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

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The equation N + S(N) = 2000 requires finding a four-digit number N, where S(N) is the sum of its digits. It is established that A, the first digit, equals 1, leading to the simplification of the equation to 999. Further analysis shows that B + C + 2D must be at least 10, and through calculations, it is determined that N equals 1981, with S(N) summing to 28. The discussion highlights the methodical approach to solving the equation through digit analysis and constraints.
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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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