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14 April, 06:06

Find four consecutive integers such that the product of the two largest is 46 more than the product of the two smallest integers

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  1. 14 April, 09:10
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    Let x, x+1, x+2 & x+3 be the four integers.

    (x+2) (x+3) = x (x+1) + 46

    x^2+5x+6=x^2+x+46

    x^2-x^2+5x-x=46-6

    4x=40

    x=40/4

    x=10 for the smallest integer.

    10=1=11 for the next integer.

    10=2=12 for the third integer.

    10=3=13 for the largest integer.

    12*13=10*11+46

    156=110+46

    156=156
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