Find N Given N+S(N)=2000

[albert391212]

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

 

[topsquark]

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 \(\displaystyle N + S(N) = 1000A + 100B + 10C + (A + B + C + 2D) = 2000\).

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

 

[albert391212]

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 #

 

[albert391212]

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

Thanks for your answer
topsquark

From Albert