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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 |
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 \(\displaystyle N = ABCD \equiv A \times 10^3 + b \times 10^2 + C \times 10 + D\) So \(\displaystyle N + S(N) = 1000A + 100B + 10C + (A + B + C + 2D) = 2000\).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 \(\displaystyle N = ABCD \equiv A \times 10^3 + b \times 10^2 + C \times 10 + D\) So \(\displaystyle N + S(N) = 1000A + 100B + 10C + (A + B + C + 2D) = 2000\).
Note that A = 1. So
\(\displaystyle N + S(N) = 100B + 10C + (B + C + 2D) = 999\)
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 \(\displaystyle B + C + 2D \geq 10\). 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
The equation is trying to solve for the value of N, where N is a number and S(N) is the sum of the digits of N.
To find the solution for N, you can use algebraic manipulation to isolate N on one side of the equation. Then, you can use trial and error or a calculator to find the specific value of N that satisfies the equation.
No, there can be multiple solutions for N in this equation. However, there is only one positive integer solution for N.
One possible solution for N is 1999. This is because when you plug in 1999 for N, the sum of its digits (1+9+9+9) equals 28, and when you add 1999 and 28, you get 2000.
The equation can be used in various fields of science, such as mathematics, physics, and computer science. It can help with solving problems related to number theory, optimization, and algorithms. It can also be used to understand patterns and relationships in data sets.