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

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The equation N + S(N) = 2000, where N is a 4-digit number and S(N) is the sum of its digits, leads to the conclusion that N must equal 1981. By analyzing the equation, it is established that A = 1, and through systematic case analysis, the digits B, C, and D are determined to be 9, 8, and 1 respectively. The calculations confirm that 1981 satisfies the equation, as 1981 + S(1981) equals 2000.

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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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