# Four integers

Fnd four consecutive integers so that the product of the first two is 70 times smaller than the product of the next two.

Result

a =  16
b =  17
c =  18
d =  19

#### Solution:

$a(a+1)=(a+2) \cdot \ (a+3) - 70 \ \\ a^2+a=a^2+5a+6-70 \ \\ -4a=-64 \ \\ a=64/4=16$
$b=a+1=16+1=17$
$c=b+1=17+1=18$
$d=c+1=18+1=19$

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