Question

Find three consecutive positive integers such that the sum of the square of the first and the product of other two is $154$.

Answer 1

Let $(x-1),x$ and $(x+1)$ be the three positive integers.

It is given that the sum of the square of the first integer and the product of other two is $154$, therefore,

$(x-1{)}^2+x(x+1)=154\Rightarrow x^2+1-2x+x^2+x=154(\because (a-b{)}^2=a^2+b^2-2ab)\Rightarrow x^2+1-2x+x^2+x-154=0\Rightarrow 2x^2-x-153=0\Rightarrow 2x^2-18x+17x-153=0\Rightarrow 2x(x-9)+17(x-9)=0\Rightarrow (2x+17)=0,(x-9)=0\Rightarrow 2x=-17,x=9\Rightarrow x=-\frac{17}{2},x=9$

Since $x$ is a positive integer thus, $x=9$.

Now, $x-1=9-1=8$ and $x+1=9+1=10$

Since the unit place of the digit is $6$ and the tens place is $2$

Hence, the three consecutive positive integers are $8,9$ and $10$.