1. ## Maximum Value

Let $x_{1},x_{2},x_{3},x_{4}$ be the four digits of a four digit number $N=x_{1}x_{2}x_{3}x_{4}.$

Find the maximum value of $\frac{N}{x_{1}+x_{2}+x_{3}+x_{4}}$

2. Wouldn't it be 1000?

$\frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} \leq N$

So, if $x_{1}+x_{2}+x_{3}+x_{4} = 1$, then we get $N=N$, which is the most that $\frac{N}{x_{1}+x_{2}+x_{3}+x_{4}}$ can be.

Since the number must have four digits, we can't have $x_{1} = 0$, so it must be at least 1, and since $x_{1}+x_{2}+x_{3}+x_{4} = 1$, we have that $x_{2}=x_{3}=x_{4} = 0$.

So, we get $N=1000$ and $\frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} = 1000$.

Actually, N=2000,3000,4000,...,9000 works also, but in all those cases we have that $\frac{N}{x_{1}+x_{2}+x_{3}+x_{4}} = 1000$

3. Originally Posted by pankaj
Let $x_{1},x_{2},x_{3},x_{4}$ be the four digits of a four digit number $N=x_{1}x_{2}x_{3}x_{4}.$

Find the maximum value of $\frac{N}{x_{1}+x_{2}+x_{3}+x_{4}}$
$N\ =\ 1000x_1+100x_2+10x_3+x_4$

$\le\ 1000x_1+1000x_2+1000x_3+1000x_4$

$\therefore\ \frac N{x_1+x_2+x_3+x_4}\ \le\ 1000$

Equality is attained when $x_2=x_3=x_4=0.$ Hence the maximum value of $N$ is 1000.

4. Originally Posted by TheAbstractionist
$N\ =\ 1000x_1+100x_2+10x_3+x_4$
$\le\ 1000x_1+1000x_2+1000x_3+1000x_4$
$\therefore\ \frac N{x_1+x_2+x_3+x_4}\ \le\ 1000$

Equality is attained when $x_2=x_3=x_4=0.$ Hence the maximum value of $N$ is 1000.
You mean the maximum value of $\frac{N}{x_{1}+x_{2}+x_{3}+x_{4}}$ is 1000

5. I do indeed.