Categories
Algebra AMC 10 Math Olympiad USA Math Olympiad

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

Try this beautiful problem from Algebra based on least possible number.AMC-10A, 2019. You may use sequential hints to solve the problem

Contents
 [hide]

    Try this beautiful problem from Algebra based on Least Possible Value.

    Least Possible Value – AMC-10A, 2019- Problem 19


    What is the least possible value of \(((x+1)(x+2)(x+3)(x+4)+2019)\)

    where (x) is a real number?

    • \((2024)\)
    • \((2018)\)
    • \((2020)\)

    Key Concepts


    Algebra

    quadratic equation

    least value

    Check the Answer


    Answer: \((2018)\)

    AMC-10A (2019) Problem 19

    Pre College Mathematics

    Try with Hints


    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 .

    Subscribe to Cheenta at Youtube


    Leave a Reply

    This site uses Akismet to reduce spam. Learn how your comment data is processed.