Calvin goes to the local "7-11" store!

• (1): Given $4$ numbers. The sum of these numbers is $12$ . What is the maximum value of these number's product? The solution is $3\times3\times3\times3=81$ . When the numbers are approximately equals, then we will get the biggest product.

• (2): Given $3$ numbers. The product of these numbers is $24$ . What is the minimum value of these number's sum? The solution is $2+3+4$ . When the numbers are approximately equals, then we will get the smallest sum.

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We can multiply each number by 100. Then: $\begin{aligned} A\times B\times C\times D&=711,000,00\\ A+B+C+D&=711 \end{aligned}$

$711000000=2^6\times3^2\times5^6\times79$ . Let's be $A$ multiples of $79$ .

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If $A=79\times5$ , then $B\approx C\approx D\approx\cfrac{711-79\times5}{3}\approx105$ . Therfore the maximum vaslue of $A\times B\times C\times D$ is $(79\times 5)\times105\times105\times106=461616750<711000000$ . Therefore $A<79\times5$

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Nove prove that: $5|B,C,D$ :

• Assume that only one price is divisible by $5$ : the sum will be bigger than $711$
• If only two prices are divisible by $5$ $(B,C)$ :
  • If $B$ , is multiples of $5^4$ or $5^5$ , then with simple logic the sum will be bigger, than $711$
  • If $B,C$ are multiples of $5^3$ (using (1) and (2) ): $\begin{array}{c|c|c|c|c|} A&B&C&D&\text{sum}\\ \hline 79&125&125&576&>711\\ 79\times2&125&125&288&<711\\ 79\times3&125&125&192<711\\ 79\times4&125&125&144&710&\text{Note: the sum will be bigger, if we continue that. So there are no solutions}\\ \vdots&125&125&\vdots&>711\\ \hline 79&250&125&288&>711\\ 79\times2&250&125&144&<711\\ 79\times3&250&125&96&708&\text{Same way, there are no solutions}\\ \vdots&250&125&\vdots&>711\\ \hline 79&375&125&192&>711\\ 79\times2&375&125&86&>711\\ \vdots&375&125&\vdots&>711&\text{B can't be bigger. So we should see only B=250,C=250.}\\ & & & & &\text{In other cases there are no solutions(with simple logic)} \\ \hline \hline 79&250&250&144&>711\\ \vdots&250&250&\vdots&>711 \end{array}$

Therefore $5|B,C,D$ .

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Now let's see the last digits:

• $A$ : $\begin{array}{c|c} A&\text{last digit}\\ \hline 79\times1&9\\ 79\times2&8\\ 79\times3&7\\ 79\times4&6\\ \end{array}$
• $B,C,D$ : $0$ or $5$ , because they are divisible by 5.

Now, the last digit of $711$ , so the last digit of $A$ should be equal to $6$ , because $5+6\rightarrow1$ . Therefore $A=79\times4=316$ . Now there are two opportunities:

• The last digits of $B,C,D$ are $5,5,5$ - but then $2\not{|}B,C,D$ , so this is impossible
• The last digits of $B,C,D$ are $0,0,5$

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Now let's see the second last digit(second digit from the end of the number):

Let's be $[\cdot]$ the second digit from the end of the number.

The carry over from the last digits is $\left \lfloor \cfrac{6+0+0+5}{10} \right \rfloor=1$ . $\begin{aligned} [A]+[B]+[C]+[D]+1\text{(carry over)}\pmod {10}=[711]&=1\\ [B]+[C]+[D]&=9 \end{aligned}$

With some calculations $[B,C,D]\in{2,5,7}$ .

$7$ is possible if $3\times5\times5$ or $3\times5\times5\times5$

With simple logic the only solution is: $\begin{aligned} [B]&=5\\ [C]&=2\\ [D]&=2 \end{aligned}$

With a few trial and error and logic:

$\begin{aligned} A&=316\\ B&=150\\ C&=125\\ D&=120 \end{aligned}$

But the problem says this should be in ascending order, so the final solution is:

$\begin{aligned} A&=1.20\\ B&=1.25\\ C&=1.50\\ D&=3.16 \end{aligned}\implies \cfrac{D}{B}+\cfrac{C}{A}=\cfrac{3.16}{1.25}+\cfrac{1.5}{1.2}=\cfrac{1889}{600}\implies M+N=2389$