Question 571549
Find four consecutive integers such that twice the first subtracted from the sum of the other three integers equals 16.


Let the first integer be F


Then other three are: F + 1, F + 2, and F + 3


Since twice the first subtracted from the sum of the other three integers equals 16, we therefore have: (F + 1 + F + 2 + F + 3) - 2F = 16


3F + 6 - 2F = 16


F + 6 = 16


F, or first integer = 16 - 6, or 10


The four integers are: {{{highlight_green(10_11_12_and_13)}}}


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2 times first = 2(10), or 20


Sum of other three = 11 + 12 + 13, or 36


36 - 20 = 16 (TRUE) 


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