## Learn how to solve a Math Olympiad question using just basic concepts

Math Olympiad Question: x(x + 2)(x +4)(x + 6) = 9. Find x. Can you solve in 5 minutes? Hint: Solve using mathematical reasoning and basic math concepts.

**Recommended time to solve: 5 minutes.**

First, try to solve all by yourself. With the reason and concept based approach, the problem shouldn't be hard to solve.

## Math Olympiad Question Solution: Use mathematical reasoning and basic concepts

### First stage: Deductive Analysis

The conclusions are drawn from fundamental mathematical concepts and deductive reasoning.

$x(x+2)(x+4)(x+6)=9$

Or, $(x^2+2x)(x^2+10x +24)=9$

Or, $x^4 +12x^3+44x^2+48x=9$

Or, $(x^4+44x^2) + (12x^3 +48x)=9$

We have separated the terms into even and odd powered groups. The reason behind this action is the knowledge that even powers of $x$ will NULLIFY NEGATIVE values of $x$ by making the final value positive and in the same way IMAGINARY values of $x$ will be converted to real negative, whereas odd powers of $x$ will not be converted to positive or real negative.

**CONCLUSION 1**

As the right hand side value is real 9, the value of $x$ CANNOT be imaginary because of the presence of odd powered terms of $x$.

**CONCLUSION 2**

the value of $x$ cannot be a positive integer (it cannot be a fraction also for obvious reasons) as even $x=1$ will far outweigh the LHS against only 9 on the RHS.

It follows then (this is the most defining group of conclusions),

**CONCLUSION 3.1**

The value of $x$ must be negative and it must be an integer.

**CONCLUSION 3.2**

$x$ must also be a small valued negative integer as otherwise, larger its value, faster the positive component of $x^4$ will grow beyond the term in $x^3$.

**CONCLUSION 3.3**

$x$ cannot be a negative even integer. Otherwise, the LHS won't balance the RHS.

So,

The value of $x$ must be a small negative integer.

### Second Stage of Trials

**Experiment with $x=-1$**

$(x^4 + 44x^2)+(12x^3+48x)=9$

Or, $(1+44)+(-12-48)=9$

Or, $(45)+(-60)=9$

Or, $-15$ $\neq$ $9$.

**Experiment next with $x=-3$**

$(x^4+44x^2)+(12x^3+48x)=9$

Or, $(81+396)+[12(-27)+48(-3)]=9$

Or, $477+(-324-144)=9$

Or, $477-468=9$

Or, $9=9$.

So, $x=-3$

## Math Olympiad Question Alternative Solution: By comparing the factors on both sides of the equation

From the nature of the equation, first conclusion:

**CONCLUSION 1: **

If $x$ is positive, LHS will be more than RHS 9. So, $x$ must be negative.

### Factor analysis: comparison of factors of both sides of the equation

Factors of 9 are 1, 3 and 9. But since there are four factors in the quartile expression in the LHS, these should be, $+1$, $-1$, $+3$ and $-3$.

**Solution**

As it has already been concluded that $x$ must be negative,

$x=-3$.

### End word

In the first solution, use of basic concept based mathematical reasoning focused the possible values of $x$ to only negative small odd integer values. Two experiments resulted in the solution. No hesitation, no uncertain steps. This is assured step by step problem solving using mathematical reasoning and basic math concepts.

But the **second solution is much faster** by the direct use of the basic math concepts on factor comparison. It is good to know that the Math Olympiad Question can be solved in two altogether different ways.

*Credit for second solution: Somnath Maity.*

### Food for thought

*What do you think about the pros and cons of the two solutions?*

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Good luck to your problem-solving!