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 |
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 |
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].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
Thanks for your answertopsquark 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