How many 4-digit numbers with all odd digits are divisible by 11

A number is divisible by 11 if the difference between the sum of the digits in odd positions and the sum of the digits in even positions is divisible by 11. This was the most important fact . Secondly every number added to another was odd.

[Ron_DeForest]

Ok seriously. Is there anything you can’t figure out mathematically? This was very clever.

[저녘놀]

what a beautiful proof!

The idea of (A+C) - (B+D) = 0 (mod 11)
I couldn't think about this.

[mhah]

By far the best explanation of how to find whether a digit is divisible by eleven or not! The rest question is then almost solved and is very easy to solve.

[nanamacapagal]

BEFORE WATCHING:

The divisibility test for 11 is to add every other digit and check if the sum is the same as if you started adding every other other digit (or is a multiple of 11 away)
So in the case of 38179, check if 3 + 1 + 9 = 8 + 7 or is a multiple of 11 away. In this case, 3 + 1 + 9 = 13 but 8 + 7 = 15. So 38179 is not a multiple of 11.

We can use this to our advantage here in the 4 digits case. If the number is ABCD, we want A + C = B + D, or a multiple of 11 away.
ODD + ODD = EVEN, so we only care about even sums. Smallest we can do is 1 + 1 = 2 and the largest we can do is 9 + 9 = 18. None of the other even sums are a multiple of 11 away from each other.

So it's just counting terms. Ordered pairs, rather, because the order of the digits matters.
For 2, we have 1 + 1.
For 4, we have 1 + 3 and 3 + 1.
For 6, we have 1 + 5, 3 + 3, and 5 + 1.
We can keep going until 10 = 1 + 9, 3 + 7, 5 + 5, 7 + 3, and 9 + 1.
Then it goes back down again. 12 = 3 + 9, 5 + 7, 7 + 5, and 9 + 3.
So the total count per sum is 1, 2, 3, 4, 5, 4, 3, 2, 1.
But since we have 2 pairs to worry about, we have to multiply.
For example:
Sum is 6. 3 ways for A + C, 3 ways for B + D, 9 ways combined.
So the total amount of numbers with only odd digits, that are divisible by 11, is 1² + 2² + 3² + 4² + 5² + 4² + 3² + 2² + 1² = 85

Turns out all 85 of these are present in the thumbnail, so I could have just counted those instead

[ArthurAmon]

not mathematics related, but today's hat is one of the best so far!

[Grecks]

Hi PN, my solution went exactly along the same lines as yours, and I did it in my mind while hiking, without access to pen or paper.
A small point where I deviated a bit was when I computed the total sum of combinations: I used the explicit formula for the "sum of first n squares": s=1/6*n*(n+1)*(2n+1).
So I went (using the inherent symmetry): S = 2 * "Sum of k^2 for k=1..4" + 5^2 = 1/3 * (4*5*9) + 5*5 = 3*4*5 + 5*5 = (3*4 + 5) * 5 = 85.

[fabiozanucoli]

I almost could see the Pascal's triangle at the end. Beautiful!

[ForestHills]

Another counting method:
Let M=11*N
N=100a+10*b+c
M=1000a+100(a+b)+10(b+c)+c
c must be odd
b must be even
(I) If b+c<10, a must be odd and a+b<10
(II) If b+c>=10, a must be even and a+b>=10

For c=1, 3, 5, 7, 9 we get a count of:
(I) 15, 14, 12, 9, 5
(II) 0, 4, 7, 9, 10
For a total of 85

[jasonryan]

How can it be so easy yet be so inconvenient at the same time. As if said on one of your earlier videos, methods matter far more than just being able to compute huge numbers mentally. This was a beautiful question, and now I see why all mathematics enthusiasts say the same thing all the time 😅

[Nothingx]

Sir if you have tried to find the least number with this condition and then the maximum number with that condition, now you will notice each number is in a sequence with common difference of 22 then use the formula easy

[Modo]

Is there a reason this pattern (1², 2², 3³... Etc) shows up or is it just a coincidence? Feels really specific to be a coincidence though

[krwada]

Again, a very good presentation and solution.

However, I cannot help but believe that there is a more general approach to this using number theory.

[RyanLewis-Johnson-wqxs]

1+4+9+16+25+16+9+4+1=85 I did it in my head.

[jusmt]

This was a fun way of looking at this problem. I was inspired by the question, and wanted to solve the same question but with 9 instead of 11. The process was very fun, and the answer wasn't what I had expected. I would love for you to try it as well.

[Nothingx]

Anyways the solution was really good I really appreciate it 😊

[ccdsah]

I wish you'd have rigorously demonstrated the divisibility by 11 criteria

[jasimmathsandphysics]

Wow amazing! I will remember this trick now

[lukaskamin]

Somehow in school we never payed attention to thus type if task( that's a numbers' theory asfaik). I have tk admit it looks overwhelming to me when you have to work with such a huge quantity of numbers and tracing certain patterns. I mean I understand that theoretically, but had I s test with such task, I'd definitely skipped it unless I solved everything, double checked it and still had some time remained

[Nothingx]

Arithmetic progression nth formula left the chat 😊

[SamyakAditya]

Sir being some amc exam questions if possible, those questions are quite interesting