# Answer to Question #6123 in Combinatorics | Number Theory for bandar

Question #6123

the sum of all decimal number 0.xyz (where x, y, and z are distinct digits) are...

Expert's answer

1 count sum of numbers (sum of arithmetic progression)

2 count sum of numbers (S1) where x=y=z.

3 count sum (S2) where exactly 2 indicate numbers.

For example: 00x, 0x0, x00; x isn't 0.

S20 = (0.001 + 0.002 + ... + 0.009) + (0.01 + 0.02 + ... + 0.09) + (0.1 + 0.2 + ... 0.9) = 0.045 + 0.45 + 4.5 = 4.995

S21 = (0.110 + 0.112 + ... + 0.119) + (0.101 + 0.121 + ... + 0.191) + (0.011 + 0.211 + ... + 0.911) = (0.11*9 + 0.044) + (0.101*9 + 0.44) + (0.011*9 + 4.4) = (0.99 + 0.044) + (0.101 + 0.44) + (0.099 + 4.4) = 1.034 + 0.541 + 4.499 = 6.074...

S29 = (0.990 + 0.991 + ... + 0.998) + (0.909 + 0.919 + ... + 0.989) + (0.099 + 0.199 + ... + 0.899) = (0.99*9 + 0.036) + (0.909*9 + 0.36) + (0.099*9 + 3.6) = (8.91 + 0.036) + (8.181 + 0.36) + (0.891 + 3.6) = 8.946 + 8.541 + 4.491 = 21.978

Answer:

**S=0.001 + 0.002 + ... + 0.999 = (0.001 + 0.999) / 2* 999 = 0.5 * 999 = 499.5**2 count sum of numbers (S1) where x=y=z.

**S1 = 0.111 + 0.222 + ... + 0.999 = (0.111 + 0.999) / 2 * 9 = 0.555 * 9 = 4.995**3 count sum (S2) where exactly 2 indicate numbers.

For example: 00x, 0x0, x00; x isn't 0.

S20 = (0.001 + 0.002 + ... + 0.009) + (0.01 + 0.02 + ... + 0.09) + (0.1 + 0.2 + ... 0.9) = 0.045 + 0.45 + 4.5 = 4.995

S21 = (0.110 + 0.112 + ... + 0.119) + (0.101 + 0.121 + ... + 0.191) + (0.011 + 0.211 + ... + 0.911) = (0.11*9 + 0.044) + (0.101*9 + 0.44) + (0.011*9 + 4.4) = (0.99 + 0.044) + (0.101 + 0.44) + (0.099 + 4.4) = 1.034 + 0.541 + 4.499 = 6.074...

S29 = (0.990 + 0.991 + ... + 0.998) + (0.909 + 0.919 + ... + 0.989) + (0.099 + 0.199 + ... + 0.899) = (0.99*9 + 0.036) + (0.909*9 + 0.36) + (0.099*9 + 3.6) = (8.91 + 0.036) + (8.181 + 0.36) + (0.891 + 3.6) = 8.946 + 8.541 + 4.491 = 21.978

**S2 = S20 + S21 + ... + S29**Answer:

**S-S1-S2 = 359.64**Need a fast expert's response?

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## Comments

Assignment Expert31.01.12, 15:05You are welcome

bandar31.01.12, 05:15thank you very much

I wish you answer my next question

it will be helpful to me to prepare my self on Mathematics Competition

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