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Let $N$ be any four digit number say $x_1 x_2 x_3 x_4$. Then maximum value of $\frac{N}{x_1+x_2+x_3+x_4}$is equal to

(A) 1000

(B) $\frac{1111}{4}$

(C) $800$

(D) none of these

My approach is as follow $N=1000x_1+100x_2+10x_3+x_4$

$T=\frac{1000x_1+100x_2+10x_3+x_4}{x_1+x_2+x_3+x_4}$

$T=\frac{999x_1+99x_2+9x_3}{x_1+x_2+x_3+x_4}+1$

Not able to proceed from here

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  • $\begingroup$ An informal (i.e. intuitive, somewhat invalid) approach is that when you start with a baseline of $~x_2 = x_3 = x_4 = 0,~$ it seems clear (at least to me) that increasing the value of (for example) $~x_2~$ has a more significant effect on the denominator, than it does on the numerator. This intuition holds across various values for $~x_1.$ $\endgroup$ Commented May 11, 2023 at 5:54

3 Answers 3

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The answer is (A). First, we can note that $$ \frac{1000}{1+0+0+0} = 1000, $$ thus we rule out (B), (C) immediately. It only remains to find out whether we can achieve a value greater than 1000. But we can note that \begin{align} \frac{N}{x_1+x_2+x_3+x_4} &> 1000\\ N &> 1000(x_1+x_2+x_3+x_4)\\ 1000x_1 + 100x_2 + 10x_3 + x_4 &> 1000(x_1+x_2+x_3+x_4) \end{align} but it is clear that the last line cannot be true since the $x_i$'s are $\geq 0$, and in particular, $x_1\geq 1$ since $N$ is four digits. Thus the answer must be (A).

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$T=\frac{1000x_1+100x_2+10x_3+x_4}{x_1+x_2+x_3+x_4}$

$T=\frac{999x_1+99x_2+9x_3}{x_1+x_2+x_3+x_4}+1$

Now $x_4$ is a free variable, we want $T$ to be max, so let $x_4=0$

$$T=\frac{999x_1+99x_2+9x_3}{x_1+x_2+x_3}+1=1+9\left(\frac{110x_1+10x_2}{x_1+x_2+x_3}+1\right)$$

Now, $x_3$ is free variable, so let $x_3=0$, and keep going, can you proceed from here?

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Hints.

  • Check that the value $1000$ can be obtained

  • Compute $1110-T$ in terms of the $x_i's$ and show it is non negative (remember that $x_i\geq 0$).

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  • $\begingroup$ > "Compute $1110-T$ in terms of the $x_i's$ and show it is non negative (remember that $x_i\geq 0$)." ~ where is the speciality in the number $1110$ ?? not getting it. $\endgroup$
    – Skdmg
    Commented May 3 at 12:58

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