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3 April, 16:29

Find four consecutive integers such that three times the sum of the first two integers exceeds the sum of the last two by 70.

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  1. 3 April, 17:24
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    we know the difference of consecutive integers is 1 so we can achieve every next integer by adding 1 in the previous integer.

    let us start with assuming the four consecutive integers as:

    x, x+1, x+2, x+3

    3 times (sum of first two) = sum of last two + 70

    3 (x + x+1) = x+2+x+3 + 70

    3 (2x + 1) = 2x + 75

    6x + 3 = 2x + 75

    6x-2x = 75 - 3

    4x = 72

    x = 72/4 = 18

    so first integer is x = 18 next is 19 next 20 and next 21

    Answer : 18,19,20,21
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