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

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 .

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