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The sum of three consecutive numbers is twelve less than six times the lowest number. What are the three numbers?
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if our 1st number is say "a", then the CONSECUTIVE number will be the next number, that is a+1 and the CONSECUTIVE from that will be (a+1)+1 so a a + 1 (a+1) + 1 ^ ^ ^ 1st number 2nd number 3rd number you add them up and they're equal to "12 less than 6 times the lowest number" well, obviously "a" is the lowest number, because the others have an extra 1 or 1+1 6 times "a" = 6a 12 less than that => 6a-12 so \(\bf a+[a+1]+[a+1]+1=6a-12\) solve for "a"
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