What is the Value of N in the Equation N + S(N) = 2000?

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Discussion Overview

The discussion revolves around finding the value of N in the equation N + S(N) = 2000, where N is a 4-digit number and S(N) represents the sum of its digits. Participants explore various approaches to solve this equation, including algebraic manipulations and case analysis.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • Some participants note that N must be a 4-digit number and express that S(N) is the sum of its digits.
  • One participant proposes that since A = 1, the equation simplifies to N + S(N) = 999, leading to further analysis of B, C, and D.
  • Another participant suggests that if B + C + 2D < 10, it leads to a contradiction, indicating that B + C + 2D must be at least 10.
  • One participant calculates specific values for C and D, concluding with N = 1981, while others continue to explore the implications of their findings.
  • There is a repetition of earlier points, indicating that participants are refining their understanding and calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the value of N, as multiple approaches and calculations are presented, leading to different interpretations and results.

Contextual Notes

Some assumptions about the values of B, C, and D are not fully explored, and the implications of carrying over in digit sums are discussed but not resolved. The discussion remains open-ended with various paths of reasoning.

albert391212
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N+S(N) =2000
N is a 4-digit number,and S(N) is the sum of each digit of N
given N+S(N)=2000
please find N
 
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Albert391212 said:
N+S(N) =2000
N is a 4-digit number,and S(N) is the sum of each digit of N
given N+S(N)=2000
please find N
Let [math]N = ABCD \equiv A \times 10^3 + b \times 10^2 + C \times 10 + D[/math] So [math]N + S(N) = 1000A + 100B + 10C + (A + B + C + 2D) = 2000[/math].

Note that A = 1. So
[math]N + S(N) = 100B + 10C + (B + C + 2D) = 999[/math]

Now start working through some cases. For example, B + C + 2D < 10 is impossible because it means B = C = 9, which is a contradiction. So [math]B + C + 2D \geq 10[/math]. Thus when adding we have to carry a 1 into the 10's place, which means that C is at most 8. etc. It will take a while.

-Dan
 
Let N=abcd , and S(N)=a+b+c+d<28 ,we have a=1, b=9
N+S(N)=1000+900+10c+d+1+9+c+d=1910+11c+2d=2000
11c+2d=90
we get c=8 , d=1
so N=1981 #
 
topsquark said:
Let [math]N = ABCD \equiv A \times 10^3 + b \times 10^2 + C \times 10 + D[/math] So [math]N + S(N) = 1000A + 100B + 10C + (A + B + C + 2D) = 2000[/math].

Note that A = 1. So
[math]N + S(N) = 100B + 10C + (B + C + 2D) = 999[/math]

Now start working through some cases. For example, B + C + 2D < 10 is impossible because it means B = C = 9, which is a contradiction. So [math]B + C + 2D \geq 10[/math]. Thus when adding we have to carry a 1 into the 10's place, which means that C is at most 8. etc. It will take a while.

-Dan
Thanks for your answer
topsquark

From Albert
 

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