Least Possible Value Problem | AMC-10A, 2019 | Quesstion19

To find out the least positive value of \((x+1)(x+2)(x+3)(x+4)+2019\), at first we have to expand the expression .\(((x+1)(x+2)(x+3)(x+4)+2019)\) \(\Rightarrow (x+1)(x+4)(x+2)(x+3)+2019=(x^2+5x+4)(x^2+5x+6)+2019)\)

Let us take \(((x^2+5x+5=m))\)

then the above expression becomes \(((m-1)(m+1)+2019)\) \(\Rightarrow m^2-1+2019\) \(\Rightarrow m^2+2018\)

Can you now finish the problem ..........

Clearly in \((m^2+2018).......(m^2)\) is positive ( squares of any number is non-negative) and least value is 0

can you finish the problem........

Therefore minimum value of \(m^2+2108\) is \(2018\) since \(m^2 \geq 0\) for all m belongs to real .

To find out the least positive value of \((x+1)(x+2)(x+3)(x+4)+2019\), at first we have to expand the expression .\(((x+1)(x+2)(x+3)(x+4)+2019)\) \(\Rightarrow (x+1)(x+4)(x+2)(x+3)+2019=(x^2+5x+4)(x^2+5x+6)+2019)\)

Let us take \(((x^2+5x+5=m))\)

then the above expression becomes \(((m-1)(m+1)+2019)\) \(\Rightarrow m^2-1+2019\) \(\Rightarrow m^2+2018\)

Can you now finish the problem ..........

Clearly in \((m^2+2018).......(m^2)\) is positive ( squares of any number is non-negative) and least value is 0

can you finish the problem........

Therefore minimum value of \(m^2+2108\) is \(2018\) since \(m^2 \geq 0\) for all m belongs to real .