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

least value

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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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

least value

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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