- #1

- 11

- 0

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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- MHB
- Thread starter albert391212
- Start date

- #1

- 11

- 0

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 |

- #2

- 1,944

- 973

Let \(\displaystyle N = ABCD \equiv A \times 10^3 + b \times 10^2 + C \times 10 + D\) So \(\displaystyle N + S(N) = 1000A + 100B + 10C + (A + B + C + 2D) = 2000\).

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

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

- #3

- 11

- 0

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 #

- #4

- 11

- 0

Thanks for your answerLet \(\displaystyle N = ABCD \equiv A \times 10^3 + b \times 10^2 + C \times 10 + D\) 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

topsquark

From Albert

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