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Three consecutive natural numbers are such that the square of the greatest is greater than the product of the other two by 19. The smallest of these numbers is
Let the three consecutive natural numbers be $$(x-1)$$ , $$(x)$$ and $$(x+1)$$
According to ques, => $$(x+1)^2 - (x-1)(x)=19$$
=> $$(x^2+2x+1) - (x^2-x)=19$$
=> $$3x+1=19$$
=> $$3x=19-1=18$$
=> $$x=\frac{18}{3}=6$$
$$\therefore$$ Smallest number = $$(x-1)=6-1=5$$
=> Ans - (A)
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