9/2

Then I put those into a + b + c = 9 and solved:
1/2 - 9/2 + c = 9 --> c = 13

So f(x) = (1/2)n^2 - (9/2)n + 13.
 
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