Order of the six digits palindromic number 969969 from 999999? Is it 31?

[Salah]

Please, I need the help in finding the order of Palindromic number 969969, starting descendingly from the last 6 digits palindromic number 999999, is it 31?

969969=3*7*11*13*17*19

I need this in my research in Philosophy of Numbers.

Thank you.

 

[Cubist]

Please, I need the help in finding the order of Palindromic number 969969, starting descendingly from the last 6 digits palindromic number 999999, is it 31?

I think you want to know how many palindromic numbers there are between 969969 and 999999 (inclusive). Please can you show us the work that you've done to find your answer 31, and then somebody here might check it? We prefer to help people to answer their own questions rather than simply providing answers.

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[Salah]

How I guessed (31)?
These 6 digits palindromic numbers have no axis, since 6 digits=even number.
3 digits=1 digit+axis+1 digit
5 digits=2 digits+axis+2 digits
7 digits=3 digits+axis+3 digits,
And so on.
So, from the last (higher) palindromic number i.e: 999999 descendingly, the order become like this:
999999
998899
997799, and so on, until: 990099.=10 numbers.

Then;
989989, 988889, until: 980089.=10 numbers.

Then, so for: 979979, until: 970079.=10 numbers.
10+10+10=30 numbers.

Then, our number, I mean: 969969=31.

Am I correct?

If I am correct, then I need other proves from other ways.

Thanks in advance.

 

[Cubist]

Your proof looks both correct and efficient to me.

I can only think of one alternative way to prove your result:- write a computer program to check every single number between 969969 and 999999 to see how many are palindromic. I think that's a question better asked on a programing forum.

 

[Cubist]

Actually I just thought of another way. We only need to consider the first three digits of the 6 digit palindrome - since the last three digits are simply the first 3 reflected (see the red digits below)

...
987789
986689
985589
...

EDIT: If the black left hand digits increase(or decrease) by 1 each time then no palindromes can be missed between consecutive lines.

Therefore how many 3 digit numbers are there between 969 and 999 (inclusive), the answer is 999 - 969 + 1 = 31

 

[pka]

Any integer in the interval set \(\left[969,999\right]\) is "one-half" of a number that you want.
Can you explain why that is the case?
And yes there are \(31\) of them.

 

[Salah]

Thank you for your helpful answer. As I said, I am a Pharmacist. I had got the final mark in the last year exam of Maths of the middle school [60/60]. I had got also [59/60] in the last year exam of Maths of the high school. This was 23 years ago. Now I need to refresh my memory and my information. Actually I am a Philosopher. Inspite of all these science and technology revolutions, I think there is a great part of Truth which is absent, but we can discover it easily.
Thanks, again.