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The sum 9 + 99 + 999 + 9999 + 99 999 + ....

where the last number to be added consists of nine digits of 9 becomes:

(10 – 1) + (100 – 1) + (1000 – 1) + (10 000 – 1) + ...

where the last multiple of 10 has nine zeros.

There are nine pairs of brackets altogether.

Add the multiples of 10 first and then subtract 9:

The digit 1 appears 9 times in the final answer.

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humany_human2020

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2y ago
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15y ago

Using exactly eight 9's : 999 + ( 99/9 ) - ( 9/ .9 ) [ 999 + 11 - 10 = 1000 ] or (999 + 9/9) x ( 99-9) [ (999 +1 ) x 1 ] --(obviously the five 9's are enough : 999 + 9/9 )

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Q: Choose a digit from 1-9 Make numbers that add to 1000 using only the digit you chose You can only use your digit eight times?
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