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Four consecutive integers have the property that the sum of 16 less the first number, 15 less than three times the second number, 4 more than the third, and half of the fourth is 680. Find the integers.
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Twice the sum of a number and three is the same as triple the difference of a number and two.
Call the numbers n, n+1, n+2 and n+3. If we translate the sentences into algebra, we get: 16-n + 3(n+1) -15 + n+2 +4 + (n+3)/2 = 16-n+3n+3-15+n+6+0.5n+1.5=680 If you solve this equation, you have n, so you also have the other three numbers...
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