Funny Boxes

There are many solutions to this problem.One of them is 9863+752+41+0.The technique is to give the largest number(9) to thousands place.Larger numbers to hundreds place(8&7)and smaller number to tens place(4&5&6) and smallest numbers to ones place(0&1&2&3).

To get the largest sum we just have to put greater numbers in higher place values.

1. We must put 9 in thousands place

2. We can put an 8 in 4-digit no. and 7 in 3-digit no. (At hundreds place).

3. We can put 6 in 4-digit, 5 in 3-digit and 4 in 2- digit (tens place)

4. At last we put 3 in 4 digit, 2 in 3-digit, 1 in 2 -digit and 0 in one-digit (ones place).

In each bullet, we can interchange position of numbers as there place value doesn't change.

The first key observation to make is that two solutions yield the same sum if digits with the same place value are switched (e.g. $143 + 57 = 153 + 47 = 157 + 43$ ). Thus we only need to determine which digits will be given which place value. Intuitively it seems reasonable that the maximum sum will be obtained with the highest numbers being assigned the highest place values, etc. For example, $9,863 + 752 + 41 + 0 = 10,656$ . To see that this is indeed the maximum possible sum, consider the effect of switching two digits with different place values. Specifically, let $m$ have place value $10^k$ and $n$ have place value $10^l$ , where $k > l$ . By construction, we have $m > n$ . Then when we switch $m$ and $n$ , the difference between the old and new sums is

$(10^km + 10^ln) - (10^kn + 10^lm) = 10^l(10^{k - l}m + n - 10^{k - l}n - m) = 10^l(m - n)(10^{k - l} - 1)$ ,

which is strictly positive, meaning the old sum was larger. Therefore our original sum of $10,656$ was indeed optimal.