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The fraction 512/999 cannot be reduced any more.

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Q: What is the lowest term of 512 and 999?
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How can 125.512512512etc be written as a fraction?

Three digits repeating the the decimal, so put over three 9s (999): 125.512512... = 125512/999 (as a mixed number) = (125 x 999 + 512)/999 = 125387/999 (as an improper fraction)


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The largest number with three digits in base two is 111, which equals seven. Other answer: largest 3 digit number = 999 base 2=binary 00000000000000000000000000=0 00000000000000000000000001=1 00000000000000000000000010=2 00000000000000000000000100=4 999/2=~500 log 500/log 2=8.9->9 ->9 zeroes after 1 2^9=512 1000000000=512 1100000000=512+256=768 1110000000=512+256+128=896 1111000000=512+256+128+64=960 1111100000=512+256+128+64+32=992 1111100111=512+256+128+64+32+4+2+1=999


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512 and 6910 are integers so their sum is an integer, not an improper fraction.


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