Proving Congruence Sum: Positive Integer Mod 9

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Homework Help Overview

The problem involves proving that every positive integer is congruent to the sum of its digits modulo 9. Participants are exploring the relationship between a number's representation in terms of its digits and the properties of modular arithmetic, specifically with respect to the modulus of 9.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss representing a number in terms of its digits and powers of 10, questioning how to show that the difference between this representation and the sum of the digits is divisible by 9.
  • There are attempts to manipulate expressions involving powers of 10 and their relationship to modular arithmetic, with some participants expressing uncertainty about how to proceed with the proof.
  • Questions arise regarding the divisibility of specific terms and the implications of congruences in the context of the problem.

Discussion Status

The discussion is ongoing, with participants providing insights and suggestions for proving specific aspects of the problem. Some have identified key points related to divisibility and congruences, while others are still grappling with the implications of their findings.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on understanding the properties of numbers in relation to modular arithmetic without providing direct solutions.

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Homework Statement



Prove that every positive integer is congruent to the sum of its digits (mod 9). (for example 38 is congruent to 11(mod 9))

Homework Equations



If a is congruent to b (mod n), then n divides (a-b)

The Attempt at a Solution



let a= {a1, a2... a9} be digits
where 0 is less than or equal to a which is less than 10

That is where i don't know where to go, i have these digits and I don't know how to show that when added together they = b(mod 9)
maybe i should do something about 9 divides some digits, but i have no clue.

Any suggestions or tips would be great =)
 
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Let anan-1...a0 represent your number, where the ai are the digits. Another way of representing your number is 10^n a_n + \cdots + 10^0 a_0. So what you want to prove is that 9 divides (10^n a_n + \cdots + 10^0 a_0) - (a_n + \cdots + a_0)
 
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I know this may sound stupid, but how am I supposed to show that. Right now I have:

9 divides a_n (10^(n)-1 )+ a_(n-1) (10^(n-1)-1)+⋯a_1 (10-1).
How do i get that 9 divides that?
 
Every term in a_n (10^n - 1) + a_{n-1} (10^{n-1} - 1) + \cdots + a_1 (10-1) is of the form a_i (10^i -1). If you can prove that 9 divides any such term, then you are done.

Can you figure out how to prove that 9 divides a_i (10^i -1) for all positive integers i?
 
well a_1(10-1)= a_1(9), and 9 divides a_1(9). So does that mean that it divides all i? I thought that only would apply for n=1
 
10 is congruent to 1 mod 9. Hence 10^n is congruent to 1 mod 9.
 
Where does the a_n go? wouldn't (a_1)10 be congruent to 1 mod 9, meaning that a_n would be congruent to 1 mod 9?
 
ok, nevermind, i just figured out that that doesn't matter because 9 divides both a_n10^n and also 9 divides just 10^n
 
well (10^n -1) that is
 
  • #10
a_n doesn't have to be congruent to anything. The point is that (10^n-1) is divisible by nine, as HitMan-2 pointed out. So a_n*10^n is congruent to a_n mod 9.
 

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