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8 November, 18:01

Find three consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third (only an algebraic solution will be accepted)

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  1. 8 November, 18:10
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    Let us call them a, b and c

    we know that

    a+1=b

    a+2=c

    a+2b=12+c

    we can deduce the following:

    a+2 (a+1) = 12 + (a+2)

    3a+2=a+14

    2a=12

    a=6

    so b would be 7 and c would be 8

    Makes sense?
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