INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

August 2, 2020

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


    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


    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.

    Cheenta. Passion for Mathematics

    Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.