Number Theory Puzzler: Proving N-S is a Multiple of 9

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Discussion Overview

The discussion revolves around a number theory problem involving a four-digit number N and the sum of its digits S. Participants are tasked with proving that N minus S is a multiple of 9, and later, a revised version of the problem asks to prove that the sum of the digits of (N - S) is divisible by 9. The scope includes mathematical reasoning and proofs related to divisibility rules.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant defines N as a four-digit number represented by its digits "abcd" and derives that N - S equals (a * 999 + b * 99 + c * 9), leading to the conclusion that (N - S)/9 is an integer, thus showing N - S is divisible by 9.
  • A later post revises the original problem to focus on proving that the sum of the digits of (N - S) is divisible by 9.
  • Another participant asserts that the revised problem follows from the previous post, stating that a number is divisible by 9 if and only if the sum of its digits is divisible by 9, providing a general proof involving modular arithmetic.

Areas of Agreement / Disagreement

Participants appear to agree on the mathematical principles involved, but the discussion introduces a revised problem that shifts the focus, leaving the original proof and its implications open to further exploration.

Contextual Notes

The discussion does not resolve potential assumptions about the properties of digit sums or the implications of modular arithmetic in the context of the revised problem.

sizzlaw
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Given: N is a four digit number. S is the sum of N's digits.

Prove: N minus S is a multiple of 9.
 
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Call the four digits of N "abcd"
Then N is (a * 1000 + b * 100 + c * 10 + d)
And N - S = N - a - b - c - d = (a * 999 + b * 99 + c * 9)
And (N - S)/9 is (a * 111 + b * 11 + c), which is an integer.
Hence N-S is divisible by 9.
 
Revised puzzler:

Given: N is a four digit number. S is the sum of N's digits.

Prove: The sum of the digits of (N - S) is divisible by 9.
 
This follows easily from Goongyae's post, since a number is divisible by 9 if and only if the sum of it's digits is divisible by 9.

This is real easy to prove: take x1...xn be an n-digit number, then

[tex]x_110^{n-1}+...+x_n10^0=x_1+...+x_n~(mod~9)[/tex].
 

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