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What least number must be subtracted from each of the numbers $$14, 17, 34$$ and $$42$$ so that the remainders may be proportional?
A
$$0$$B
$$1$$C
$$2$$D
$$ 14-x:17-x=34-x:42-x$$
$$\Rightarrow \cfrac { 14-x }{ 17-x } =\cfrac { 34-x }{ 42-x } $$
$$\Rightarrow (14-x)(42-x)=(34-x)(17-x)$$
$$\Rightarrow { x }^{ 2 }-56x+588={ x }^{ 2 }-51x+578$$
$$\Rightarrow 5x=10$$
$$\Rightarrow x=2$$