Question 826231: Find three consecutive integers such that the sum of twice the second and three times the third is five less than six times the first. Can you please help me?
Answer by Fermat(136)   (Show Source):
You can put this solution on YOUR website! Let the three integers be n1,n2 and n3 such that
n1 = n
n2 = n+1
n3 = n+2
We are told that the sum of twice the second and three times the third is five less than six times the first.
putting that into numbers,
2*n2 + 3*n3 = 6*n1 - 5
i.e. 2(n+1) + 3(n+2) = 6n - 5
2n + 2 + 3n + 6 = 6n - 5
rearranging the eqn gives us,
13 = n
n = 13
Answer: the three numbers are: 13, 14, 15
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