What is this mess?

Algebra Level 5

$f(x)=\dfrac{103^{x}}{x!}, \quad x\in \mathbb{N}$

If the values of $x$ for which $f(x)$ is maximum are $x_{1},x_{2},\ldots, x_{n}$ , then find $n+\displaystyle \sum^{n}_{j=1}x_{j}$ .

The answer is 207.

1 solution

Zee Ell
Sep 5, 2016

It is easy to see, that for x > 1:

$f(x) = \frac {103^x}{x!} = \frac {103 × 103^{x - 1}}{x × (x - 1)!} = \frac {103}{x} f(x - 1)$

Now, our discrete f(x) is increasing, when:

f(x - 1) < f(x)

Since f(x) is always positive, this is equivalent to:

$1 < \frac {f(x)}{f(x - 1)}$

$1 < \frac {103}{x}$

$x < 103$

We can also find (by substituting x = 103 into our first formula), that:

$f(103) = \frac {103}{103} f(102) = f(102)$

This means, that f(x) has its maximum at x = 102 and since f(103) = f(102) (and f(x - 1) > f(x) for x > 103 (can be proven by simply exchanging"<" for ">" at our steps above), therefore we have:

2 maximum values at x = 102 and x = 103

Hence, the solution is:

$2 + 102 + 103 = \boxed {207}$

Pretty easy but damn complex looking question.

Shreyash Rai - 4 years, 9 months ago

Nice solution but please explain me how you got this ?

"f(103) = f(102)"

Priyanshu Mishra - 4 years, 9 months ago

$f(x) = \frac {103^x}{x!} = \frac {103 × 103^{x - 1}}{x × (x - 1)!} = \frac {103}{x} f(x - 1)$

$f(103) = \frac {103}{103} f(102) = f(102)$

I also adjusted the solution to include this.

Zee Ell - 4 years, 9 months ago

Oh i should have rather noted that earlier.

Anyways, thanks for your reply. i Understood that.

Priyanshu Mishra - 4 years, 9 months ago

What is this mess?

The answer is 207.

1 solution

0 pending reports