## Friday, April 24, 2015

### Evaluate √((1000000)(1000001)(1000002)(1000003)+1)

Evaluate $\sqrt{(1000000)(1000001)(1000002)(1000003)+1}$ without the help of calculator.

At first glance, you might notice $1000000,\,1000001,\,1000002,\,1000003$ are the terms from arithmetic sequence and now, we have to compute the product of these four figures, but not to find the sum of the first four terms from that sequence.

But, let us look at it a lot more closely, we can actually now discern a pattern if we employ the heuristic strategy by using the substitution method which helps in solving this intriguing problem.

We first let [MATH]\color{yellow}\bbox[5px,purple]{x=1000000}[/MATH], then

$\sqrt{\color{yellow}\bbox[5px,green]{(x)(x+1)(x+2)(x+3)}\color{black}{+1}}$

Now, things might not be sufficiently clear to you how this could be perceived as any sort of positive help and a move in the right direction, but deep inside our heart, we know we could easily remove the square root sign if the expression inside it is a perfect square. So, $\color{yellow}\bbox[5px,green]{(x)(x+1)(x+2)(x+3)}\color{black}{+1}$ has to be a perfect square.

What matters now is how we are going to do the grouping for the $\color{yellow}\bbox[5px,green]{(x)(x+1)(x+2)(x+3)}$:

Attempt I:

If we group the first two terms and last two terms together, we have:

$\color{yellow}\bbox[5px,orange]{((x)(x+1))((x+2)(x+3))}\color{black}{+1}$

$=(x^2+x)(x^2+5x+6)+1$

Hmm...that won't help as it is not clear how $(x^2+x)(x^2+5x+6)+1$ can be rewritten as the square of a quantity.

Attempt II:

What if we group the first and third and then second and last together? Let's see:

$\color{yellow}\bbox[5px,blue]{((x)(x+2))((x+1)(x+3))}\color{black}{+1}$

$=(x^2+2x)(x^2+4x+3)+1$

Okay, we end up with an impasse as well, as $(x^2+2x)(x^2+4x+3)+1\ne (\text{some quantity})^2$

Attempt III:

We will try the other combination of grouping, that is to group the first and last and then the middle terms together:

$\color{yellow}\bbox[5px,purple]{((x)(x+3))((x+1)(x+2))}\color{black}{+1}$

$=(x^2+3x)(x^2+3x+2)+1$

Ah! This works because

\begin{align*}\color{yellow}\bbox[5px,purple]{((x)(x+3))((x+1)(x+2))}\color{black}{+1}&=(x^2+3x)(x^2+3x+2)+1\\&=(x^2+3x)^2+2(x^2+3x)+1\\&=u^2+2u+1\,\,\text{if we letu=x^2+3x}\\&=(u+1)^2\\&=(x^2+3x+1)^2\end{align*}

Therefore, when $x=1000000$,

$(x)(x+1)(x+2)(x+3)+1=(x^2+3x+1)^2$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=((1000000)^2+3(1000000)+1)^2$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=(1000003000001)^2$

$\therefore \sqrt{(1000000)(1000001)(1000002)(1000003)+1}=(1000003000001)^2$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=1000003000001$

As a student, you might want to ask if we really need to try out all the three trials only to realize the last one works?

Yes, you asked the right question because someone who is smarter than you might "foresee" the third grouping would only work, that saved them precious time in working out all the cases. But, you need to remember too that we can train ourselves in order to be as smart and intelligent as the born genius, the only question is, are we willing to work hard, and be an independent and mature learner? After all, to truly understand any heuristic problem solving skills, we must be willing to embrace all it's aspect by trying each case out.

All in all, this problem serves as another heuristic practice that can help students to practice and be trained so they could make smart choices quickly in the future and be independent problem solver.