How Cheenta works to ensure student success?
Explore the Back-Story
Problems and Solutions from CMI Entrance 2022.  Learn More 

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

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


    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.

      Knowledge Partner

      Cheenta is a knowledge partner of Aditya Birla Education Academy
      Cheenta

      Cheenta Academy

      Aditya Birla Education Academy

      Aditya Birla Education Academy

      Cheenta. Passion for Mathematics

      Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
      JOIN TRIAL
      support@cheenta.com
      magic-wand